LINE AND 17 SURFACE INTEGRALS 17.1 Vector Fields (ET Section 16.1) Preliminary Questions 1. Which of the following is a unit vector field in the plane? (a) F = y, x x y , (b) F = x 2 + y2 x 2 + y2 y x , (c) F = 2 x + y2 x 2 + y2 SOLUTION
(a) The length of the vector y, x is y, x =
y2 + x 2
This value is not 1 for all points, hence it is not a unit vector field. (b) We have
2
2
y x x y , + = x 2 + y2 x 2 + y2 x 2 + y2 x 2 + y2 y2 y2 + x 2 x2 = + 2 = =1 2 2 2 x +y x +y x 2 + y2 Hence the field is a unit vector field, for (x, y) = (0, 0). (c) We compute the length of the vector: 2 2 2
y + x2 x y 1 x y + = 2 = x 2 + y2 , x 2 + y2 = 2 + y2 2 2 x 2 + y2 x 2 + y2 x x +y Since the length is not identically 1, the field is not a unit vector field. 2. Sketch an example of a nonconstant vector field in the plane in which each vector is parallel to 1, 1. SOLUTION
The non-constant vector x, x is parallel to the vector 1, 1. y
x
−→ 3. Show that the vector field F = −z, 0, x is orthogonal to the position vector O P at each point P. Give an example of another vector field with this property. SOLUTION
The position vector at P = (x, y, z) is x, y, z. We must show that the following dot product is zero: x, y, z · −z, 0, x = x · (−z) + y · 0 + z · x = 0
S E C T I O N 17.1
Vector Fields
(ET Section 16.1)
563
Therefore, the vector field F = −z, 0, x is orthogonal to the position vector. Another vector field with this property is F = 0, −z, y, since 0, −z, y · x, y, z = 0 · x + (−z) · y + y · z = 0 4. Give an example of a potential function for yz, x z, x y other than ϕ (x, y, z) = x yz. Since any two potential functions of a gradient vector field differ by a constant, a potential function for the given field other than φ (x, y, z) = x yz is, for instance, φ1 (x, y, z) = x yz + 1. SOLUTION
Exercises
1. Compute and sketch the vector assigned to the points P = (1, 2) and Q = (−1, −1) by the vector field F = x 2 , x .
SOLUTION
The vector assigned to P = (1, 2) is obtained by substituting x = 1 in F, that is, F(1, 2) = 12 , 1 = 1, 1
Similarly, F(−1, −1) = (−1)2 , −1 = 1, −1 y
F(P) = 〈1, 1〉
1
x 1 −1
F(Q) = 〈1, −1〉
and sketch the vector assigned to the points P = (0, 1, 1) and Q = (2, 1, 0) by the vector field F = 3. Compute Compute and sketch the vector assigned to the points P = (1, 2) and Q = (−1, −1) by the vector field F = x y, z 2 , x . −y, x. SOLUTION To find the vector assigned to the point P = (0, 1, 1), we substitute x = 0, y = 1, z = 1 in F = x y, z 2 , x. We get F(P) = 0 · 1, 12 , 0 = 0, 1, 0 Similarly, F(Q) is obtained by substituting x = 2, y = 1, z = 0 in F. That is, F(Q) = 2 · 1, 02 , 2 = 2, 0, 2 z
F(Q) = 〈2, 0, 2〉
F(P) = 〈0, 1, 0〉 y
x
In Exercises 5–13, sketch the following planar vector fields by drawing the vectors attached to points witheinteger coorer r , and 2 . to the vector assigned points = (1, 1, 0)vectors and Q with = (2,their 1, 2)true by the vector fields er , if necessary dinates Compute in the rectangle −3 ≤ x, y ≤ to 3. the Instead of Pdrawing the lengths, scale them r r avoid overlap. 5. F = 1, 0 SOLUTION
The constant vector field 1, 0 is shown in the figure: y 3 2 1 x
−3 −2 −1
1 −1 −2 −3
2
3
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7. F = 0, x F = xi SOLUTION We sketch the vector field F(x, y) = 0, x: y
x
9. F = 1, 1 F = yj SOLUTION We sketch the graph of the constant vector field F(x, y) = 1, 1: y
x
x y , 2 11. F =F = 2x 2 i +2 yj x + y x + y2 SOLUTION y
x
−x y , 13. F = 2 −y x 2 2 2 F =x +2y x2 , +2y 2 x +y x +y SOLUTION y
x
Vector Fields
S E C T I O N 17.1
(ET Section 16.1)
565
In Exercises 14–17, match the planar vector field with the corresponding plot in Figure 7. y
y 2
2 x
0
x
0 −2
−2 −2
0 (A)
−2
2
0
2
(B) y
y 2
2 x
0
x
0 −2
−2 −2
0 (C)
−2
2
0 (D)
2
FIGURE 7
15. F = 2x + 2, y F = 2, x SOLUTION We compute the images of the point (0, 2), for instance, and identify the corresponding graph accordingly: F(x, y) = 2x + 2, y
⇒
F(0, 2) = 2, 2
⇒
Plot(C)
17. F = x + y, x − y F = y, cos x SOLUTION We compute the images of the point (0, 2), for instance, and identify the corresponding graph accordingly: F(x, y) = x + y, x − y
⇒
F(0, 2) = 2, −2
⇒
Plot(A)
In Exercises 18–21, match the three-dimensional vector field with the corresponding plot in Figure 8.
(A)
(B)
(C)
(D)
FIGURE 8
19. F = x, 0, z F = x, y, z SOLUTION This vector field is shown in (A) (by process of elimination). 21. F = 1, 1, 1
x y z constant vector , field 1, 1, 1 ,isshown in plot (C). The x 2 + y2 + z2 x 2 + y2 + z2 x 2 + y2 + z2 In Exercises 22–25, find a potential function for the vector field F by inspection. 23. F = ye x y , xe x y F = x, y F= SOLUTION
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∂ϕ ∂ϕ The function ϕ (x, y) = e x y satisfies ∂ x = ye x y and ∂ y = xe x y , hence ϕ is a potential function for the given vector field. 2 2 25. F = 2x ze x2 , 0, 2e x F = yz , x z , 2x yz SOLUTION
2 2 2 ∂ϕ ∂ϕ ∂ϕ The function ϕ (x, y, z) = ze x satisfies ∂ y = 0, ∂ x = 2x ze x and ∂z = e x , hence ϕ is a potential function for the given vector field.
SOLUTION
27. Which of (A) or (B) in Figure 9 is the contour plot eof a potentiale function for the vector field F? Recall that the r r potential functions fortothe gradientFind vectors are perpendicular thevector level fields curves.F = 3 and G = 4 in R3 . r r 2 1 0 −1 −2 −2
−1
0
2
2
1
1
0
0
−1
−1
−2
1
2
−2
−1
−2 −2
−1
0 (A)
1
2
0 (B)
1
2
FIGURE 9 SOLUTION By the equality ∇ ϕ = F and since the gradient vectors are perpendicular to the level curves, it follows that the vectors F are perpendicular to the corresponding level curves of ϕ . This property is satisfied in (B) and not satisfied in (A). Therefore (B) is the contour plot of ϕ . 2 . Express ∇ ϕ in terms of the unit radial vector e in R2 . ϕ = lnof r , (A) where r =in Figure x 2 + y10 29. LetWhich or (B) is the contour plot of a potential function of ther vector field F?
Since r = (x 2 + y 2 + z 2 ) partial derivatives: SOLUTION
1/2
, we have ϕ = ln (x 2 + y 2 + z 2 )
1/2
= 12 ln(x 2 + y 2 + z 2 ). We compute the
1 2x ∂ϕ x = = 2 ∂x 2 x 2 + y2 + z2 r 1 2y ∂ϕ y = = 2 ∂y 2 x 2 + y2 + z2 r 1 2z ∂ϕ z = = 2 ∂z 2 x 2 + y2 + z2 r Therefore, the gradient of ϕ is the following vector: ∂ϕ ∂ϕ ∂ϕ x y z 1 x y z er , , , , = ∇ϕ = = 2, 2, 2 = ∂ x ∂y ∂z r r r r r r r r Match the force fields (a)–(c) with (A)–(C) in Figure 11. Note that the vectors (A)–(C) indicate direction but are Further Insights and Challenges not drawn to indicate magnitude.
31. Show that any vector field of the form F = f (x), g(y), h(z) has a potential function. Assume that f , g, and h are (a) One positive and one negative charge continuous. (b) Two positive charges SOLUTION Let F(x), G(y), and H (z) be antiderivatives of f (x), g(y), and h(z), respectively. That is, F (x) = f (x), (c) Two positive charges and one negative charge G (y) = g(y), and H (y) = h(z). We define the function
ϕ (x, y, z) = F(x) + G(y) + H (z) Then, ∂ϕ = F (x) = f (x), ∂x
∂ϕ = G (y) = g(y), ∂x
∂ϕ = H (z) = h(z) ∂z
Therefore, ∇ ϕ = F, or ϕ is a potential function for F. In this exercise, we show that the vector field F in Figure 12 is not a gradient vector field. Suppose that a potential function ϕ did exist.
S E C T I O N 17.2
Line Integrals
(ET Section 16.2)
567
33. Show that if ∇ ϕ (x, y) = 0 for all (x, y) in a disk D in R2 , then ϕ is constant on D. Hint: Given points P = (a, b) and Q = (c, d) in D, let R = (c, b) (Figure 13). Use single-variable calculus to show that ϕ is constant along the segments P R and R Q and conclude that ϕ (P) = ϕ (R). y Q = (c, d)
P = (a, b) R = (c, b)
Disk D x
FIGURE 13 SOLUTION
Given any two points P = (a, b) and Q = (c, d) in D, we must show that
ϕ (P) = ϕ (Q) −→ We consider the point R = (c, b) and the segments P R and R Q. (We assume that (c, b) is in D; if not, just use
R = (a, d).) y Q = (c, d)
P = (a, b) R = (c, b)
D x
∂ϕ
∂ϕ
Since ∂ x (x, y) = 0 in D, in particular ∂ x (x, b) = 0 for a ≤ x ≤ c. Therefore, for a ≤ x ≤ c we have x x ∂ϕ (u, b) du + ϕ (a, b) = ϕ (x, b) = 0 du + ϕ (a, b) = k + ϕ (a, b) a ∂u a Substituting x = a determines k = 0. Hence,
ϕ (x, b) = ϕ (a, b) for a ≤ x ≤ c In particular,
ϕ (c, b) = ϕ (a, b) ∂ϕ
⇒
ϕ (R) = ϕ (P)
(1)
∂ϕ
Similarly, since ∂ y (x, y) = 0 in D, we have ∂ y (c, y) = 0 for b ≤ y ≤ d. Therefore for b ≤ y ≤ d we have y y ∂ϕ (c, v) dv + ϕ (c, b) = ϕ (c, y) = 0 dv + ϕ (c, b) = k + ϕ (c, b) b ∂v b Substituting y = b gives ϕ (c, b) = k + ϕ (c, b) or k = 0. Therefore,
ϕ (c, y) = ϕ (c, b) for b ≤ y ≤ d In particular,
ϕ (c, d) = ϕ (c, b)
⇒
ϕ (Q) = ϕ (R)
(2)
Combining (1) and (2) we obtain the desired equality ϕ (P) = ϕ (Q). Since P and Q are any two points in D, we conclude that ϕ is constant on D.
17.2 Line Integrals
(ET Section 16.2)
Preliminary Questions 1. What is the line integral of the constant function f (x, y, z) = 10 over a curve C of length 5?
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(ET CHAPTER 16)
Since the length of C is the line integral C 1 ds = 5, we have 10 ds = 10 1 ds = 10 · 5 = 50 C
2. (a) (c) (e)
C
Which of the following have a zero line integral over the vertical segment from (0, 0) to (0, 1)? f (x, y) = x (b) f (x, y) = y F = x, 0 (d) F = y, 0 F = 0, x (f) F = 0, y
SOLUTION
The vertical segment from (0, 0) to (0, 1) has the parametrization c(t) = (0, t),
0≤t ≤1
Therefore, c (t) = 0, 1 and c (t) = 1. The line integrals are thus computed by 1 f (x, y) ds = f (c(t)) c (t) dt C
(1)
0
C
F · ds =
1 0
F (c(t)) · c (t) dt
(2)
(a) We have f (c(t)) = x = 0. Therefore by (1) the line integral is zero. (b) By (1), the line integral is 1 1 1 1 f (x, y) ds = t · 1 dt = t 2 = = 0 2 2 C 0 0 (c) This vector line integral is computed using (2). Since F (c(t)) = x, 0 = 0, 0, the vector line integral is zero. (d) By (2) we have 1 1 t, 0 · 0, 1 dt = F · ds = 0 dt = 0 C
0
0
(e) The vector integral is computed using (2). Since F (c(t)) = 0, x = 0, 0, the line integral is zero. (f) For this vector field we have 1 1 1 t 2 1 1 0, t · 0, 1 dt = F · ds = F (c(t)) · c (t) dt = t dt = = = 0 2 0 2 C 0 0 0 So, we conclude that (a), (c), (d), and (e) have an integral of zero. 3. State whether true or false. If false, give the correct statement. (a) The scalar line integral does not depend on how you parametrize the curve. (b) If you reverse the orientation of the curve, neither the vector nor the scalar line integral changes sign. SOLUTION
(a) True: It can be shown that any two parametrizations of the curve yield the same value for the scalar line integral, hence the statement is true. (b) False: For the definition of the scalar line integral, there is no need to specify a direction along the path, hence reversing the orientation of the curve does not change the sign of the integral. However, reversing the orientation of the curve changes the sign of the vector line integral. 4. Let C be a curve of length 5. What is the value of F · ds if C
(a) F(P) is normal to C at all points P on C? (b) F(P) = T(P) at all points P on C, where T(P) is the unit tangent vector pointing in the forward direction along the curve? SOLUTION
(a) The vector line integral is the integral of the tangential component of the vector field along the curve. Since F(P) is normal to C at all points P on C, the tangential component is zero, hence the line integral C F · ds is zero. (b) In this case we have F(P) · T(P) = T(P) · T(P) = T(P)2 = 1 Therefore,
C
F · ds =
C
(F · T) ds =
C
1 ds = Length of C = 5.
S E C T I O N 17.2
Line Integrals
(ET Section 16.2)
569
Exercises 1. Let f (x, y, z) = x + yz and let C be the line segment from P = (0, 0, 0) to (6, 2, 2). (a) Calculate f (c(t)) and ds = c (t) dt for the parametrization c(t) = (6t, 2t, 2t) for 0 ≤ t ≤ 1. f (x, y, z) ds. (b) Evaluate C
SOLUTION
(a) We substitute x = 6t, y = 2t, z = 2t in the function f (x, y, z) = x + yz to find f (c(t)): f (c(t)) = 6t + (2t)(2t) = 6t + 4t 2 We differentiate the vector c(t) and compute the length of the derivative vector: c (t) =
d 6t, 2t, 2t = 6, 2, 2 dt
c (t) =
⇒
√ √ 62 + 22 + 22 = 44 = 2 11
Hence, √ ds = c (t) dt = 2 11 dt (b) Computing the scalar line integral, we obtain C
f (x, y, z) ds =
1 0
f (c(t)) c (t) dt =
1 0
√ (6t + 4t 2 ) · 2 11 dt
√ √ √ 4 3 1 4 26 11 2 = 2 11 3t + t = 2 11 3 + = 3 3 3 0
3. LetRepeat F = y 2Exercise C be y = x −1 for 1c(t) , x 2 and1 let ≤ x=≤(3t, 2, oriented with thethe parametrization t, t) for 0from ≤ t left ≤ 2.to right. (a) Calculate F(c(t)) and ds = c (t) dt for the parametrization c(t) = (t, t −1 ). F · ds. (b) Calculate the dot product F(c(t)) · c (t) dt and evaluate C
SOLUTION
(a) We calculate F (c(t)) by substituting x = t, y = t −1 in F = y 2 , x 2 . We get 2 F(c(t)) = (t −1 ) , t 2 = t −2 , t 2
We compute c (t): c (t) =
d −1 t, t = 1, −t −2 dt
ds = 1, −t −2 dt
⇒
y
x 1
2
(b) We compute the dot product: F(c(t)) · c (t) = t −2 , t 2 · 1, −t −2 = t −2 · 1 + t 2 · (−t −2 ) = t −2 − 1 Computing the vector line integral, we obtain C
F · ds =
2 1
F (c(t)) · c (t) dt =
2 1 1 (t −2 − 1) dt = −t −1 − t = − − 2 − (−1 − 1) = − 2 2 1 1
2
the integral of the given scalar In Exercises 5–8, calculate function or vector field over the curve c(t) = (cos t, sin t, t) Let F = z 2 , x, y and let C be the path c(t) = 3 + 5t 2 , 3 − t 2 , t for 0 ≤ t ≤ 2. for 0 ≤ t ≤ π . (a) Calculate F(c(t)) and ds = c (t) dt. 5. f (x, y, z) = x 2 + y 2 + z 2 (b) Calculate the dot product F(c(t)) · c (t) dt and evaluate F · ds. C
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(ET CHAPTER 16)
SOLUTION
Step 1. Compute c (t). We differentiate c(t): c (t) =
d cos t, sin t, t = − sin t, cos t, 1 dt
Hence, √ (− sin t)2 + cos2 t + 12 = sin2 t + cos2 t + 1 = 2 √ ds = c (t) dt = 2 dt
c (t) =
Step 2. Write out f (c(t)). We substitute x = cos t, y = sin t, z = t in f (x, y, z) = x 2 + y 2 + z 2 to obtain f (c(t)) = cos2 t + sin2 t + t 2 = 1 + t 2 Step 3. Compute the line integral. Using the Theorem on Scalar Line Integrals we obtain
π π √ √ π3 t 3 π √ (x 2 + y 2 + z 2 ) ds = f (c(t)) c (t) dt = (1 + t 2 ) 2 dt = 2 t + = 2 π + 3 0 3 C 0 0 7. F = x, y, z f (x, y, z) = x y + z SOLUTION
Step 1. Calculate the integrand. We write out the vectors: c(t) = (cos t, sin t, t) F (c(t)) = x, y, z = cos t, sin t, t c (t) = − sin t, cos t, 1 The integrand is the dot product: F (c(t)) · c (t) = cos t, sin t, t · − sin t, cos t, 1 = − cos t sin t + sin t cos t + t = t Step 2. Evaluate the integral. We use the Theorem on Vector Line Integrals to evaluate the integral: π π π2 1 π F ds = F (c(t)) · c (t) dt = t dt = t 2 = 2 0 2 C 0 0 9. Calculate the total mass of a circular piece of wire of radius 4 cm centered at the origin whose mass density is y, 2, z 3 ρ(x, y) F==x 2 xg/cm. SOLUTION
The total mass is the following integral: M=
C
x 2 ds
We use the following parametrization of the wire: c(t) = (4 cos t, 4 sin t),
0 ≤ t ≤ 2π
Hence, c (t) = −4 sin t, 4 cos t
⇒
c (t) =
(−4 sin t)2 + (4 cos t)2 = 4
We compute the line integral using the Theorem on Scalar Line Integrals. We get M=
2π 0
= 64
ρ (c(t)) c (t) dt =
2π 0
cos2 t dt = 64
2π 0
(4 cos t)2 · 4 dt
t sin 2t + 2 4
2π = 64 · 2π = 64π g 2 0
11. The values of a function f (x, y, z) and vector field F(x, y, z) are given at six sample points along the path ABC in Calculate thethe total of a metal tubeFinalong the helical shape c(t) = (cos t, sin t, t 2 ) (distance in centimeters) for Figure 11. Estimate linemass integrals of f and √ ABC. 0 ≤ t ≤ 2π if the mass density is ρ(x, y, z) = z g/cm.
Line Integrals
S E C T I O N 17.2
Point
f (x, y, z)
F(x, y, z)
(1, 16 , 0) (1, 12 , 0) (1, 56 , 0) (1, 1, 16 ) (1, 1, 12 ) (1, 1, 56 )
3
1, 0, 2
3.3
1, 1, 3
3.6
2, 1, 5
4.2
3, 2, 4
4.5
3, 3, 3
4.2
5, 3, 3
(ET Section 16.2)
571
z
C = (1, 1, 1) A = (1, 0, 0)
y B = (1, 1, 0)
x
FIGURE 11 SOLUTION z C = (1, 1, 1) P6 = (1, 1, 5/6) P5 = (1, 1, 1/2) 4/6 y A = (1, 0, 0) 2/6 2/6 4/6 P = (1, 1, 1/6) 4 x B = (1, 1, 0) P1 = (1, 1/6, 0) P3 = (1, 5/6, 0) P2 = (1, 1/2, 0)
We write the integrals as sum of integrals and estimate each integral by a Riemann Sum. That is,
ABC
f (x, y, z) ds =
AB
f (x, y, z) ds +
ABC
F · ds =
BC
f (x, y, z) ds ≈
AB
F · ds +
BC
f (Pi ) si +
i=1
F · ds =
3
AB
f (Pi )si
(1)
i=4
(F · T)ds +
6
BC
(F · T)ds
On AB, the unit tangent vector is T = 1, 0, 0, hence F · T = F1 . On BC, the unit tangent vector is T = 0, 0, 1, hence F · T = F3 . Therefore,
ABC
F ds =
AB
F1 ds +
BC
F3 ds ≈
3
F1 (Pi ) si +
i=1
6
F3 (Pi ) si
(2)
i=4
We consider the partitions of AB and BC to three subarcs with equal length si = 13 , therefore (1) and (2) give 1 f (x, y, z) ds ≈ ( f (P1 ) + f (P2 ) + f (P3 ) + f (P4 ) + f (P5 ) + f (P6 )) 3 ABC 1 Fds = (F1 (P1 ) + F1 (P2 ) + F1 (P3 ) + F3 (P4 ) + F3 (P5 ) + F3 (P6 )) 3 ABC We now substitute the values of the functions at the sample points to obtain the following approximations: 1 f (x, y, z) ds ≈ (3 + 3.3 + 3.6 + 4.2 + 4.5 + 4.2) = 7.6 3 ABC 14 2 1 =4 F · ds ≈ (1 + 1 + 2 + 4 + 3 + 3) = 3 3 3 ABC 13. Figure 13 shows three vector fields. In each case, determine whether the line integral around the circle (oriented Estimate the line integrals of f (x, y) and F(x, y) along the quarter circle (oriented counterclockwise) in Figure counterclockwise) is positive, negative, or zero. 12 using the values at the three sample points along each path. Point
f (x, y)
F(x, y)
A B C
1 −2 4
1, 2 1, 3 −2, 4
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(ET CHAPTER 16)
(A)
(B)
(C)
FIGURE 13 SOLUTION The vector line integral of F is the integral of the tangential component of F along the curve. The positive direction of a curve is counterclockwise. T T
T
For the vector field in (A), the line integral around the circle is zero because the contribution of the negative tangential components from the upper part of the circle is the same as the contribution of the positive tangential components from the lower part. For the vector in (B) the contribution of the negative tangential component appear to dominate over the positive contribution, hence the line integral is negative. In (C), the vector field is orthogonal to the unit tangent vector at each point, hence the line integral is zero. In Exercises 15–22,whether computethethe line integralaround of the scalar function over counterclockwise) the curve. Determine line integrals the curves (oriented in Figures 14(A) and (B) are positive or negative. What is the value of the line integral in Figure 14(C)? Explain. 15. f (x, y, z) = z 2 , c(t) = (2t, 3t, 4t) for 0 ≤ t ≤ 2 SOLUTION
Step 1. Compute c (t) We have c (t) =
d 2t, 3t, 4t = 2, 3, 4 dt
⇒
c (t) =
√ 22 + 32 + 42 = 29
Step 2. Write out f (c(t)) We substitute z = 4t in f (x, y, z) = z 2 to obtain: f (c(t)) = 16t 2 Step 3. Compute the line integral. By the Theorem on Scalar Line Integrals we have C
f (x, y, z) ds =
2 0
f (c(t)) c (t) dt =
2 0
16t 2 ·
√
29 dt =
√
29 ·
√ 16 3 2 128 29 t = ≈ 229.8 3 3 0
√ 17. f (x,f (x, y) = 9x− y, 2yy+=z,x 3 c(t) for 0=≤(2x + ≤ t, 1 2 − t, 2t) for −2 ≤ t ≤ 1 y, z)1=+3x SOLUTION The curve is parametrized by c(t) = t, t 3 for 0 ≤ t ≤ 1 Step 1. Compute c (t). We have c (t) =
d 3 t, t = 1, 3t 2 dt
⇒
c (t) =
1 + 9t 4
√ Step 2. Write out f (c(t)). We substitute x = t, y = t 3 in f (x, y) = 1 + 9x y to obtain f (c(t)) = 1 + 9t · t 3 = 1 + 9t 4 Step 3. Compute the line integral. We use the Theorem on Scalar Line Integrals to write C
f (x, y) ds =
1 0
=t+ 2
f (c(t)) c (t) dt = 9t 5 1
1 1 4 4 1 + 9t 1 + 9t dt = 1 + 9t 4 dt 0
= 14 = 2.8 5 0 5
19. f (x, y, z) = xe z3 , piecewise linear path from (0, 0, 1) to (0, 2, 0) to (1, 1, 1). y f (x, y) = 7 , y = 14 x 4 for 1 ≤ x ≤ 2 x
0
Line Integrals
S E C T I O N 17.2
(ET Section 16.2)
573
SOLUTION Let C1 be the segment joining the points (0, 0, 1) and (0, 2, 0) and C2 be the segment joining the points (0, 2, 0) and (1, 1, 1). We parametrize C1 and C2 by the following parametrization:
C1 : c1 (t) = (0, 2t, 1 − t), 0 ≤ t ≤ 1 C2 : c2 (t) = (t, 2 − t, t), 0 ≤ t ≤ 1 For C = C1 + C1 we have
f (x, y, z) ds =
C
f (x, y, z) ds +
C1
C2
f (x, y, z) ds
(1)
We compute the integrals on the right hand side. • The integral over C1 : We have
d 0, 2t, 1 − t = 0, 2, −1 dt
c 1 (t) =
√
c 1 (t) =
⇒
√ 5
0+4+1=
f (c(t)) = xe z = 0 · e(1−t) = 0 2
2
Hence, C1
f (x, y, z) ds =
1 0
f (c1 (t)) c 1 (t) dt =
1 0
0 dt = 0
(2)
• The integral over C2 : We have
c 2 (t) =
d t, 2 − t, t = 1, −1, 1 dt 2
f (c2 (t)) = xe z = tet
c 2 (t) =
⇒
√
1+1+1=
√ 3
2
Hence, C2
f (x, y, z) ds =
1
tet
2√
3 dt
0
Using the substitution u = t 2 we find that
√ 1√ 3 u 3 e du = (e − 1) ≈ 1.488 f (x, y, z) ds = 2 2 C2 0
Hence, C
f (x, y, z) ds ≈ 1.488
t , t 2 , t), 0 ≤ t ≤ 1 21. f (x, y, z) = 2x 2 + 8z, c(t) = (e√ f (x, y, z) = x 2 z, c(t) = (et , 2t, e−t ) for 0 ≤ t ≤ 1 SOLUTION
Step 1. Compute c (t). c (t) =
d t 2 t e , t , t = e , 2t, 1 dt
⇒
c (t) =
e2t + 4t 2 + 1
Step 2. Write out f (c(t)). We substitute x = et , y = t 2 , z = t in f (x, y, z) = 2x 2 + 8z to obtain: f (c(t)) = 2e2t + 8t Step 3. Compute the line integral. We have C
f (x, y, z) ds =
1 0
f (c(t)) c (t) dt =
1 0
(2e2t + 8t) e2t + 4t 2 + 1 dt
We compute the integral using the substitution u = e2t + 4t 2 + 1, du = 2e2t + 8t dt. We get: C
f (x, y, z) ds =
e2 +5 2
u 1/2 du =
2
3
2 2 3/2 e +5 2 2 3/2 u (e + 5) = − 23/2 3 3 2
f (x, y, z) = 3zy−1 + 12x z, c(t) = (t, √t , t3 ), 0 ≤ t ≤ 2 2
(3)
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In Exercises 23–35, compute the line integral of the vector field over the oriented curve. 23. F = x 2 , x y , line segment from (0, 0) to (2, 2) SOLUTION
The oriented line segment is parametrized by c(t) = (t, t),
Therefore,
t varies from 0 to 2.
F (c(t)) = x 2 , x y = t 2 , t · t = t 2 , t 2 c (t) =
The integrand is the dot product:
d t, t = 1, 1 dt
F (c(t)) · c (t) = t 2 , t 2 · 1, 1 = t 2 + t 2 = 2t 2
We now use the Theorem on vector line integral to compute C F · ds: 2 2 2t 3 2 16 F · ds = F (c(t)) · c (t) dt = 2t 2 dt = = 3 0 3 C 0 0 clockwise 25. F = x 2 , x y , circle x 2 + y 2 =2 9 oriented F = 4, y, quarter circle x + y 2 = 1 with x, y ≤ 0 oriented counterclockwise SOLUTION
y
3
x
The oriented path is parametrized by c(t) = (3 cos t, 3 sin t);
t is changing from 2π to 0.
We compute the integrand:
F (c(t)) = x 2 , x y = 9 cos2 t, 9 cos t sin t
c (t) = −3 sin t, 3 cos t F (c(t)) · c (t) = 9 cos2 t, 9 cos t sin t · −3 sin t, 3 cos t = −27 cos2 t sin t + 27 cos2 t sin t = 0 Hence, C
F · ds =
0 2π
F (c(t)) · c (t) dt =
0 2π
0 dt = 0
27. F = x y, xy−x + y,2x c(t) = (1 + t −1 , t 2 ) for 1 ≤ t ≤ 2 F= e , e , piecewise linear path from (1, 1) to (2, 2) to (0, 2) SOLUTION
Step 1. Calculate the integrand. We write the vectors and compute the integrand: c(t) = 1 + t −1 , t 2 F (c(t)) = x y, x + y = 1 + t −1 t 2 , 1 + t −1 + t 2 = t 2 + t, 1 + t −1 + t 2 c (t) =
d 1 + t −1 , t 2 = −t −2 , 2t dt
The integrand is the dot product: F (c(t)) · c (t) = t 2 + t, 1 + t −1 + t 2 · −t −2 , 2t = t 2 + t −t −2 + 1 + t −1 + t 2 · 2t = −1 − t −1 + 2t + 2 + 2t 3 = 2t 3 + 2t + 1 − t −1
Line Integrals
S E C T I O N 17.2
(ET Section 16.2)
575
Step 2. Evaluate the integral. The vector line integral is 2 2 t4 + t 2 + t − ln t 2t 3 + 2t + 1 − t −1 dt = 2 1 1 1 1 16 + 4 + 2 − ln 2 − + 1 + 1 − ln 1 = 11.5 − ln 2 = 2 2
C
F · ds =
2
F (c(t)) · c (t) dt =
29. F = 3zy −1 , 4x, −y , c(t) = (et , et2, t)1 for −1 ≤ t ≤ 1 F = y + z, x, y, c(t) = (t + t , 3 t 3 , 2 + t) for 0 ≤ t ≤ 2 SOLUTION
Step 1. Calculate the integrand. We write out the vectors and compute the integrand: c(t) = et , et , t F (c(t)) = 3zy −1 , 4x, −y = 3te−t , 4et , −et c (t) = et , et , 1 The integrand is the dot product: F (c(t)) · c (t) = 3te−t , 4et , −et · et , et , 1 = 3te−t · et + 4et · et − et · 1 = 3t + 4e2t − et Step 2. Evaluate the integral. The vector line integral is: C
F · ds =
1 −1
F (c(t)) · c (t) dt =
1 −1
3t + 4e2t − et
dt = 0 +
1 −1
4e2t − et
1 dt = 2e2t − et
= 2e2 − e − 2e−2 − e−1 = 2 e2 − e−2 − e − e−1 ≈ 12.157
−1
−y x segment from 0,center 0) to (1, 4, 4)origin oriented counterclockwise , line circle of radius R (0, with at the 31. F =F = 2x − 2y,, y 2− z, z, x + y x + y2 SOLUTION y
R
x
The path has the following parametrization: c(t) = R cos t, R sin t ,
0 ≤ t ≤ 2π
Step 1. Calculate the integrand. We have −y −R sin t x R cos t 1 F (c(t)) = 2 , , = − sin t, cos t = R x + y2 x 2 + y2 R 2 cos2 t + R 2 sin2 t R 2 cos2 t + R 2 sin2 t c (t) =
d R cos t, R sin t = −R sin t, R cos t = R − sin t, cos t dt
The integrand is the dot product: F (c(t)) · c (t) =
1 − sin t, cos t · R − sin t, cos t = sin2 t + cos2 t = 1 R
Step 2. Evaluate the integral. We obtain the following integral: C
F · ds =
2π 0
F (c(t)) · c (t) dt =
2π 0
1 dt = 2π
y , c(t) =x (cost, tan t, t) for 0 ≤ t ≤ π4 33. F = z 2 , x, −y F= 2 , , the square with vertices (1, 1), (1, −1), (−1, −1), and (−1, 1) in the counterclockx + y2 x 2 + y2 wise direction
576
C H A P T E R 17
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(ET CHAPTER 16)
SOLUTION
Step 1. Calculate the integrand. The vectors are: c(t) = (cos t, tan t, t) F (c(t)) = z 2 , x, y = t 2 , cos t, tan t 1 d cos t, tan t, t = − sin t, , 1 c (t) = dt cos2 t The integrand is the dot product F (c(t)) · c (t) = t 2 , cos t, tan t · − sin t,
1 1 + tan t , 1 = −t 2 sin t + 2 cos t cos t
Step 2. Evaluate the integral. The Theorem on vector line integrals gives the following integral: π /4 π /4 1 + tan t dt F · ds = F (c(t)) · c (t) dt = −t 2 sin t + cos t C 0 0 π /4 π /4 π /4 1 + tan t = t 2 cos t − 2t sin t − 2 cos t + ln − ln(cos t) cos t 0 0 0
1 √ π2 1 π 1 1 · √ −2· · √ −2· √ = 2 + 1 − ln 1 − ln √ − ln 1 − (−2) + ln 16 4 2 2 2 2 =
√ √ √ √ √ π2 π π (π − 8) + ln 2 + 2 + 2 − 2 ≈ 1.139 √ − √ − 2 + 2 + ln 1 + 2 + ln 2 = √ 16 2 2 2 16 2
35. F = z 3 , yz,1x , circle yz-plane with center at the origin oriented clockwise when viewed from 1 of radius 2 in the , 1 , c(t) = (t 3 , 2t, t 2 ) for 0 ≤ t ≤ 1 F = , the positive x-axis 3 y +1 z+1 SOLUTION z
y x
The oriented path has the following parametrization: c(t) = (0, 2 cos t, 2 sin t) t is changing from 2π to 0. Step 1. Calculate the integrand. We write out the vectors and compute the integrand: c(t) = (0, 2 cos t, 2 sin t) F (c(t)) = z 3 , yz, x = 8 sin3 t, 4 cos t sin t, 0 c (t) = 0, −2 sin t, 2 cos t The integrand is the dot product: F (c(t)) · c (t) = 8 sin3 t, 4 cos t sin t, 0 · 0, −2 sin t, 2 cos t = −8 cos t sin2 t Step 2. Evaluate the integral. We obtain the following vector line integral: C
F · ds =
0 2π
F (c(t)) · c (t) dt =
0 2π
−8 cos t sin2 t dt =
2π 0
8 sin2 t cos t dt = 8
sin3 t 2π =0 3 0
2 Let f (x, y, z) = x −1 yz and let C be the curve parametrized by c(t) = (ln t, t, t ) for 2 ≤ t ≤ 4. Use a
S E C T I O N 17.2
37.
Use a CAS to calculate
0 ≤ x ≤ π , oriented from left to right. SOLUTION
C
Line Integrals
(ET Section 16.2)
577
x−y x+y ,e e · ds to four decimal places, where C is the curve y = sin x for
Using the parameterization c(t) = t, sin t, our integral becomes
which is calculated to be −4.5088.
π et−sin t , et+sin t · 1, cos t dt, 0
In Exercises 38–39, calculate the line integral of F = e z , e x−y , e y over the given path. 39. The path ABC in Figure 16 The path from P to Q in Figure 15 z C = (0, 0, 6)
A = (2, 0, 0)
B = (0, 4, 0) y
x
FIGURE 16 SOLUTION z C = (0, 0, 6) C3 A = (2, 0, 0) x
C2
C1
B = (0, 4, 0) y
We denote by C1 , C2 , C3 the oriented segments from A to B, from B to C and from C to A. We parametrize these paths by, c 1 (t) = −2, 4, 0
C1 :c1 (t) = (1 − t)(2, 0, 0) + t (0, 4, 0) = (2 − 2t, 4t, 0), 0 ≤ t ≤ 1 C2 :c2 (t) = (1 − t)(0, 4, 0) + t (0, 0, 6) = (0, 4 − 4t, 6t), 0 ≤ t ≤ 1
⇒
C3 :c3 (t) = (1 − t)(0, 0, 6) + t (2, 0, 0) = (2t, 0, 6 − 6t), 0 ≤ t ≤ 1
c 2 (t) = 0, −4, 6 c 3 (t) = 2, 0, −6
Since C = C1 + C2 + C3 we have, C
F · ds =
3 i=1 Ci
F · ds
(1)
We compute the integrals on the right-hand side: C1
C2
C3
F · ds =
1 e0 , e2−6t , e4t · −2, 4, 0 dt = 1, e2−6t , e4t · −2, 4, 0 dt 0
1 1 2 2 2 = −2 + 4e2−6t dt = −2t − e2−6t = e2 − e−4 − 2 3 3 3 0 0 1 1 F · ds = e6t , e−4+4t , e4−4t · 0, −4, 6 dt = −4e−4+4t + 6e4−4t dt 0
0
1 3 3 5 = −e−4+4t − e4−4t = e4 + e−4 − 2 2 2 0 1 1 F · ds = e6−6t , e2t , e0 · 2, 0, −6 dt = 2e6−6t − 6 dt 0
0
1 1 1 19 = − e6−6t − 6t = e6 − 3 3 3 0
We substitute these values in (1) to obtain the solution: 2 2 2 −4 3 4 1 6 19 5 e − e −2 + e + e−4 − e − F · ds = + 3 3 2 2 3 3 C
578
C H A P T E R 17
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(ET CHAPTER 16)
1 6 3 4 2 2 65 1 −4 e + e + e − + e 3 2 3 6 3
=
In Exercises F be thePvortex vector field it swirls the origin as shownSuppose in Figure 18) Let C41–44, be the let path from to Q in Figure 17 (so-called that traces because C1 , C2 , and C3 inaround the orientation indicated. that −y x F · ds = F 5,= 2 F 2· ,ds 2= 8, 2 F · ds = 8 x C+ y x + y C C 1
Determine: F · ds (a)
3
(b)
−C3
C2
F · ds
(c)
−C1 −C3
F · ds
(d) What is the value of the line integral of F over the path that traverses the loop C2 four times in the clockwise direction? −y x FIGURE 18 Vector field F = , . x 2 + y2 x 2 + y2 F · ds, where C is the circle of radius 2 centered at the origin oriented counterclockwise (Figure 18). 41. Let I = C
(a) Do you expect I to be positive, negative, or zero? (b) Evaluate I . (c) Verify that I changes sign when C is oriented in the clockwise direction. SOLUTION
(a) When the circle is oriented counterclockwise, the dot product of F with the unit tangent vector at each point along the circle is positive. Therefore, we expect the vector line integral I to be positive. (b) y P
2
x
The circle of radius 2 oriented counterclockwise has the parametrization: c(t) = (2 cos t, 2 sin t),
0 ≤ t ≤ 2π
Hence, F (c(t)) =
−2 sin t
2 cos t
1 , = − sin t, cos t 2 4 cos2 t + 4 sin2 t 4 cos2 t + 4 sin2 t
c (t) = −2 sin t, 2 cos t Therefore, the integrand is the dot product, F (c(t)) · c (t) =
1 − sin t, cos t · −2 sin t, 2 cos t = sin2 t + cos2 t = 1 2
We obtain the following integral: C
F · ds =
2π 0
F (c(t)) · c (t) dt =
2π 0
1 dt = 2π
(c) When C is oriented in the clockwise direction, the parameter t is changing from 2π to 0, therefore, the line integral is, 0 2π F · ds = F (c(t)) · c (t) dt = − 1 dt = −2π C
2π
0
S E C T I O N 17.2
Line Integrals
(ET Section 16.2)
579
y
x
2
43. Calculate F · ds, where A is the arc of angle θ0 on the circle of radius R centered at the origin oriented counterComputeA F · ds, where C R is the circle of radius R centered at the origin oriented counterclockwise. Show clockwise. Note: ACbegins at (R, 0) and ends at (R cos θ0 , R sin θ0 ). R that the result is independent of R. SOLUTION We use the following parametrization for A: c(θ ) = (R cos θ , R sin θ ),
0 ≤ θ ≤ θ0
y (R cos q 0, R sin q 0 ) A
q0
x
(R, 0)
Hence, c (θ ) = −R sin θ , R cos θ −y −R sin θ x R cos θ F (c(θ )) = 2 , , = x + y2 x 2 + y2 R 2 cos2 θ + R 2 sin2 θ R 2 cos2 θ + R 2 sin2 θ −R sin θ R cos θ − sin θ cos θ , = , = R R R2 R2 The integrand is the dot product: F (c(θ )) · c (θ ) =
− sin θ cos θ , · −R sin θ , R cos θ = sin2 θ + cos2 θ = 1 R R
We now compute the line integral using the Theorem on vector line integrals. We get: θ0 θ0 θ F · ds = F (c(θ )) · c (θ ) d θ = 1 d θ = θ 00 = θ0 A
45. (a) (b) (c)
0
0
Calculate the line integral of the constant vector field F = 2, −1, 4 along the segment P Q, where: Let a > 0, b < c. Show that the integral of F along the segment from P = (a, b) to Q = (a, c) is equal to the P = (0, Q =is (1, 0) P0,O0), angle Q (O the0, origin). P = (0, 0, 0), Q = (4, 3, 5) P = (3, 2, 3), Q = (4, 8, 12)
SOLUTION
(a) The segment P Q, where P = (0, 0, 0) and Q = (1, 0, 0) is parametrized by c(t) = (t, 0, 0),
0≤t ≤1
Therefore, c (t) = 1, 0, 0 and we obtain the integral, 1 1 1 2, −1, 4 · 1, 0, 0 dt = F · ds = F (c(t)) · c (t) dt = 2 dt = 2 C
0
0
0
(b) The segment P Q, where P = (0, 0, 0) and Q = (4, 3, 5) has the parametrization, c(t) = (4t, 3t, 5t),
0≤t ≤1
Therefore, c (t) = 4, 3, 5 and we obtain the following integral: 1 1 1 2, −1, 4 · 4, 3, 5 dt = F · ds = F (c(t)) · c (t) dt = 25 dt = 25 C
0
0
0
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C H A P T E R 17
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(ET CHAPTER 16)
(c) The segment P Q, where P = (3, 2, 3) and Q = (4, 8, 12) has the parametrization, c(t) = (3 + t, 2 + 6t, 3 + 9t),
0≤t ≤1
Therefore, c (t) = 1, 6, 9 and we obtain the line integral C
F · ds =
1 0
F (c(t)) · c (t) dt =
1
2, −1, 4 · 1, 6, 9 dt =
0
1 0
32 dt = 32
47. Figure 19 shows the vector field F(x, y) = x, x. Show that if F is a constant vector field and C is any oriented path from P to Q, then (a) Are F · ds and F · ds equal? If not, which is larger? −→ AB DC F · ds = F · P Q (b) Which is smaller: the line integral of F over the path ADC or ABC? C y B
C
A
D x
FIGURE 19 SOLUTION
(a) Since x is constant on AB and DC, F(x, y) = x, x is also constant on these segments. y
B
l
l A
C l
l
a1
D a2
x
Let a1 and a2 denote the constant values of x on the segments AB and DC respectively, and l denote the lengths of these segments. By Exercise 46 we have F · ds = a1 , a1 · 0, l = a1 · 0 + a1 · l = a1 l
AB
DC
F · ds = a2 , a2 · 0, l = a2 · 0 + a2 · l = a2 l
Since a1 < a2 we have AB F · ds < DC F · ds. (b) We compute the integral over BC. This segment is parametrized by: c(t) = (a1 + lt, b) , 0 ≤ t ≤ 1. Hence, F (c(t)) = x, x = a1 + lt, a1 + lt , c (t) = l, 0 F (c(t)) · c (t) = a1 + lt, a1 + lt · l, 0 = a1 l + l 2 t Thus, BC
F · ds =
1 l 2 t 2 1 l2 = a1 l + a1 l + l 2 t dt = a1 lt + 2 t=0 2 0
We see that the line integral does not depend on b, therefore, F · ds = AD
BC
F · ds
(1)
S E C T I O N 17.2
Line Integrals
(ET Section 16.2)
581
In part (a) we showed that:
AB
Combining (1) and (2) gives: F · ds = ABC
F · ds <
AB
F · ds +
DC
F · ds
BC
F · ds <
(2)
DC
F · ds +
AD
F · ds =
ADC
F · ds
49. Calculate the work done by the force field F = x, y, z along the path (cos t, sin t, t) for 0 ≤ t ≤ 3π . Calculate the work done by a field F = x + y, x − y when an object moves from (0, 0) to (1, 1) along each of SOLUTION the paths The y = work x 2 anddone x =byy 2the . force field F is the line integral: F · ds W = C
We compute the integrand: F (c(t)) = x, y, z = cos t, sin t, t c (t) =
d cos t, sin t, t = − sin t, cos t, 1 dt
F (c(t)) · c (t) = cos t, sin t, t · − sin t, cos t, 1 = − cos t sin t + sin t cos t + t = t We obtain the following integral: W =
3π 0
F (c(t)) · c (t) dt =
3π 0
t dt =
t 2 3π 9π 2 = 2 0 2
Further Let Insights and Challenges f (x, y, z) ds. Assume that I =
f (x, y, z) ≥ m for some number m and all points (x, y, z) on C. Which Cthe text, the value of a scalar line integral does not depend on the choice of parametrization. Prove 51. of Asthe observed in following conclusions is correct? Explain. this directly. Namely, suppose that c1 (t) and c(t) are two parametrizations of C and that c1 (t) = c(ϕ (t)), where ϕ (t) is (a) I ≥ m an increasing function. Use the Change of Variables Formula to verify that (b) I ≥ m L, where L is the length of C b d f (c1 (t))c 1 (t) dt = f (c(t))c (t) dt c
a
where a = ϕ (c) and b = ϕ (d). SOLUTION
We compute the integral ab f (c(t)) c (t) dt using the substitution t = ϕ (u), a = ϕ (c), b = ϕ (d). We
get: b
f (c1 (t)) c (t) dt =
a
ϕ −1 (b) ϕ −1 (a)
f c ϕ (t) c ϕ (t) ϕ (u) du
Since ϕ is an increasing function, ϕ (u) > 0 for all u, therefore: c ϕ (u) ϕ (u) = c ϕ (u) ϕ (u)
(1)
(2)
By the Chain Rule for vector valued functions, we have, d c ϕ (u) = ϕ (u)c ϕ (u) du
(3)
Combining (2) and (3) gives: c ϕ (u) ϕ (u) = d c ϕ (u) = d c1 (u) = c (u) 1 du du
(4)
We substitute (4) in (1) to obtain: b d d f (c(t)) c (t) dt = f (c1 (u)) c 1 (u) du = f (c1 (t)) c 1 (t) dt a
c
c
The last step is simply replacing the dummy variable of integration u by t. 53. Use Eq. (8) to calculate the average value of f (x, y) = x − y along the segment from P = (2, 1) to Q = (5, 5). Suppose we wish to compute the average value of a continuous function f (x, y, z) along a curve C of length L. Given a large number N , divide C into N consecutive arcs C1 , . . . , C N , each of length L/N , and let Pi be a sample 1 f (Pi ) point in Ci (Figure 20). Then the points Pi are somewhat evenly spaced along the curve and the sum N i=1
582
C H A P T E R 17
L I N E A N D S U R FA C E I N T E G R A L S SOLUTION
(ET CHAPTER 16)
We can parametrize this line segment by c(t) = (2 + 3t, 1 + 4t),
0≤t ≤1
Therefore, c (t) = 3, 4
c (t) =
⇒
√ 9 + 16 = 5
We compute the length of the curve, L=
1 0
c (t) dt =
1√ 0
5 dt =
√
5
Thus, using our values for x and y given above, we find that Av( f ) =
1 1 1 1 1 1 x − y dt = √ (2 + 3t) − (1 + 4t) dt = √ 1 − t dt = √ L C 5 0 5 0 2 5
55. The temperature (in degrees centigrade) at a point P on a circular wire of radius 2 cm centered at the origin is equal x ≤ 1. Use Eq. (8) to calculate thePaverage of Compute x y along the the average curve y = x 2 for 0 ≤along to the square of the distance from to P0 =value (2, 0). temperature the wire. SOLUTION y P = (x, y)
P0 = (2, 0) x
The temperature at a point P(x, y) on the wire is given by the function, T (x, y) = (x − 2)2 + y 2 The length of the wire is the length of the circle of radius 2, L = 2π · 2 = 4π . Therefore, the average temperature along the wire is, 1 1 Av(T ) = T ds = (x − 2)2 + y 2 ds L C 4π C To compute the line integral, we parametrize the circle by: c(t) = (2 cos t, 2 sin t),
0 ≤ t ≤ 2π .
Then, c(t) = −2 sin t, 2 cos t
⇒
c (t) =
4 sin2 t + 4 cos2 t = 2
We express T in terms of the parameter: T (c(t)) = (x − 2)2 + y 2 = (2 cos t − 2)2 + (2 sin t)2 = 4 cos2 t − 8 cos t + 4 + 4 sin2 t = 4 cos2 + sin2 t + 4 − 8 cos t = 8(1 − cos t) We obtain the integral, 2π 2π 2π 4 · 2π 1 1 4
t − sin t = Av(T ) = T (c(t)) c (t) dt = 16(1 − cos t) dt = =8 4π 0 4π 0 π π 0 57. Let F = y, x. Prove that if C is any path from (a, b) to (c, d), then Let F = x, 0. Prove that if C is any path from (a, b) to (c, d), then F · ds = cd 1 −2 ab 2 C F · ds = (c − a ) 2 C SOLUTION
S E C T I O N 17.3 (c, d)
c
(a, b)
We denote a parametrization of the path by: c(t) = (x(t),
Conservative Vector Fields
(ET Section 16.3)
583
584
C H A P T E R 17
L I N E A N D S U R FA C E I N T E G R A L S
(ET CHAPTER 16)
SOLUTION
(a) This statement is always true, since every gradient vector field is conservative. (b) If F is conservative on a connected domain D, then F has a potential function D and consequently the cross partials of F are equal in D. (c) If the cross partials of F are equal in a simply-connected region D, then F is a gradient vector field in D. F · ds = 4. 4. Let C, D, and E be the oriented curves in Figure 15 and let F = ∇ ϕ be a gradient vector field such that C
What are the values of the following integrals? F · ds (a)
(b)
D
E
F · ds
y Q
C D E P
x
FIGURE 15 SOLUTION Since F is a gradient vector field the integrals over closed paths are zero. Therefore, by the equivalent conditions for path independence we have: (a) D F · ds = C F · ds = 4 (b) E F · ds = −C F · ds = − C F · ds = −4
Exercises
1. Let ϕ (x, y, z) = x y sin(yz). Evaluate SOLUTION
c
∇ ϕ · ds, where c is any path from (0, 0, 0) to (1, 1, π ).
By the Fundamental Theorem for Gradient Vector Fields, we have: ∇ ϕ · ds = ϕ (1, 1, π ) − ϕ (0, 0, 0) = 1 · 1 sin π − 0 = 0 c
In Exercises 3–8, verify that F = ∇ ϕ and evaluate the line integral of F over the given path. Let F be the gradient of ϕ (x, y, z) = x y + z 2 . Compute the line integral of F along: 2 ; (1, 2,c(t) (a)= The line segment 2) = (t, 2t −1 ) for 1 ≤ t ≤ 4 3, 6y, ϕ (x, y,from z) =(1, 3x 1, + 1) 3yto 3. F 2 (b) A circle the x z-plane SOLUTION Theingradient of ϕ = 3x + 3y is: (c) The upper half of a circle of radius 4 with its center(0, 1, 1) in the yz-plane, oriented clockwise ∂ϕ ∂ϕ , ∇ϕ = = 3, 6y = F ∂ x ∂y Using the Fundamental Theorem for Gradient Vector Fields, we have: 9 1 1 − ϕ (1, 2) = 3 · 4 + 3 · − (3 · 1 + 3 · 4) = − F · ds = ϕ (c(4)) − ϕ (c(1)) = ϕ 4, 2 4 4 c upper half of the unit circle centered at the √ origin oriented counterclock5. F = x y2 , x 2 2y , ϕ (x, y)2 = 12 x 2 y 2 ; F = sin y , 2x y cos y , ϕ (x, y) = x sin y 2 ; line segment from (2, 0) to (3, π ) wise SOLUTION
We compute the gradient of ϕ (x, y) = 12 x 2 y 2 : ∂ϕ ∂ϕ , ∇ϕ = = x y2, x 2 y = F ∂ x ∂y
We now use the Fundamental Theorem of Gradient Vector Fields. The terminal point is Q = (−1, 0) and the initial point is P = (1, 0), therefore: 1 1 F · ds = ϕ (Q) − ϕ (P) = ϕ (−1, 0) − ϕ (1, 0) = · (−1)2 · 02 − · 12 · 02 = 0 2 2 c y
c Q = (−1, 0)
P = (1, 0)
x
S E C T I O N 17.3
Conservative Vector Fields
(ET Section 16.3)
585
7. F = ye z , xe z , x ye z , 2 ϕ (x, y, z) = x ye z ; c(t) = (t 2 , t 3 , t − 1) for 1 ≤ t ≤ 2 F = y + 2x z, x, x , ϕ (x, y, z) = x y + x 2 z; any path from (1, 2, 1) to (3, 1, 0) SOLUTION We verify that F is the gradient of ϕ : ∂ϕ ∂ϕ ∂ϕ , , ∇ϕ = = ye z , xe z , x ye z = F ∂ x ∂y ∂z We use the Fundamental Theorem for Gradient Vectors with the initial point c(1) = (1, 1, 0) and terminal point c(2) = (4, 8, 1), to obtain: F · ds = ϕ (4, 8, 1) − ϕ (1, 1, 0) = 32e − 1 c
9. Find a potential function for F = 2x y + 5, x 2 − 4z, −4y and evaluate x z , , ln(x − y) , ϕ (x, y, z) = z ln(x − y); F= x−y y−z F · ds 2 ellipse 2x + 3(y − 4)2 = 12 in the clockwise direction c 2 where c(t) = (t 2 , sin(π t), et −2t ) for 0 ≤ t ≤ 2.
SOLUTION
We find a potential function ϕ (x, y, z) for F, using the following steps. ∂ϕ
Step 1. Use the condition ∂ x = F1 . ϕ is an antiderivative of F1 = 2x y + 5 when y and z are fixed, therefore, ϕ (x, y, z) = (2x y + 5) d x = x 2 y + 5x + g(y, z)
(1)
∂ϕ
Step 2. Use the condition ∂ y = F2 . We have, ∂ 2 x y + 5x + g(y, z) = x 2 − 4z ∂y x 2 + g y (y, z) = x 2 − 4z
⇒
g y (y, z) = −4z
We integrate with respect to y, holding z fixed: g(y, z) =
−4z d y = −4zy + h(z)
Combining with (1) gives:
ϕ (x, y, z) = x 2 y + 5x − 4zy + h(z) ∂ϕ
Step 3. Use the condition ∂z = F3 . We have, ∂ 2 x y + 5x − 4zy + h(z) = −4y ∂z −4y + h (z) = −4y h (z) = 0
⇒
h(z) = c
Substituting in (2) we obtain the general potential function:
ϕ (x, y, z) = x 2 y + 5x − 4zy + c To compute the line integral we need one of the potential functions. We choose c = 0 to obtain the function,
ϕ (x, y, z) = x 2 y + 5x − 4zy We now use the Fundamental Theorem for Gradient Vector Fields to evaluate the line integral: F · ds = ϕ γ (2) − ϕ γ (0) = ϕ (4, 0, 1) − ϕ (0, 0, 1) = 20 − 0 = 20 c
2 y over any path from (0, 0, 0) to (3, 2, 1). 11. Find the line of F =condition 2x yz, x 2and z, xfind Verify theintegral cross-partials a potential function for SOLUTION We first show that the cross partials condition is satisfied: F = 2x yz −1 + yz, x 2 z −1 + x z, −x 2 yz −2 + x y ∂ ∂ F1 = (2x yz) = 2x z on the domain {z = 0}. ∂ F2 ∂ F1 ∂y ∂y = ⇒ ∂y ∂x ∂ F2 ∂ 2 = x z = 2x z ∂x ∂x
(2)
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C H A P T E R 17
L I N E A N D S U R FA C E I N T E G R A L S
(ET CHAPTER 16)
∂ 2 ∂ F2 = x z = x2 ∂z ∂z ∂ 2 ∂ F3 = x y = x2 ∂y ∂y ∂ 2 ∂ F3 = x y = 2x y ∂x ∂x ∂ F1 ∂ = (2x yz) = 2x y ∂z ∂z
⇒
∂ F3 ∂ F2 = ∂z ∂y
⇒
∂ F1 ∂ F3 = ∂x ∂z
Since the cross partials condition is satisfied at all points, F is conservative. We find a potential function for F. ∂ϕ
Step 1. Use the condition ∂ x = F1 . ϕ is an antiderivative of F1 = 2x yz when y and z are fixed. Therefore: ϕ (x, y, z) = 2x yz d x = x 2 yz + g(y, z)
(1)
∂ϕ
Step 2. Use the condition ∂ y = F2 . By (1) we have: ∂ϕ = x 2 yz + g(y, z) = x 2 z ∂y x 2 z + g y (y, z) = x 2 z
⇒
g y (y, z) = 0
It follows that g(y, z) = h(z). Substituting in (1) gives
ϕ (x, y, z) = x 2 yz + h(z)
(2)
∂ϕ
Step 3. Use the condition ∂z = F3 . This condition along with (2) gives: ∂ = x 2 yz + h(z) = x 2 y ∂z x 2 y + h (z) = x 2 z h (z) = 0
⇒
h(z) = C
Substituting in (2) we get
ϕ (x, y, z) = x 2 yz + C Since only one potential function is needed, we choose the one corresponding to C = 0. That is,
ϕ (x, y, z) = x 2 yz Using the Fundamental Theorem for Gradient Vectors we obtain: F · ds = ϕ (3, 2, 1) − ϕ (0, 0, 0) = 32 · 2 · 1 − 0 = 18 c
In Exercises 12–17, determine whether the vector field is conservative and, if so, find a potential function. 13. F = 0, x, y F = z, 1, x ∂F ∂F ∂F ∂F SOLUTION Since ∂ y1 = ∂∂y (0) = 0 and ∂ x2 = ∂∂x (x) = 1, we have ∂ y1 = ∂ x2 . Therefore F does not satisfy the cross-partial condition, hence F is not conservative. 15. F = y, x,2z 3 F = y , 2x y + e z , ye z SOLUTION We examine whether the field F = y, x, z 3 satisfies the cross partials condition. ∂ ∂ F1 = (y) = 1 ∂y ∂y ∂ F2 ∂ = (x) = 1 ∂x ∂x ∂ ∂ F2 = (x) = 0 ∂z ∂z ∂ F3 ∂ 3 = z =0 ∂y ∂y
⇒
∂ F1 ∂ F2 = ∂y ∂x
⇒
∂ F3 ∂ F2 = ∂z ∂y
Conservative Vector Fields
S E C T I O N 17.3
∂ 3 ∂ F3 = z =0 ∂x ∂x ∂ F1 ∂ = (y) = 0 ∂z ∂z
(ET Section 16.3)
587
∂ F1 ∂ F3 = ∂x ∂z
⇒
Since F satisfies the cross partials condition everywhere, F is conservative. We find a potential function for F. ∂ϕ
Step 1. Use the condition ∂ x = F1 . ϕ is an antiderivative of F1 = y when y and z are fixed. Therefore: ϕ (x, y, z) = y d x = yx + g(y, z)
(1)
∂ϕ
Step 2. Use the condition ∂ y = F2 . By (1) we have: ∂ (yx + g(y, z)) = x ∂y x + g y (y, z) = x
⇒
g y (y, z) = 0
Therefore, g(y, z) = g(z). Substituting in (1) gives:
ϕ (x, y, z) = yx + g(z)
(2)
∂ϕ
Step 3. Use the condition ∂z = F3 . Using (2) we have: ∂ (yx + g(z)) = z 3 ∂z g (z) = z 3
⇒
g(z) =
1 4 z +c 4
Substituting in (2) gives the following general potential function: 1 ϕ (x, y, z) = yx + z 4 + c 4 Choosing c = 0 we obtain the potential:
ϕ (x, y, z) = yx +
z4 . 4
17. F = cos z, 2y, −x sin z F = cos(x z), sin(yz), x y sin z SOLUTION We examine whether F satisfies the cross partials condition: ∂ F1 ∂ = (cos z) = 0 ∂y ∂y ∂ F2 ∂ = (2y) = 0 ∂x ∂x ∂ ∂ F2 = (2y) = 0 ∂z ∂z ∂ ∂ F3 = (−x sin z) = 0 ∂y ∂y ∂ ∂ F3 = (−x sin z) = − sin z ∂x ∂x ∂ F1 ∂ = (cos z) = − sin z ∂z ∂z
⇒
∂ F1 ∂ F2 = ∂y ∂x
⇒
∂ F3 ∂ F2 = ∂z ∂y
⇒
∂ F1 ∂ F3 = ∂x ∂z
We see that the conditions are satisfied, therefore F is conservative. We find a potential function for F. ∂ϕ
Step 1. Use the condition ∂ x = F1 . ϕ (x, y, z) is an antiderivative of F1 = cos z when y and z are fixed, therefore: ϕ (x, y, z) = cos z d x = x cos z + g(y, z) (1) ∂ϕ
Step 2. Use the condition ∂ y = F2 . Using (1) we get: ∂ (x cos z + g(y, z)) = 2y ∂y
588
C H A P T E R 17
L I N E A N D S U R FA C E I N T E G R A L S
(ET CHAPTER 16)
g y (y, z) = 2y We integrate with respect to y, holding z fixed: 2y d y = y 2 + g(z)
g(y, z) = Substituting in (1) gives
ϕ (x, y, z) = x cos z + y 2 + g(z)
(2)
∂ϕ
Step 3. Use the condition ∂z = F3 . By (2) we have ∂ x cos z + y 2 + g(z) = −x sin z ∂z −x sin z + g (z) = −x sin z g (z) = 0
⇒
g(z) = c
Substituting in (2) we obtain the general potential function:
ϕ (x, y, z) = x cos z + y 2 + c Choosing c = 0 gives the potential function:
ϕ (x, y, z) = x cos z + y 2 . 1 −1 , the work 19. LetCalculate F= . Calculate the when work against F is required move object from (1, 1) to (3,and 4) along path 16 in expended moved to from O an to Q along segments OP P Q inany Figure a2 particle x y 2 . How much work is expended moving in a complete circuit around the in the presence of the force field F = x , y the first quadrant. square? SOLUTION F is a conservative force, since F = −∇ ϕ with potential energy ϕ (x, y) = ln y − ln x. The work required to move an object from (1, 1) to (3, 4) along any path C is equal to the change in potential energy: Work against F = − F · ds = ϕ (3, 4) − ϕ (1, 1) = (ln 4 − ln 3) − (ln 1 − ln 1) = ln 4 − ln 3 C
21. vector fieldfield F ininFigure horizontal appears to the depend on only the figure x-coordinate. that P Q, Qand R, and P R be segments in the with theSuppose orientations Let FThe be the vector Figure17 17.is Let g(x), 0. Prove that F = indicated. F · ds positive, negative, or zero? (a) Is F · ds = F · ds PR PR PQ F · ds? (b) What is the value of b
QR
by showing that both integrals are equal to
g(x) d x. a y
R
P
Q a
b
x
FIGURE 17 SOLUTION
The vector field F has the form: F = g(x), 0
Since ∂∂Fy1 = ∂∂y (g(x)) = 0 and ∂∂Fx2 = 0, we have ∂∂Fy1 = ∂∂Fx2 consequently F is conservative. Therefore: F · ds = F · ds + F · ds PR
PQ
(1)
QR
The vector field F is orthogonal to Q R, therefore the tangential component of F at each point along Q R is zero. We conclude that: F · ds = 0 QR
Conservative Vector Fields
S E C T I O N 17.3
Substituting in (1) we get:
(ET Section 16.3)
589
F · ds =
PR
PQ
F · ds
Let’s try it again, this time using the hint. We compute the integral along P Q, using the parametrization P Q : c(t) = (t, y0 ),
a≤t ≤b
we get: PQ
F · ds =
b a
F (c(t)) · c (t) dt =
b a
g(t), 0 · 1, 0 dt =
b a
g(t) dt =
b g(x) d x a
If we now compute the integral along P R with the parameterization P R : c(t) = (t, y(t)),
a ≤ t ≤ b,
we get the same thing: b b b b g(t), 0 · 1, y dt = F · ds = F (c(t)) · c (t) dt = g(t) dt = g(x) d x PR
a
a
a
a
We see, again, that these are equal. 23. How much energy joules) does it take to carry a 2-kg object from sea level along any path to the top of a hill that −1 2(in −1 2 , 2z log(x y) . F = Assume x z , ythatzthe is 1,000Let m high? force of gravity F is constant −mg in the vertical direction, where g = 9.8 m/s2 . Hint: Find(a) a potential function for ϕ ,F. where ϕ (x, y, z) = z 2 log(x y). Verify that F=∇ SOLUTION The force of gravity is F = 0, e2t−mg, , t 2 fortherefore 1 ≤ t ≤ F3.= −∇ ϕ for ϕ (x, y, z) = mgz. The work performed (b) Evaluate F · ds, where c(t) = et , 0, by the gravitationalcfield is the line integral of F over the path. Since F is conservative, the energy is independent of the path connecting the two points. Using the Fundamental1Theorem for Gradient Vector Fields we have: (c) Evaluate F · ds for any path c from P = ( 2 , 4, 2) to Q = (2, 3, 3) contained in the region x > 0, y > 0, c
z > 0. W = − F · ds = ϕ (z = 1000) − ϕ (z = 0) = mg · 1000 = 2 · 9.8 · 103 = 19,600 joules c (d) Why was it necessary to specify that the path lie in the region where x, y, and z are positive?
−y x Further , Challenges be the vortex vector field. Determine LetInsights F = 2 and 2 2 2
F · ds for each of the paths in Figure 18. x x+ y c y , is defined on the domain D = {(x, y) = (0, 0)}. x 2 + y2 x 2 + y2 Show that F satisfies the cross-partials condition on D. Show that ϕ (x, y) = 12 ln(x 2 + y 2 ) is a potential function for F. Is D simply connected? Do these results contradict Theorem 4?
x +y 25. The vector field F = (a) (b) (c) (d)
SOLUTION
(a) We compute the partials of F: ∂ ∂ F2 = ∂x ∂x ∂ ∂ F1 = ∂y ∂y
y x 2 + y2 x x 2 + y2
= =
−2x y
2 x 2 + y2 −2yx
2 x 2 + y2
The cross partials are equal in D.
(b) We compute the gradient of ϕ (x, y) = 12 ln x 2 + y 2 : ∇ϕ =
∂ϕ ∂ϕ 2y y x 1 2x , , , = =F = ∂x ∂x 2 x 2 + y2 x 2 + y2 x 2 + y2 x 2 + y2
(c) D is not simply-connected since it has a “hole” at the origin. (d) The requirement in Theorem 4 (that the domain be simply connected) is a sufficient condition for a vector field with equal cross-partials to have a potential function. It is not necessary, since as in our example, even if the domain is not simply-connected the field may have a gradient function. Moreover, for any closed curve in D, ϕ have the same value after completing one round along c. This is perhaps best seen by noting that ϕ = log(r ) in polar coordinates, which will be independent of θ . Therefore, F · ds = 0 c
Hence, F is conservative.
590
C H A P T E R 17
L I N E A N D S U R FA C E I N T E G R A L S
(ET CHAPTER 16) PQ
17.4 Parametrized Surfaces and Surface Integrals
(ET Section 16.4)
Preliminary Questions 1. What is the surface integral of the function f (x, y, z) = 10 over a surface of total area 5? SOLUTION
Using Surface Integral and Surface Area we have: f (x, y, z) d S = f ((u, v)) n(u, v) du dv = D
S
= 10
D
D
10n(u, v) du dv
n(u, v) du dv = 10 Area(S) = 10 · 5 = 50
2. What interpretation can we give to the length n of the normal vector for a parametrization (u, v)? SOLUTION
The approximation: Area Si j ≈ n u i j , vi j Area Ri j
tells that n is a distortion factor that indicates how much the area of a small rectangle Ri j is altered under the map φ .
j + 0.02 Φ
Rij
Sij
j ui ui + 0.01
u
3. A parametrization maps a rectangle of size 0.01 × 0.02 in the uv-plane onto a small patch S of a surface. Estimate Area(S) if Tu × Tv = 1, 2, 2 at a sample point in the rectangle. SOLUTION
We use the estimation Area(S) ≈ n(u, v)Area(R)
where n(u, v) = Tu × Tv at a sample point in R. We get: Area(S) ≈ 1, 2, 2 · 0.01 · 0.02 =
12 + 22 + 22 · 0.0002 = 0.0006
+ 0.02 Φ
R
u
u + 0.01
S
u
4. A small surface S is divided into three small pieces, each of area 0.2. Estimate the values 0.9, 1, and 1.1 at sample points in these three pieces. SOLUTION
S
f (x, y, z) d S if f (x, y, z) takes
We use the approximation obtained by the Riemann Sum: f (x, y, z) d S ≈ f Pi j Area Si j = 0.9 · 0.2 + 1 · 0.2 + 1.1 · 0.2 = 0.6 S
ij
Parametrized Surfaces and Surface Integrals
S E C T I O N 17.4
(ET Section 16.4)
591
5. A surface S has a parametrization whose domain is the square 0 ≤ u, v ≤ 2 such that n(u, v) = 5 for all (u, v). What is Area(S)? Writing the surface area as a surface integral where D is the square [0, 2] × [0, 2] in the uv-plane, we have: Area(S) = n(u, v) du dv = 5 du dv = 5 1 du dv = 5Area(D) = 5 · 22 = 20
SOLUTION
D
D
D
6. What is the outward-pointing unit normal to the sphere of radius 3 centered at the origin at P = (2, 2, 1)? SOLUTION
The outward-pointing normal to the sphere of radius R = 3 centered at the origin is the following vector: (1) cos θ sin φ , sin θ sin φ , cos φ z 1
2
P = (2, 2, 1)
y
2 x
We compute the values in (1) corresponding to P = (2, 2, 1): x = y = 2, z = 1 hence 0 ≤ θ ≤ π2 and 0 < φ < π2 . We get: √ 2 1 z 1 2 2 ⇒ sin φ = 1 − = cos φ = = ρ 3 3 3 1 x 2 1 1 √ = √ ⇒ sin θ = 1 − = √ = cos θ = 2 2 ρ sin φ 2 2 2 3· 3
Substituting in (1) we get the following unit normal: √ √ 1 2 2 1 2 2 1 2 2 1 ,√ · , √ · = , , 3 3 3 3 3 3 2 2
Exercises 1. Match the parametrization with the surface in Figure 16. (a) (u, cos v, sin v) (b) (u, u + v, v) (c) (u, v 3 , v) (d) (cos u sin v, 3 cos u sin v, cos v) (e) (u, u(2 + cos v), u(2 + sin v)) z
x (i)
(ii)
y (iii)
z
x
y y
x
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C H A P T E R 17
L I N E A N D S U R FA C E I N T E G R A L S
(ET CHAPTER 16)
Show that (u, v) = (2u + 1, u − v, 3u + v) parametrizes the plane 2x − y − z = 2. Then: Show that θ) = (r v). cos θ , r sin θ , 1 − r 2 ) parametrizes the paraboloid z = 1 − x 2 − y 2 . Describe the grid Calculate Tu , (r, Tv , and n(u, curves of this parametrization. Find the area of S = (D), where D = {(u, v) : 0 ≤ u ≤ 2, 0 ≤ v ≤ 1}. f (x, y, z) d S. (c) Express f (x, y, z) = yz in terms of u and v and evaluate
3. (a) (b)
S
SOLUTION
We show that x = 2u + 1, y = u − v, and z = 3u + v satisfy the equation of the plane, 2x − y − z = 2(2u + 1) − (u − v) − (3u + v) = 4u + 2 − u + v − 3u − v = 2
Moreover, for any x, y, z satisfying 2x − y − z = z, there are values of u and v such that x = 2u + 1, y = u − v, and z = 3u + v, since the following equations can be solved for u and v: x = 2u + 1 y =u−v
⇒
z = 3u + v
u=
x −1 , 2
v=
x −1 −y 2
2x − y − z = 2 We conclude that (u, v) parametrizes the whole plane 2x − y − z = 2. (a) The tangent vectors Tu and Tv are: ∂ ∂φ = (2u + 1, u − v, 3u + v) = 2, 1, 3 ∂u ∂u ∂ ∂φ = (2u + 1, u − v, 3u + v) = 0, −1, 1 Tv = ∂v ∂v
Tu =
The normal vector is the following cross product: i j 1 n(u, v) = Tu × Tv = 2 0 −1
k 3 1
1 = −1
2 3 i − 0 1
2 3 j + 0 1
1 k −1
= 4i − 2j − 2k = 4, −2, −2 (b) That area of S = (D) is the following surface integral: √ n(u, v) du dv = 4, −2, −2 du dv = 24 Area(S) = =
√
D
D
24 Area(D) =
√ √ 24 · 2 · 1 = 4 6
D
1 du dv
(c) We express f (x, y, z) = yz in terms of the parameters u and v: f φ (u, v) = (u − v)(3u + v) = 3u 2 − 2uv − v 2 Using the Theorem on Surface Integrals we have: f (x, y, z) d S = f φ (u, v) n(u, v) du dv = S
D
D
3u 2 − 2uv − v 2 4, −2, −2 du dv
2 √ 1 2 2 √ 1 3 = 24 3u − 2uv − v 2 du dv = 24 u − u 2 v − v 2 u 0
=
√
0
0
√ 1 √ 2 3 1 32 6 2 2 24 8 − 4v − 2v dv = 24 8v − 2v − v = 3 3 0 0
dv
u=0
5. Let (x, z) = (x, y, x y). Let S = where = {(u, v) : u 2 + v 2 ≤ 1, u ≥ 0, v ≥ 0} and is as defined in Exercise 3. , T y , and n(x,Dy). (a) Calculate Tx(D), (a) Calculatethe area of S. surface (b) Let S be the part of the surface with parameter domain D = {(x, y) : x 2 + y 2 ≤ 1, x ≥ 0, y ≥ 0}. Verify the following formula and evaluate polar coordinates: (b) Evaluate (x − y) dusing S. Hint: Use polar coordinates. S 1 dS = 1 + x 2 + y2 d x d y D
S
(c) Verify the following formula and evaluate: S
z dS =
π /2 1 0
0
(sin θ cos θ )r 3 1 + r 2 dr d θ
Parametrized Surfaces and Surface Integrals
S E C T I O N 17.4
(ET Section 16.4)
593
SOLUTION
(a) The tangent vectors are: ∂ ∂φ = (x, y, x y) = 1, 0, y ∂x ∂x ∂ ∂φ = (x, y, x y) = 0, 1, x Ty = ∂y ∂y Tx =
The normal vector is the cross product: i n(x, y) = Tx × T y = 1 0
j 0 1
k y x
0 = 1
1 y i − x 0
1 y j + x 0
0 k 1
= −yi − xj + k = −y, −x, 1 (b) Using the Theorem on evaluating surface integrals we have: 1 dS = n(x, y) d x d y = −y, −x, 1 d x d y = D
S
D
D
y2 + x 2 + 1 d x d y
y 1
D
x 0
1
We convert the integral to polar coordinates x = r cos θ , y = r sin θ . The new region of integration is: 0 ≤ r ≤ 1,
0≤θ≤
π . 2
We get:
π /2 1 π /2 1 2 2 1 dS = r + 1 · r dr d θ = r + 1 · r dr d θ
0
S
0
0
0
√
π /2 2 √ π /2 √ 2 2−1 π u 2 2−1 du d θ = dθ = = 3 6 0 1 2 0 (c) The function z expressed in terms of the parameters x, y is f ((x, y)) = x y. Therefore, z dS = x y · n(x, y) d x d y = x y 1 + x 2 + y2 d x d y D
S
D
We compute the double integral by converting it to polar coordinates. We get: S
z dS =
π /2 1 0
=
0
0
π /2
π /2 1 (r cos θ )(r sin θ ) 1 + r 2 · r dr d θ = (sin θ cos θ )r 3 1 + r 2 dr d θ
(sin θ cos θ ) d θ
0
1
r 3 1 + r 2 dr
0
0
We compute each integral in (1). Using the substitution u = 1 + r 2 , du = 2r dr we get: √ 2 1 2 du 5/2 3/2 2 + 1 2 u u = − r 3 1 + r 2 dr = r 2 1 + r 2 · r dr = = u 3/2 − u 1/2 2 5 3 1 15 0 0 1
1
Also, π /2 0
sin θ cos θ d θ =
π /2 sin 2θ cos 2θ π /2 1 dθ = − = 2 4 2 0 0
(1)
594
C H A P T E R 17
L I N E A N D S U R FA C E I N T E G R A L S
(ET CHAPTER 16)
We substitute the integrals in (1) to obtain the following solution: √ √ 2 + 1 2 2+1 1 = z dS = · 2 15 15 S In Exercises 7–10,Scalculate Tu , Tv , and n(u, v)v)forwhose the parametrized at the Then findthat theequation A surface has a parametrization (u, domain D issurface the square in given Figurepoint. 17. Suppose has the of the tangent plane the surface at that point. following normaltovectors: 7. (u, v) = (2u + v, u − 4v, 3u);
un( =A) 1, = v2, =10 4 , n(B) = 1, 3, 0 SOLUTION The tangent vectors are the following vectors, n(C) = 3, 0, 1 , n(D) = 2, 0, 1 ∂ ∂ = (2u + v, u − 4v, 3u) = 2, 1, 3 Tu =f is a function f (x, y, z) d S, where Estimate ∂u ∂u such that f ((u, v)) = u + v. S
Tv =
∂ ∂ = (2u + v, u − 4v, 3u) = 1, −4, 0 ∂v ∂v
The normal is the cross product: i n(u, v) = Tu × Tv = 2 1
j 1 −4
k 3 0
1 = −4
2 3 i − 1 0
2 3 j + 1 0
1 k −4
= 12i + 3j − 9k = 3 4, 1, −3 The equation of the plane passing through the point P : (1, 4) = (6, −15, 3) with the normal vector 4, 1, −3 is: x − 6, y + 15, z − 3 · 4, 1, −3 = 0 or 4(x − 6) + y + 15 − 3(z − 3) = 0 4x + y − 3z = 0
θ = π2 , φ = π4 9. (θ , φ ) = (cos θ sin φ , sin θ sin φ , cos φ ); (u, v) = (u 2 − v 2 , u + v, u − v); u = 2, v = 3 SOLUTION We compute the tangent vectors: ∂ ∂ = (cos θ sin φ , sin θ sin φ , cos φ ) = − sin θ sin φ , cos θ sin φ , 0 ∂θ ∂θ ∂ ∂ = (cos θ sin φ , sin θ sin φ , cos φ ) = cos θ cos φ , sin θ cos φ , − sin φ Tφ = ∂φ ∂φ Tθ =
The normal vector is the cross product: i j k 0 n(θ , φ ) = Tθ × Tφ = − sin θ sin φ cos θ sin φ cos θ cos φ sin θ cos φ − sin φ = − cos θ sin2 φ i − sin θ sin2 φ j + − sin2 θ sin φ cos φ − cos2 θ cos φ sin φ k = − cos θ sin2 φ i − sin θ sin2 φ j − (sin φ cos φ )k The tangency point and the normal at this point are, √ √ π π π π π π π 2 2 , = cos sin , sin sin , cos = 0, , P= 2 4 2 4 2 4 4 2 2 n
π π 1 1 1 1 , = − j − k = − (j + k) = − 0, 1, 1 2 4 2 2 2 2
√ √ The equation of the plane orthogonal to the vector 0, 1, 1 and passing through P = 0, 22 , 22 is: √ √ 2 2 x, y − ,z − · 0, 1, 1 = 0 2 2
Parametrized Surfaces and Surface Integrals
S E C T I O N 17.4
or
(ET Section 16.4)
595
√ 2 2 +z− =0 y− 2 2 √ y+z = 2 √
11. Use the normal vector computed in Exercise 8 to estimate the area of the small patch of the surface (u, v) = 2, u + θ )v,=u(r−cos θ , r sin θby, 1 − r 2 ); r = 12 , θ = π4 (u 2 − v(r, v) defined 2 ≤ u ≤ 2.1,
3 ≤ v ≤ 3.2
SOLUTION We denote the rectangle D = {(u, v) : 2 ≤ u ≤ 2.1, 3 ≤ v ≤ 3.2}. Using the sample point corresponding to u = 2, v = 3 we obtain the following estimation for the area of S = (D):
Area(S) ≈ n(2, 3)Area(D) = n(2, 3) · 0.1 · 0.2 = 0.02n(2, 3)
(1)
In Exercise 8 we found that n(2, 3) = 2 −1, −1, 5. Therefore, √ n(2, 3) = 2 12 + 12 + 52 = 2 27 Substituting in (1) gives the following estimation: Area(S) ≈ 0.02 · 2 ·
√
27 ≈ 0.2078.
13. A surface S has a parametrization (u, v) with domain 0 ≤ u ≤ 2, 0 ≤ v ≤ 4 such that the following partial Sketch the small patch of the sphere whose spherical coordinates satisfy derivatives are constant: π π π π − 0.15∂ ≤θ≤ + 0.15, ∂ −4, 0.1 ≤ φ ≤ + 0.1 = 2,20, 1 , 0, 3 2 4= 4 ∂u ∂v Use the normal vector computed in Exercise 9 to estimate its area. What is the surface area of S? SOLUTION Since the partial derivatives are constant, the normal vector is also constant. We find it by computing the cross product: i j k ∂ ∂ × = 2 0 1 = −2j = 0, −2, 0 ⇒ n = 2 n = Tu × Tv = ∂u ∂v 4 0 3
We denote the rectangle D = {(u, v) : 0 ≤ u ≤ 2, 0 ≤ v ≤ 4}, and use the surface area to compute the area of S = (D). We obtain: n du dv = 2 du dv = 2 1 du dv = 2 · Area(D) = 2 · 2 · 4 = 16 Area(S) = D
D
D
be ellipsoid the surface with parametrization ShowLet thatSthe cos z+ sin x+ 2sin v) (u, v) = (3 2 v) sin u, v y 2u, (3 + + =1 a b c for 0 ≤ u, v ≤ 2π . Using a computer algebra system: 15.
is parametrized by different viewpoints. Is S best described as a “vase that holds water” or a “bottomless vase”? (a) Plot S from several (b) Calculate the normal vector n(u, v). (φ , θ ) = (a cos θ sin φ , b sin θ sin φ , c cos φ ) (c) Calculate the surface area of S to four decimal places. Express the surface area of the ellipsoid as an integral but do not attempt to evaluate it. SOLUTION (a) We show the graph of S here. 4 −4 2 6
0 −2
4 2 4
2
2 0
−2
−4 −4
−2
0
0 4
−4
−2
0
2
4 6 4 2 0
596
C H A P T E R 17
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(ET CHAPTER 16)
Note that it is best described as a “bottomless vase”. (b) We compute the tangent and normal vectors: ∂ = (3 + sin v)(− sin u), (3 + sin v)(cos u), 0 ∂u ∂ = cos v cos u, cos v sin u, 1 Tv = ∂v
Tu =
The normal vector is the cross product: i n(u, v) = Tu × Tv = (3 + sin v)(− sin u) cos v cos u
j (3 + sin v)(cos u) cos v sin u
k 0 1
= ((3 + sin v) cos u)i + ((3 + sin v) sin u)j − ((3 + sin v) cos v)k Hence, n(u, v) = (3 + sin v) 1 + cos2 v We obtain the following area: Area(S) =
D
n du dv =
2π 2π 0
0
(3 + sin v) 1 + cos2 v du dv ≈ 144.0181
17. Use spherical coordinates to compute the surface area of a sphere of radius R. Let S be the surface z = ln(5 − x 2 − y 2 ) for 0 ≤ x, y ≤ 1. Using a computer algebra system: SOLUTION The sphere of radius R centered at the origin has the following parametrization in spherical coordinates: (a) Calculate the surface area of S to four decimal places. (x θ2 y, 3φ )d S=to (Rfour cos decimal θ sin φ , R sin θ sin φ , R cos φ ), 0 ≤ θ ≤ 2π , 0 ≤ φ ≤ π (b) Calculate places. S
The length of the normal vector is: n = R 2 sin φ Using the integral for surface area gives: Area(S) =
D
n d θ d φ =
2π π 0
0
R 2 sin φ d φ d θ =
2π 0
R2 d θ
π
0
sin φ d φ
π = 2π R 2 · − cos φ = 2π R 2 · 2 = 4π R 2
0
2 over the octant of the unit sphere centered at the origin, where x, y, z ≥ 0. 19. Compute the the integral of xof Compute integral z over the upper hemisphere of a sphere of radius R centered at the origin. SOLUTION The octant of the unit sphere centered at the origin, where x, y, z ≥ 0 has the following parametrization in spherical coordinates:
(θ , φ ) = (cos θ sin φ , sin θ sin φ , cos φ ),
0≤θ≤
π , 2
0≤φ≤
π 2
The length of the normal vector is: n = sin φ The function x 2 expressed in terms of the parameters is cos2 θ sin2 φ . Using the theorem on computing surface integrals we obtain, π /2 π /2 π /2 π /2 x2 d S = cos2 θ sin3 φ d φ d θ cos2 θ sin2 φ (sin φ ) d φ d θ = S
0
= =
0
0
π /2
cos2 θ d θ
π /2 0
0
sin3 φ d φ
=
0
sin 2θ θ + 2 4
π /2 π /2 2 sin2 φ cos φ − cos · − φ 3 3 θ =0 φ =0
π 2 π · = 4 3 6
2+ Show21–32, that thecalculate hemisphere x 2f + , z given ≥ 0 issurface parametrized by In Exercises (x,yy, z)zd2S=forR 2the and function. S (r, θ ) = (r cos θ , r sin θ , R 2 − r 2 )
Parametrized Surfaces and Surface Integrals
S E C T I O N 17.4
21. (u, v) = (u cos v, u sin v, u),
(ET Section 16.4)
597
f (x, y, z) = z(x 2 + y 2 )
0 ≤ u, v ≤ 1;
SOLUTION
Step 1. Compute the tangent and normal vectors. We have: ∂ ∂ = (u cos v, u sin v, u) = cos v, sin v, 1 ∂u ∂u ∂ ∂ = (u cos v, u sin v, u) = −u sin v, u cos v, 0 Tv = ∂v ∂v
Tu =
The normal vector is the cross product: i n = Tu × Tv = cos v −u sin v
j sin v u cos v
k 1 0
= (−u cos v)i − (u sin v)j + u cos2 v + u sin2 v k = (−u cos v)i − (u sin v)j + uk = −u cos v, −u sin v, u
We compute the length of n: √ √ 2 2 2 n = (−u cos v) + (−u sin v) + u = u 2 cos2 v + sin2 v + 1 = u 2 · 2 = 2|u| = 2u Notice that in the region of integration u ≥ 0, therefore |u| = u.
Step 2. Calculate the surface integral. We express the function f (x, y, z) = z x 2 + y 2 in terms of the parameters u, v: f (, (u, v)) = u u 2 cos2 v + u 2 sin2 v = u · u 2 = u 3 We obtain the following integral: 1 1 1 1 √ f (x, y, z) d S = f (, (u, v)) n du dv = u 3 · 2u du dv S
0
=
0
1√
1
2 dv
0
u 4 du
0
0
0
1 √ √ u 5 2 = 2· = 5 5 0
23. x 2 + y 2 = 4, 0 ≤ z ≤ 4; f (x, y, z) = e−z (r, θ ) = (r cos θ , r sin θ , θ ), 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π ; f (x, y, z) = x 2 + y 2 SOLUTION The cylinder has the following parametrization in cylindrical coordinates: (θ , z) = (2 cos θ , 2 sin θ , z), 0 ≤ θ ≤ 2π , 0 ≤ z ≤ 4 Step 1. Compute the tangent and normal vectors. The tangent vectors are the partial derivatives: ∂ ∂ = (2 cos θ , 2 sin θ , z) = −2 sin θ , 2 cos θ , 0 ∂θ ∂θ ∂ (2 cos θ , 2 sin θ , z) = 0, 0, 1 Tz = ∂z
Tθ =
The normal vector is their cross product: i n(θ , z) = Tθ × Tz = −2 sin θ 0
j 2 cos θ 0
k 0 1
= (2 cos θ )i + (2 sin θ )j = 2 cos θ , 2 sin θ , 0
The length of the normal vector is thus √ n(θ , z) = (2 cos θ )2 + (2 sin θ )2 + 0 = 4 cos2 θ + sin2 θ = 4 = 2 Step 2. Calculate the surface integral. The surface integral equals the following double integral: 2π 4 f (x, y, z) d S = f ((θ , z)) n d θ dz = e−z · 2 d θ dz D
S
=
2π 0
2 dθ
4 0
e−z dz
0
0
−z 4 = 4π 1 − e−4 = 4π · −e 0
598
C H A P T E R 17
L I N E A N D S U R FA C E I N T E G R A L S
(ET CHAPTER 16)
25. z = 4 − x 2 −2y 2 , 2z ≥ 0; f (x, y, z) = z(x 2 + y 2 ) z = 4 − x − y , z ≥ 0; f (x, y, z) = z SOLUTION We use the formula for the surface integral over a graph: f (x, y, z) d S = f (x, y, g(x, y)) 1 + gx2 + g 2y d x d y
(1)
D
S
Since z = g(x, y) = 4 − x 2 − y 2 , we have: f (x, y, g(x, y)) = 4 − x 2 − y 2 x 2 + y 2 = 4 − x 2 + y 2 x 2 + y2 1 + gx2 + g 2y = 1 + (−2x)2 + (−2y)2 = 1 + 4 x 2 + y 2 The domain of integration D is determined by the inequality: D : z = 4 − x 2 − y2 ≥ 0 ⇒ x 2 + y2 ≤ 4 y 2
D x
−2
0
2
−2
By (1) we obtain:
S
f (x, y, z) d S =
D
4 − (x 2 + y 2 ) · (x 2 + y 2 ) 1 + 4(x 2 + y 2 ) d x d y
We convert the integral to polar coordinates x = r cos θ , y = r sin θ to obtain: S
f (x, y, z) d S =
2π 2 4 − r 2 r 2 1 + 4r 2 · r dr d θ 0
0
= 2π
2
4r 3
0
2 1 + 4r 2 dr − r 5 1 + 4r 2 dr
4r 3
0
2 1 + 4r 2 dr = r 2 1 + 4r 2 · 4r dr = 0
= 2 0
√17 4 u − u2 4
1
du =
1 4
⎛
√
2u 5 u 3 1 ⎝ u7 − + = 64 7 5 3
√
1 + 4r 2 , du = 4r u dr . This gives:
17 u 2 − 1 u 2 du
4
√ √ u5 u 3 17 391 17 + 1 − = 5 3 1 30
2 r 5 1 + 4r 2 dr = r 4 1 + 4r 2 · r dr = 0
(2)
0
We compute the integrals using the (somewhat unusual) substitution u = 2
1
√ 1
17
1
⎞
17
2 u2 − 1 16
√17 u2 1 · du = u 6 − 2u 4 + u 2 du 4 64 1
√ ⎠ = 7769 17 − 1 840
Substituting the integrals in (2) gives:
S
f (x, y, z) d S = 2π
√
√ √ 3179 17 + 29 π 391 17 + 1 7769 17 − 1 − ≈ 98.26 = 30 840 420
27. y = 9 − z 2 , 20 ≤ x, z ≤ 3; f (x, y, z) = z f (x, y, z) = 1 y = 9 − z , 0 ≤ x ≤ z ≤ 3; SOLUTION We use the formula for the surface integral over a graph y = g(x, z): f (x, y, z) d S = f (x, g(x, z), z) 1 + gx2 + gz2 d x dz S
D
(1)
S E C T I O N 17.4
Parametrized Surfaces and Surface Integrals
(ET Section 16.4)
599
Since y = g(x, z) = 9 − z 2 , we have gx = 0, gz = −2z, hence: 1 + gx2 + gz2 = 1 + 4z 2 f (x, g(x, z), z) = z The domain of integration is the square [0, 3] × [0, 3] in the x z-plane. By (1) we get:
3 3 3 3 3 f (x, y, z) d S = z 1 + 4z 2 dz d x = 1 dx z 1 + 4z 2 dz = 3 z 1 + 4z 2 dz S
0
0
0
0
0
We use the substitution u = 1 + 4z 2 , du = 8z dz to compute the integral. This gives: S
f (x, y, z) d S = 3
√ 3 37 1/2 37 37 − 1 u du = ≈ 56 z 1 + 4z 2 dz = 3 8 4 0 1
29. Part of the plane x + y + z = 1, where x, y, z ≥ 0; f (x, y, z) = z (u, v) = (u, v 3 , u + v), 0 ≤ u, v ≤ 1; f (x, y, z) = y SOLUTION We let z = g(x, y) = 1 − x − y and use the formula for the surface integral over the graph of z = g(x, y), where D is the parameter domain in the x y-plane. That is: f (x, y, z) d S = f (x, y, g(x, y)) 1 + gx2 + g 2y d x d y (1) D
S
We have, gx = −1 and g y = −1 therefore: √ 1 + gx2 + g 2y = 1 + (−1)2 + (−1)2 = 3 We express the function f (x, y, z) = z in terms of the parameters x and y: f (x, y, g(x, y)) = z = 1 − x − y The domain of integration is the triangle D in the x y-plane shown in the figure. z y 1
x+y=1 D
y x S: x + y + z = 1, x ≥ 0, y ≥ 0, z ≥ 0
x 0
1
By (1) we get: 1−y √ √ 1 x2 − yx (1 − x − y) 3 d x d y = 3 x− dy 2 0 0 0 x=0
√ √ 1 3 1 (1 − y)2 2 1 − 2y + y 2 d y (1 − y) − = 3 dy = 2 2 0 0
√ √ 3 3 y 3 1 y − y2 + = = 2 3 0 6
S
f (x, y, z) d S =
1 1−y
31. Part of the surface x = z 3 , where 0 ≤ x, y ≤ 1; f (x, y, z) = x 2 f (x, y, z) = x z Region in the plane x + y + z = 0 contained in the cylinder x + y 2 = 1; SOLUTION We let z = g(x, y) = x 1/3 and use the formula for the surface integral over a graph: f (x, y, z) d S = f (x, y, g(x, y)) 1 + gx2 + g 2y d x d y S
D
where D is the square [0, 1] × [0, 1] in the x y-plane. We compute the integrand in (1): 1 −2/3 1 2 2 gx = x , gy = 0 ⇒ 1 + gx + g y = 1 + x −4/3 3 9
(1)
600
C H A P T E R 17
L I N E A N D S U R FA C E I N T E G R A L S
(ET CHAPTER 16)
f (x, y, g(x, y)) = x Substituting in (1) we get: 1 1 1 1 −4/3 1 f (x, y, z) d S = x 1+ x dx dy = x 1 + x −4/3 d x 9 9 S 0 0 0
We compute the integral using the substitution 13 x −2/3 = tan θ . Then: 1+
1 −4/3 1 x = 1 + tan2 θ = 9 cos θ
2 1 − x −5/3 d x = dθ 9 cos2 θ Hence:
1+
⇒
2 1 9 8/3 d θ = − 1 cos θ d θ x dx = − · x 2 cos2 θ 18 sin4 θ
1 −4/3 1 cos θ x · x dx = − dθ 9 18 sin4 θ
We obtain the following integral, which we compute by substituting u = sin θ , du = cos θ d θ : 10 1
()
tan−1
1 3
3
S
f (x, y, z) d S =
tan−1 1 3 π /2
sin tan−1 1 = √1 3 1 cos θ 1 10 1 − d θ = − du 4 18 sin θ 18 1 u4
√ 1 1/( 10) 1 1 √ · 3 10 10 − 1 ≈ 0.567 = = 54 u 1 54
33. Let S be the sphere of radius R centered at the origin. Explain the following equalities using symmetry: sphere centered Part of theunit at the origin, where x ≥ 0 and |y| ≤ x; f (x, y, z) = x x dS = y dS = z dS = 0 (a) S S S (b) x2 d S = y2 d S = z2 d S S S S 4 x 2 d S = π R4 . Then show, by adding the three integrals in part (b), that 3 S SOLUTION
(a) Since the sphere is symmetric with respect to the yz-plane, the surface integrals of x over the hemispheres on the two sides of the plane cancel each other and the result is zero. The two other integrals are zero due to the symmetry of the sphere with respect to the x z and x y-planes. (b) the sphere is symmetric with respect to the x y, x z and yz-planes, Since 2 interchanging x and y in2 the integral for 2 S x d S does not change the value of the integral and the result is S y d S. The equality for S z d S is explained similarly. On the sphere, we have x 2 + y 2 + z 2 = R 2 so, using properties of integrals, the integral for surface area, and the surface area of the sphere of radius R we obtain: x2 d S + y2 d S + z2 d S = 1 dS x 2 + y 2 + z2 d S = R2 S
S
S
S
S
= R 2 · Area(S) = R 2 · 4π R 2 = 4π R 4 Combining with (b) we conclude that the value of each of the integrals is 43 π R 4 . That is: 4 x2 d S = y2 d S = z2 d S = π R4 . 3 S S S 35. Find the area ofthe portion of the plane 2x + 3y + 4z = 28 lying above the rectangle 1 ≤ x ≤ 3, 2 ≤ y ≤ 5 in the x y-plane. Calculate (x y + e z ) d S, where S is the triangle in Figure 18 with vertices (0, 0, 3), (1, 0, 2), and (0, 4, 1). S
Hint: Find the equation of the plane containing the triangle.
Parametrized Surfaces and Surface Integrals
S E C T I O N 17.4 SOLUTION
(ET Section 16.4)
601
We rewrite the equation of the plane as: z = g(x, y) = −
3 x − y+7 2 4
(1)
The domain of the parameters is the rectangle D = [1, 3] × [2, 5] in the x y-plane. Using the integral for surface area and the surface integral over a graph we have: Area(S) = 1 dS = 1 + gx2 + gz2 d x d y (2) D
S
y 5 D 2
x 1
0
3
By (1) we have: 1 3 gx = − , g y = − ⇒ 2 4
√ 29 9 1 = 1 + gx2 + gz2 = 1 + + 4 16 4
We substitute in (2) to obtain: √
Area(S) =
D
29 dx dy = 4
√
29 4
√
D
29 Area(D) = 4
1 dx dy =
√ √ 3 29 29 ·3·2= 4 2
2 2 + y 2 + z 2 )−1 over the cap of the sphere x 2 + y 2 + z 2 = 4 defined by 37. Compute thethe integral f (x, y, z) = What is area ofofthe portion of zthe(xplane 2x + 3y + 4z = 28 lying above the domain D in the x y-plane in z ≥ Figure 1. 19 if Area(D) = 5? SOLUTION
We use spherical coordinates to parametrize the cap S. (θ , φ ) = (2 cos θ sin φ , 2 sin θ sin φ , 2 cos φ ) D : 0 ≤ θ ≤ 2π , 0 ≤ φ ≤ φ0
The angle φ0 is determined by cos φ0 = 12 , that is, φ0 = π3 . The length of the normal vector in spherical coordinates is: n = R 2 sin φ = 4 sin φ −1 We express the function f (x, y, z) = z 2 x 2 + y 2 + z 2 in terms of the parameters: f (θ , φ ) = (2 cos φ )2 4−1 = cos2 φ Using the theorem on computing the surface integral we get:
S
f (x, y, z) d S = =
D 2π
0
= 8π
f (θ , φ ) n d φ d θ =
4 dθ 1 1 − 8 6
π /3 0
2π π /3 cos2 φ · 4 sin φ d φ d θ 0
cos2 φ sin φ d φ
1 − − 3
= 8π ·
0
cos φ π /3 = 8π − 3 0
7π 7 = 24 3
39. Let S be the portion of the sphere x 2 + y 2 + z 2 = 9, where 1 ≤ x 2 + y 2 ≤ 4 and z ≥ 0 (Figure 20). Find a f (x, y, z) = z(x 2 and + y 2use ) d itS,towhere S is the hemisphere x 2 + y 2 + z 2 = R 2 , z ≥ 0. Calculateof S in polar parametrization coordinates compute: S z −1 d S (a) The area of S (b) S
602
C H A P T E R 17
L I N E A N D S U R FA C E I N T E G R A L S
(ET CHAPTER 16) z
1 2
y
x
FIGURE 20 SOLUTION y
D 1
2
3
x
We parametrize S by spherical coordinates as follows: (θ , φ ) = (3 cos θ sin φ , 3 sin θ sin φ , 3 cos φ ) D : 0 ≤ θ ≤ 2π , φ0 ≤ φ ≤ φ1 The angles φ0 and φ1 are determined by, 1 3 2 sin φ1 = 3
sin φ0 =
1 3 2 φ1 = sin−1 3
φ0 = sin−1
⇒ ⇒ z
f0
f1
x y
The length of the normal is: n = R 2 sin φ = 9 sin φ (a) Using the integral for the surface area we have, Area(S) =
D
n d φ d θ =
2π sin−1 (2/3)
0
sin−1 (1/3)
φ1 =sin−1 (2/3) = 18π − cos φ −1 φ0 =sin
2
(1/3)
9 sin φ d φ d θ =
2π 0
√
9 dφ
sin−1 (2/3) sin−1 (1/3)
sin φ d φ
√ √ √ 5 8 + 8 − 5 ≈ 11.166 = 18π − = 6π 3 3
3 f1 5
1
3 f0 8
(b) We express the function f (x, y, z) = z −1 in the terms of the parameters: sec φ f (θ , φ ) = (3 cos φ )−1 = 3
Parametrized Surfaces and Surface Integrals
S E C T I O N 17.4
(ET Section 16.4)
603
Using the surface integral as a double integral we obtain: 2π φ 1 2π φ 1 sec φ · 9 sin φ d φ d θ = z −1 d S = f (θ , φ ) n d φ d θ = 3 tan φ d φ d θ 3 S D 0 φ0 0 φ0
φ1 =sin−1 (2/3) 2π φ1 = 3 dθ tan φ d φ = 6π ln(sec φ ) −1 φ0
0
3 3 = 6π ln √ − ln √ 5 8
= 6π ln
φ0 =sin
(1/3)
8 = 3π ln 1.6 ≈ 4.43 5
2 = x 2 + y 2 , where z ≥ 0, contained within the cylinder 41. Find the surface area of the portion S of the cone 2 + zy 2 = z 2 between the planes z = 2 and z = 5. Find the surface area of the part of the cone x 2 2 y + z ≤ 1. SOLUTION We rewrite the equation of the cone as x = ± z 2 − y 2 . The projection of the cone onto the yz-plane is obtained by setting x = 0 in the equation of the cone, that is, x = 0 = z 2 − y 2 ⇒ z = ±y
Since on S, z ≥ 0, we get z = |y|. We conclude that the projection of the upper part of the cone x 2 + y 2 = z 2 onto the yz-plane is the region between the lines z = y and z = −y on the upper part of the yz-plane. Therefore, the projection D of S onto the yz-plane is the region shown in the figure: z y2 + z2 = 1
z=
1 − y2
D z = −y
z=y y
− 1
1
2
2
The area of S is the surface integral: Area(S) =
dS S
We compute the integral using a surface integral over a graph. Since x = g(y, z) = ± z 2 − y 2 we have, gz = ± Hence, (notice that z ≥ 0 on S): 1 + g 2y + gz2 =
z z2 − y2
y , gy = ± 2 z − y2
z2 y2 1+ 2 + = z − y2 z2 − y2
√ 2z 2 z 2 = z2 − y2 z2 − y2
We obtain the following integral: Area(S) =
D
1 + g 2y + gz2 d y dz =
D
√ z 2 z2 − y2
dz d y
Using symmetry gives: √ √
√ 1/( 2) 1−y 2 z dz dz d y = 2 2 dy Area(S) = 2 0 0 y y z2 − y2 z2 − y2 We compute the inner integral using the substitution u = z 2 − y 2 , du = uz dz. We get: √1−2y 2 √1−y 2 √1−y 2 z dz u du = = du = 1 − 2y 2 u y 0 0 z2 − y2 √ We substitute in (1) and compute the resulting integral using the substitution t = 2y. We get:
1/(√2) √1−y 2
√ z 2
1 1/(√2) √ 1 dt π π 1 − 2y 2 d y = 2 2 1 − t2 √ = 2 1 − t 2 dt = 2 · = Area(S) = 4 2 2 0 0 0
(1)
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43. Prove a famous result of Archimedes: The surface area of the portion of the sphere of radius r between two horizontal e2x+yztoover the following of the cube of portion side 2 inofFigure 21. planes zFind = athe andintegral z = b of is equal the surface area offaces the corresponding the circumscribed cylinder (Figure (b) The face PQRS 22). (a) The top face
z
b a
FIGURE 22 SOLUTION We compute the area of the portion of the sphere between the planes a and b. The portion S1 of the sphere has the parametrization,
(θ , φ ) = (r cos θ sin φ , r sin θ sin φ , r cos φ ) where, D1 : 0 ≤ θ ≤ 2π , φ0 ≤ φ ≤ φ1 If we assume 0 < a < b, then the angles φ0 and φ1 are determined by, b r a cos φ1 = r
cos φ0 =
b
b r a −1 φ1 = cos r
φ0 = cos−1
⇒ ⇒
a
r
r f1
f0
The length of the normal vector is n = r 2 sin φ . We obtain the following integral: Area (S1 ) =
D1
nd φ d θ =
cos−1 = 2π r 2 − cos φ
2π
φ0
0
a r
φ =cos−1 br
φ1
r 2 sin φ d φ d θ =
2π
0
r 2d φ
φ2 φ1
sin φ d φ
b a = 2π r 2 − + = 2π r (b − a) r r
The area of the part S2 of the cylinder of radius r between the planes z = a and z = b is: Area (S2 ) = 2π r · (b − a) We see that the two areas are equal: Area (S1 ) = Area (S2 )
Further Insights and Challenges 45. Use Eq. (13) to compute the surface area of z = 4 − y 2 for 0 ≤ y ≤ 2 rotated about the z-axis. Surfaces of Revolution Let S be the surface formed by revolving the region underneath the graph z = g(y) in SOLUTION Since y 2 , wethe have g (y) = −2y. Eq. (13) wec obtain the yz-plane forg(y) c≤= y ≤4 − d about z-axis (Figure 23).By Assume that ≥ 0. the following integral, (a) Show that the circle generated byrotating z-axis a point (0, a, b) aboutthe is parametrized by 2 2 2 |y| 1 + (−2y) d y = 2π y · 1 + 4y 2 d y Area(S) = 2π 0 (a cos θ , a sin θ , b), 0 ≤ θ 0≤ 2π (b) Show that S is parametrized by (y, θ ) = (y cos θ , y sin θ , g(y))
Parametrized Surfaces and Surface Integrals
S E C T I O N 17.4
(ET Section 16.4)
605
We compute the integral using the substitution u = 1 + 4y 2 , du = 8y d y. We get: Area(S) = 2π
17 1
u 1/2 ·
2 u 3/2 17 π √ du = 2π · 17 17 − 1 ≈ 36.18 = 8 3 8 1 6
47. Area of a Torus Let T be the torus obtained by rotating the circle in the yz-plane of radius a centered at (0, b, 0) 2 + y 2 = z 2 for 0 ≤ z ≤ d as a surface of revolution (Figure 2) and use Describe the upper the cone about the z-axis (Figure 24).half Weof assume thatxb > a > 0. Eq. (13) to compute surface area. (a) Use Eq. (13) to showits that Area(T) = 4π
b+a b−a
ay dy a 2 − (b − y)2
z
z b−a
b+a b
y
y
x
x
FIGURE 24 The torus obtained by rotating a circle of radius a.
(b) Show that Area(T) = 4π 2 ab. Hint: Rewrite the integral using substitution. SOLUTION
(a) Using symmetry, the area of the surface obtained by rotating the upper part of the circle is half the area of the torus. z b−a
b+a b
y
x
The rotated graph is z = g(y) = a 2 − (y − b)2 , b − a ≤ y ≤ b + a. So, we have, −2(y − b) y−b = − g (y) = 2 a 2 − (y − b)2 a 2 − (y − b)2 a 2 − (y − b)2 + (y − b)2 (y − b)2 a 1 + g (y)2 = 1 + = = 2 2 2 2 a − (y − b) a − (y − b) a 2 − (y − b)2 We now use symmetry and Eq. (13) to obtain the following area of the torus (we assume that b − a > 0, hence y > 0): b+a b+a ay Area (T) = 2 · 2π |y| 1 + g (y)2 d y = 4π dy (1) b−a b−a a 2 − (y − b)2 1 (b) We compute the integral using the substitution u = y−b a , du = a d y. We get: b+a 1 1 2 1 1 ay a 2 u + ab a u + ab a2 u ab dy = a du = du = du + du 2 2 2 2 2 b−a −1 −1 −1 −1 2 2 a −a u 1−u 1−u 1 − u2 a − (y − b)
The first integral is zero since the integrand is an odd function. We get: b+a b−a
1 π −1 −0 = dy = 2 du = 2ab sin u = 2ab 2 0 1 − u2 0 a 2 − (y − b)2 ay
1
ab
π ab
Substituting in (1) gives the following area: Area (T) = 4π · π ab = 4π 2 ab 49. Compute the surface area of the torus in Exercise 47 using Pappus’s Theorem. Pappus’s Theorem Also called Guldin’s Rule, Pappus’s Theorem states that the area of a surface of revolution S is equal to the length L of the generating curve times the distance traversed by the center of mass. Use Eq. (13) to prove Pappus’s Theorem. If C is the graph z = g(y) for c ≤ y ≤ d, then the center of mass is defined as the point
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SOLUTION The generating curve is the circle of radius a in the (y, z)-plane centered at the point (0, b, 0). The length of the generating curve is L = π a. z
2πb L = 2πb b−a
b b+a
y
x
The center of mass of the circle is at the center (y, z) = (b, 0), and it traverses a circle of radius b centered at the origin. Therefore, the center of mass makes a distance of 2π b. Using Pappus’ Theorem, the area of the torus is: L · 2π a = 2π a · 2π b = 4π 2 ab. 51. Calculate the gravitational potential ϕ for a hemisphere of radius R with uniform mass distribution. Potential due to a Uniform Sphere Let S be a hollow sphere of radius R with center at the origin with a SOLUTION In Exercise 50(b) we expressed the potential ϕ for a sphere of2radius R. To find the potential for 2a hemiuniform mass distribution of total mass m [since S has surface area 4π R , the mass density is ρ = m/(4π R )]. The sphere of radius R, we needϕ only to modify limitsP of φ equal to 0 ≤toφ ≤ π2 . This gives the following integral: gravitational potential (P) due to S atthe a point =the (a,angle b, c) is π /2 sin φ d θ d φ ρ d S sin φ d φ Gm π /2 2π Gm ϕ (0, 0, r ) = ϕ (r ) = − =− · 2π −G 4π 0 4 π 2 2 2 0 SR 2 +(xr 2−−a)2Rr + cos (y − φ b) + (z − c) 0 R 2 + r 2 − 2Rr cos φ /2 the potential depends only on the distance r from P to the center of the sphere. (a) Use symmetry to conclude sin φ d φ Gm πthat for a point P = (0, 0, r ) on the z-axis (with r = R). =to −compute ϕ (P) Therefore, it suffices 4π 0 2 + r 2 − 2Rr cos φ (b) Use spherical coordinates to showRthat ϕ (0, 0, r ) is equal to π 22π 2 We compute the integral using the substitution −Gm u = sin φcos d θ φd,φ du = 2Rr sin φ d φ . We obtain: R + r − 2Rr 4π 0 0 2R 2 2+ r 2 − 2Rr cos φ R 2 +r 2 2 2 du Gm R +r 2Rr Gm Gm R +r −1/2 · 2u 1/2 ϕ (r ) = − u du = − √ =− 2 2 2 4Rr 4Rr u 2 2 ) r − 2Rr cos φ to(R−r u=(R−r )2 (c) Use the substitution u =(R−r R + show) that 1/2 1/2 Gm Gm −mG =− − r )2 |R + r=| − − |R − r | R 2 + r 2 − |R − r | R 2 + r 2ϕ (0, 0,−r ) (R = 2Rr 2Rr 2Rr (d) Verify formula (11) for ϕ . 53. Let S be the part of the graph z = g(x, y) lying over a domain D in the x y-plane. Let φ = φ (x, y) be the angle The surface of a cylinder of radius R and length L has a uniform mass distribution ρ (the top and bottom of between the normal to S and the vertical. Prove the formula the cylinder are excluded). Use Eq. (10) to find the gravitational potential at a point P located along the axis of the dA cylinder. Area(S) = D | cos φ | SOLUTION z
n f = f(x, y)
S (x, y, g(x, y))
y
D x
Using the Surface Integral over a Graph we have: Area(S) =
S
1 dS =
D
1 + gx2 + g 2y d A
In parametrizing the surface by φ (x, y) = (x, y, g(x, y)), (x, y) = D, we have: ∂ = 1, 0, gx ∂x ∂ = 0, 1, g y Ty = ∂y
Tx =
(1)
Surface Integrals of Vector Fields
S E C T I O N 17.5
(ET Section 16.5)
607
Hence, i j k n = Tx × T y = 1 0 gx = −gx i − g y j + k = −gx , −g y , 1 0 1 gy n = gx2 + g 2y + 1 k
n
−k
There are two adjacent angles between the normal n and the vertical, and the cosines of these angles are opposite numbers. Therefore we take the absolute value of cos φ to obtain a positive value for Area(S). Using the Formula for the cosine of the angle between two vectors we get: | −gx , −g y , 1 · 0, 0, 1 | 1 |n · k| = = | cos φ | = nk 2 2 1+g +g ·1 1 + g2 + g2 x
y
x
y
Substituting in (1) we get: Area(S) =
dA D | cos φ |
17.5 Surface Integrals of Vector Fields (ET Section 16.5) Preliminary Questions 1. Let F be a vector field and (u, v) a parametrization of a surface S, and set n = Tu × Tv . Which of the following is the normal component of F? (a) F · n (b) F · en The normal component of F is F · en rather than F · n. F · dS is equal to the scalar surface integral of the function (choose the correct 2. The vector surface integral
SOLUTION
S
answer): (a) F (b) F · n, where n is a normal vector (c) F · en , where en is the unit normal vector SOLUTION The vector surface integral dS is defined as the scalar surface integral of the normal component of S F · F on the oriented surface. That is, S F · dS = S (F · en ) d S as stated in (c). F · dS is zero if (choose the correct answer): 3. S
(a) F is tangent to S at every point. (b) F is perpendicular to S at every point. SOLUTION Since S F · dS is equal to the scalar surface integral of the normal component of F on S, this integral is zero when the normal component is zero at every point, that is, when F is tangent to S at every point as stated in (a). 4. If F(P) = en (P) at each point on S, then F · dS is equal to (choose the correct answer): S
(a) Zero SOLUTION
(b) Area(S) If F(P) = en (P) at each point on S, then,: F · dS = (en · en ) d S = S
Therefore, (b) is the correct answer.
S
(c) Neither S
en 2 d S =
S
1 d S = Area(S)
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5. Let S be the disk x 2 + y 2 ≤ 1 in the x y-plane oriented with normal in the positive z-direction. Determine for each of the following vector constant fields: (a) F = 1, 0, 0 (b) F = 0, 0, 1 (c) F = 1, 1, 1
S
F · dS
The unit normal vector to the oriented disk is en = 0, 0, 1. (a) Since F · en = 1, 0, 0 · 0, 0, 1 = 0, F is perpendicular to the unit normal vector at every point on S, therefore S F · dS = 0. (b) Since F = en at every point on S, we have: F · dS = en 2 d S = 1 d S = Area(S) = π (en · en ) d S = SOLUTION
S
S
(c) For F = 1, 1, 1 we have: F · dS = S
S
S
S
(F · en ) d S =
S
1, 1, 1 · 0, 0, 1 d S =
S
1 d S = Area(S) = π
6. Estimate
S
F · dS, where S is a tiny oriented surface of area 0.05 and the value of F at a sample point in S is a
vector of length 2 making an angle π4 with the normal to the surface. SOLUTION en
F
π 4
S
P
Since S is a tiny surface, we may assume that the dot product F · en on S is equal to the dot product at the sample point. This gives the following approximation: F · dS = 1dS = F(P) · en Area(S) (F · en ) dS ≈ (F(P) · en (P)) dS = F(P) · en (P) S
S
S
S
That is, F · dS ≈ F(P) · en (P)Area(S)
S
(1)
We are given that Area(S) = 0.05. We compute the dot product: F(P) · en (P) = F(P)en (P) cos
√ 1 π =2·1· √ = 2 4 2
Combining with (1) gives the following estimation: √ F · dS ≈ 0.05 2 ≈ 0.0707. S
7. A small surface S is divided into three pieces of area 0.2. Estimate
S
F · dS if F is a unit vector field making
angles of 85, 90, and 95◦ with the normal at sample points in these three pieces. SOLUTION
en
en
F
F
en
F P1 S1
P2
S2
P3 S3
S E C T I O N 17.5
Surface Integrals of Vector Fields
(ET Section 16.5)
609
We estimate the vector surface integral by the following sum: F · dS = F (P1 ) · en (P1 ) Area (S1 ) + F (P2 ) · en (P2 ) Area (S2 ) + F (P3 ) · en (P3 ) Area (S3 ) S
= 0.2 (F (P1 ) · en (P1 ) + F (P2 ) · en (P2 ) + F (P3 ) · en (P3 )) We compute the dot product. Since F and en are unit vectors, we have: F (P1 ) · en (P1 ) = cos 85◦ ≈ 0.0872 F (P2 ) · en (P2 ) = cos 90◦ = 0 F (P3 ) · en (P3 ) = cos 95◦ ≈ −0.0872 Substituting gives the following estimation: F · dS ≈ 0.2(0.0872 + 0 − 0.0872) = 0. S
Exercises 1. Let F = y, z, x and let S be the oriented surface parametrized by (u, v) = (u 2 − v, u + v, v 2 ) for 0 ≤ u ≤ 2, −1 ≤ v ≤ 1. Calculate: (a) n and F · n as functions of u and v (b) The normal component of F to the surface at P = (3, 3, 1) = (2, 1) (c) F · dS S
SOLUTION
(a) The tangent vectors are, ∂ 2 ∂ = u − v, u + v, v 2 = 2u, 1, 0 ∂u ∂u ∂ 2 ∂ = u − v, u + v, v 2 = −1, 1, 2v Tv = ∂v ∂v
Tu =
The normal vector is their cross product: i j n = Tu × Tv = 2u 1 −1 1
k 0 2v
= (2v)i − (4uv)j + (2u + 1)k = 2v, −4uv, 2u + 1
We write F = y, z, x in terms of the parameters x = u 2 − v, y = u + v, z = v 2 and then compute F · n: F ((u, v)) = y, z, x = u + v, v 2 , u 2 − v F ((u, v)) · n(u, v) = u + v, v 2 , u 2 − v · 2v, −4uv, 2u + 1 = 2v(u + v) − 4uv · v 2 + (2u + 1) u 2 − v = 2u 3 − 4uv 3 + 2v 2 + u 2 − v (b) At the point P = (3, 3, 1) = (2, 1) we have: F(P) = 3, 1, 3 n(P) = 2 · 1, −4 · 2 · 1, 2 · 2 + 1 = 2, −8, 5 en (P) =
2, −8, 5 n(P) 1 = √ = √ 2, −8, 5 n(P) 4 + 64 + 25 93
Hence, the normal component of F to the surface at P is the dot product: 1 1 13 F(P) · en (P) = 3, 1, 3 · √ 2, −8, 5 = √ (6 − 8 + 15) = √ 93 93 93
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(ET CHAPTER 16)
(c) Using the definition of the vector surface integral and the dot product in part (a), we have: 2 1 F · dS = F φ (u, v) · n(u, v) du dv = 2u 3 − 4uv 3 + 2v 2 + u 2 − v dv du S
=
2 0
D
0
2 1 1 2u 3 v − uv 4 + v 3 + u 2 v − v 2 3 2
2
−1
v=−1
du
2 1 2 1 + u2 − − −2u 3 − u − − u 2 − du 3 2 3 2 0 2 4 2 4u 2 = 24 4u 3 + 2u 2 + du = u 4 + u 3 + = 3 3 3 0 0 =
2u 3 − u +
3. Let S be the square in the x y-plane shown in Figure 13, oriented with the normal pointing in the positive z-direction. Compute the surface integral of the vector field F = x, y, x + y over the portion S of the paraboloid z = Estimate x 2 + y 2 lying over the disk x 2 + y 2 ≤ 1. S
F · dS
where F is a vector field whose values at the labeled points are F( A) = 2, 6, 4,
F(B) = 1, 1, 7
F(C) = 3, 3, −3,
F(D) = 0, 1, 8
The unit normal vector to S is en = 0, 0, 1. We estimate the vector surface integral division and sample points given in Figure 12. SOLUTION
S F · dS using the
y 1 A
B
C
D 1
x
Each subsquare has area 14 , therefore we obtain the following estimation: 1 F · dS ≈ (F( A) · en + F(B) · en + F(C) · en + F(D) · en ) · 4 S = (2, 6, 4 · 0, 0, 1 + 1, 1, 7 · 0, 0, 1 + 3, 3, −3 · 0, 0, 1 + 0, 1, 8 · 0, 0, 1) · = (4 + 7 − 3 + 8) ·
1 4
1 =4 4
In Exercises 5–17, compute the surface integral over the given oriented surface. Suppose that S is a surface in R3 with a parametrization whose domain D is the square in Figure 13. The values function , a vector 1, and the normal vector n = Tu × Tv at (P) are given for the four sample 5. F = y,ofz,ax, planef 3x − 4y +field z =F, following table. Estimate the surface integrals of f and F over S. 0points ≤ x, yin≤D1,in the upward-pointing normal SOLUTION
We rewrite the equation of the plane as z = 1 − 3x + 4y, and parametrize the plane by: Point P(x, in D y) = f (x, y, 1F− 3x + 4y) n
Here, the parameter domain is the square D =A{(x, y)3: 0 ≤2, x,6,y4 ≤ 1} in1, the1,x1 y-plane. 1, 1, 7 1, 1, 0 B 1 Step 1. Compute the tangent and normal vectors. C 2 3, 3, −3 1, 0, −1 ∂5 ∂ 0, 1, 8 2, 1, 0 D = (x, y, 1 − 3x + 4y) = 1, 0, −3 Tx = ∂x ∂x ∂ ∂ = (x, y, 1 − 3x + 4y) = 0, 1, 4 Ty = ∂y ∂y i j k Tx × T y = 1 0 −3 = 3i − 4j + k = 3, −4, 1 0 1 4 Since the plane is oriented with upward pointing normal, the normal vector n is: n = 3, −4, 1
Surface Integrals of Vector Fields
S E C T I O N 17.5
(ET Section 16.5)
611
Step 2. Evaluate the dot product F · n. We write F in terms of the parameters: F ((x, y)) = y, z, x = y, 1 − 3x + 4y, x The dot product F · n is thus F ((x, y)) · n = y, 1 − 3x + 4y, x · 3, −4, 1 = 3y − 4(1 − 3x + 4y) + x = 13x − 13y − 4 Step 3. Evaluate the surface integral. The surface integral is equal to the following double integral: 1 1 F · dS = F ((x, y)) · n(x, y) d x d y = (13x − 13y − 4) d x d y S
D
0
0
1 1 1 13 5y 13y 2 1 13x 2 − 13yx − 4x − 13y − 4 d y = − dy = = −4 = 2 2 2 2 0 0 0 x=0
zx 2 , hemisphere x 2 + y 2 + z 2 = 9, 7. F = 0, 3, F = e , z, x , (r, s) = (r s, r + s, r ), z ≥ 0, outward-pointing normal 0 ≤ r ≤ 1, 0 ≤ s ≤ 1, oriented by Tr × Ts SOLUTION We parametrize the hemisphere S by: (θ , φ ) = (3 cos θ sin φ , 3 sin θ sin φ , 3 cos φ ), 0 ≤ θ ≤ 2π , 0 ≤ φ ≤
π 2
Step 1. Compute the normal vector. As seen in the text, the normal vector that points to the outside of the hemisphere is: n = Tφ × Tθ = sin φ cos θ sin φ , sin θ sin φ , cos φ For 0 ≤ φ ≤ π2 we have sin φ cos φ ≥ 0, therefore n points to the outside of the hemisphere. z
y x
Step 2. Evaluate the dot product F · n. We express the vector field in terms of the parameters: F (θ , φ ) = 0, 3, x 2 = 0, 3, 9 cos2 θ sin2 φ Hence:
F (θ , φ ) · n(θ , φ ) = 0, 3, 9 cos2 θ sin2 φ · sin φ cos θ sin φ , sin θ sin φ , cos φ = sin φ 3 sin θ sin φ + 9 cos2 θ sin2 φ cos φ = 3 sin θ sin2 φ + 9 cos2 θ sin3 φ cos φ
Step 3. Evaluate the surface integral. The surface integral is equal to the following double integral: F · dS = F (θ , φ ) · n(θ , φ ) d θ d φ S
D
π /2 2π = 3 sin θ sin2 φ + 9 cos2 θ sin3 φ cos φ d θ d φ 0
0
= =
π /2 2π 0
π /2 0
=
0
3 sin θ sin2 φ d θ d φ +
sin2 φ d φ
2π 0
π /2 2π 0
3 sin θ d θ +
0
9 cos2 θ sin3 φ cos φ d θ d φ
π /2 0
9 sin3 φ cos φ d φ
0
2π
cos2 θ d θ
⎞ 2π 4 φ π /2 sin 2φ π /2 sin 2 φ θ θ 9 sin 2 π ⎠ − + −3 cos θ |θ =0 + ⎝ 2 4 φ =0 4 2 4 θ =0 φ =0
⎛
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C H A P T E R 17
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=0+
(ET CHAPTER 16)
9 9π ·π = 4 4
9. F = e z , z, x , z = 9 − x 2 − y 2 , z ≥ 0, upward-pointing normal √ 3 1 , inward-pointing normal F = x, y, z, part of sphere x 2 + y 2 + z 2 = 1, where ≤ z ≤ SOLUTION 2 2 Step 1. Find a parametrization. We use x and y as parameters and parametrize the surface by: (x, y) = x, y, 9 − x 2 − y 2 The parameter domain D is determined by the condition z = 9 − x 2 − y 2 ≥ 0, or x 2 + y 2 ≤ 9. That is: # " D = (x, y) : x 2 + y 2 ≤ 9 Step 2. Compute the tangent and normal vectors. We have: ∂ ∂ = x, y, 9 − x 2 − y 2 = 1, 0, −2x ∂x ∂x ∂ ∂ = x, y, 9 − x 2 − y 2 = 0, 1, −2y Ty = ∂y ∂y
Tx =
We compute the cross product of the tangent vectors: i j k Tx × T y = 1 0 −2x 0 1 −2y
= (2x)i + (2y)j + k = 2x, 2y, 1
Since the z-component is positive, the vector points upward, and we have: n = 2x, 2y, 1 Step 3. Evaluate the dot product F · n. We first express the vector field in terms of the parameters x and y, by setting z = 9 − x 2 − y 2 . We get: 2 2 F ((x, y)) = e z , z, x = e9−x −y , 9 − x 2 − y 2 , x We now compute the dot product: 2 2 2 2 F ((x, y)) · n(x, y) = e9−x −y , 9 − x 2 − y 2 , x · 2x, 2y, 1 = 2xe9−x −y + 2y(9 − x 2 − y 2 ) + x Step 4. Evaluate the surface integral. y
D 3
The surface integral is equal to the following double integral: F · dS = F ((x, y)) · n(x, y) d x d y = D
S
D
x
2 2 2xe9−(x +y ) + 2y 9 − (x 2 + y 2 ) + x d x d y
We convert the integral to polar coordinates to obtain: 3 2π 2 F · dS = 2r cos θ e9−r + 2r sin θ (9 − r 2 ) + r cos θ r d θ dr S
0
0
2π 3 3 2 dr = 0 dr = 0 2r 2 e9−r + r 2 sin θ − 2r 2 (9 − r 2 ) cos θ = 0
θ =0
0
2 i + 2j − xk, portion of the plane x + y + z = 1 in the octant x, y, z ≥ 0, upward-pointing normal 11. F =F y= sin y, sin z, yz, rectangle 0 ≤ y ≤ 2, 0 ≤ z ≤ 3 in the (y, z)-plane, normal pointing in negative x-direction SOLUTION
S E C T I O N 17.5
Surface Integrals of Vector Fields
(ET Section 16.5)
z
1
n
S
y
1 1 x
We parametrize the surface by: (x, y) = (x, y, 1 − x − y), using the parameter domain D shown in the figure. y
1
D
0
1
x
Step 1. Compute the tangent and normal vectors. We have: ∂ ∂ = (x, y, 1 − x − y) = 1, 0, −1 ∂x ∂x ∂ ∂ = (x, y, 1 − y) = 0, 1, −1 Ty = ∂y ∂y i j k Tx × T y = 1 0 −1 = i + j + k = 1, 1, 1 0 1 −1 Tx =
Since the normal points downward, the z-component must be negative, hence: n = −1, −1, −1 Step 2. Evaluate the dot product F · n. y
1 0≤x≤1−y D
0
1
x
We compute the dot product: F ((x, y)) · n = y 2 , 2, −x · −1, −1, −1 = −y 2 − 2 + x Step 3. Evaluate the surface integral. The surface integral is equal to the following double integral: 1 x 2 1−y −y 2 x − 2x + dy −y 2 − 2 + x d x d y = 2 x=0 D 0 0 0
1 1 (1 − y)2 (y − 1)2 2 3 2 −y (1 − y) − 2(1 − y) + y − y + 2(y − 1) + = dy = dy 2 2 0 0
S
F · dS =
=
F ((x, y)) · n d x d y =
1 1−y
1 1 y3 (y − 1)3 1 y4 1 11 − + (y − 1)2 + − = − 1 − =− 4 3 6 4 3 6 12 0
613
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C H A P T E R 17
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, disk of radius 13. F = x z, yz, z −1 3 at height 4 parallel to the x y-plane, upward-pointing normal z F = x, y, e , cylinder x 2 + y 2 = 4, 1 ≤ z ≤ 5, outward-pointing normal SOLUTION z n 3 4
y x
We parametrize the surface S by: (θ , r ) = (r cos θ , r sin θ , 4) with the parameter domain: D = {(θ , r ) : 0 ≤ θ ≤ 2π , 0 ≤ r ≤ 3} Step 1. Compute the tangent and normal vectors. We have: ∂ ∂ = (r cos θ , r sin θ , 4) = −r sin θ , r cos θ , 0 ∂θ ∂θ ∂ ∂ = (r cos θ , r sin θ , 4) = cos θ , sin θ , 0 Tr = ∂r ∂r i j k Tθ × Tr = −r sin θ r cos θ 0 = −r sin2 θ − r cos2 θ k = −r k = 0, 0, −r cos θ sin θ 0 Tθ =
Since the orientation of S is with an upward pointing normal, the z-coordinate of n must be positive. Hence: n = 0, 0, r Step 2. Evaluate the dot product F · n. We first express F in terms of the parameters: 1 F ((θ , r )) = x z, yz, z −1 = r cos θ · 4, r sin θ · 4, 4−1 = 4r cos θ , 4r sin θ , 4 We now compute the dot product: 1 r F ((θ , r )) · n(θ , r ) = 4r cos θ , 4r sin θ , · 0, 0, r = 4 4 Step 3. Evaluate the surface integral. The surface integral is equal to the following double integral:
S
F · dS =
D
F ((θ , r )) · n(θ , r ) dr d θ =
3 2π 3 r r r 2 3 9π dr d θ = 2π dr = 2π · = 4 4 8 4 0 0 0 0
15. F = 0, 0, e y+z , boundary of unit cube 0 ≤ x, y, z ≤ 1, outward-pointing normal F = x y, y, 0, cone z 2 = x 2 + y 2 , x 2 + y 2 ≤ 4, z ≥ 0, downward-pointing normal SOLUTION
z E F
D G O x
A
C B
y
Surface Integrals of Vector Fields
S E C T I O N 17.5
(ET Section 16.5)
We denote the faces of the cube by: S1 = Face O ABC
S2 = Face DG E F
S3 = Face ABG F
S4 = Face OC D E
S5 = Face BC DG
S6 = Face O AF E
• On S1
1 (x, y) = (x, y, 0) and n1 = 0, 0, −1. Thus,
F (1 (x, y)) · n1 = 0, 0, e y · 0, 0, −1 = −e y
• On S2
2 (x, y) = (x, y, 1) and n2 = 0, 0, 1. Thus,
F (2 (x, y)) · n2 = 0, 0, e y+1 · 0, 0, 1 = e y+1
• On any other surface Si , 3 ≤ i ≤ 6, we have
F (1 (x, y)) · ni = 0, because the z-component of ni = 0 and the x, y components of F equal 0. Thus, 1 1 1 1 F · dS = F · dS + F · dS = −e y d x d y + e y+1 d x d y S
S1
0
S2
0
0
0
1 1 1 = e y+1 − e y d x d y = e y+1 − e y d y 0
=
1 0
0
0
1 y y e (e − 1) d y = (e − 1)e = (e − 1)2 0
17. F = y, z, 0, 2(u, v) = (u 3 − v, u + v, v 2 ), 0 ≤ u ≤ 2, 0 ≤ v ≤ 3, downward-pointing normal F = 0, 0, z , (u, v) = (u cos v, u sin v, v), 0 ≤ u ≤ 1, 0 ≤ v ≤ 2π , upward-pointing normal SOLUTION
Step 1. Compute the tangent and normal vectors. We have, ∂ 3 ∂ = u − v, u + v, v 2 = 3u 2 , 1, 0 Tu = ∂u ∂u ∂ 3 ∂ = u − v, u + v, v 2 = −1, 1, 2v Tv = ∂v ∂v i j k Tu × Tv = 3u 2 1 0 = (2v)i − 6u 2 v j + 3u 2 + 1 k = 2v, −6u 2 v, 3u 2 + 1 −1 1 2v Since the normal is pointing downward, the z-coordinate is negative, hence, n = −2v, 6u 2 v, −3u 2 − 1 Step 2. Evaluate the dot product F · n. We first express F in terms of the parameters: F ((u, v)) = y, z, 0 = u + v, v 2 , 0 We compute the dot product:
F ((u, v)) · n(u, v) = u + v, v 2 , 0 · −2v, 6u 2 v, −3u 2 − 1 = −2v(u + v) + 6u 2 v · v 2 + 0 = −2vu − 2v 2 + 6u 2 v 3
Step 3. Evaluate the surface integral. The surface integral is equal to the following double integral: 3 2 F · dS = F ((u, v)) · n(u, v) du dv = −2uv − 2v 2 + 6u 2 v 3 du dv S
=
3 0
D
2 −u 2 v − 2v 2 u + 2u 3 v 3
u=0
0
dv =
0
3 0
3 4 16v 3 − 4v 2 − 4v dv = 4v 4 − v 3 − 2v 2 = 270 3 0
615
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C H A P T E R 17
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(ET CHAPTER 16)
Let er be the unit radial vector and r = x 2 + y 2 + z 2 . Calculate the integral of F = e−r er over: Let S be the oriented in 9, Figure 14. In (a)–(f),normal determine whether F · dS is positive, negative, The upper-hemisphere of x 2half-cylinder + y2 + z2 = outward-pointing S or zero. Explain your reasoning. The octant x, y, z ≥ 0 of the unit sphere centered at the origin (a) F = i (b) F = j SOLUTION (c) F = k (d) F = yi (a) We parametrize the upper-hemisphere by, (e) F = −yj (f) F = xj : x = 3 cos θ sin φ , y = 3 sin θ sin φ , z = 3 cos φ
19. (a) (b)
with the parameter domain: " π# D = (θ , φ ) : 0 ≤ θ < 2π , 0 ≤ φ < 2 The outward pointing normal is (see Eq. (4) in sec. 17.4): n = 9 sin φ er We compute the dot product F · n on the sphere. On the sphere r = 3, hence, F · n = e−r er · n = e−3 er · 9 sin φ er = 9e−3 sin φ er · er = 9 e−3 sin φ We obtain the following integral:
S
F · dS =
D
(F · n) d φ d θ =
= 18π e−3
π /2 0
sin φ d φ
2π π /2 0
0
9e−3 sin φ d φ d θ
= 18π e−3
π /2 − cos φ = 18π e−3 0
(b) We parametrize the first octant of the sphere by, : x = cos θ sin φ , y = sin θ sin φ , z = cos φ with the parameter domain: " π π# D = (θ , φ ) : 0 ≤ θ < , 0 ≤ φ < 2 2 The outward pointing normal is (as seen above): n = 1 sin φ er We compute the dot product F · n on the sphere. On the sphere r = 1, hence, F · n = e−r er · n = e−1 er · sin φ er = e−1 sin φ er · er = e−1 sin φ We obtain the following integral:
S
F · dS =
D
(F · n) d φ d θ =
π /2 π /2 0
0
e−1 sin φ d φ d θ
π /2 π −1 π /2 π −1 π − cos φ = e−1 = e sin φ d φ = e 2 2 2 0 0
er 21. The electric field due to a point er charge located at the origin is E = k 2 , where k is a constant. Calculate the flux of Show that the flux of F = 2 through a sphere centered at the origin r does not depend on the radius of the sphere. r E through the disk D of radius 2 parallel to the x y-plane with center (0, 0, 3). SOLUTION Let r = x 2 + y 2 + z 2 and rˆ = x 2 + y 2 . We parametrize the disc by: (ˆr, θ ) = (ˆr cos θ , rˆ sin θ , 3) ∂ = cos θ , sin θ , 0 ∂ rˆ ∂ = −ˆr sin θ , rˆ cos θ , 0 Tθ = ∂θ i j k sin θ 0 = 0, 0, rˆ n = Trˆ × Tθ = cos θ −ˆr sin θ rˆ cos θ 0 Trˆ =
Surface Integrals of Vector Fields
S E C T I O N 17.5
(ET Section 16.5)
617
Now, k rˆ zk rˆ er E · n = k 2 · 0, 0, rˆ = 3 x, y, z · 0, 0, 1 = 3 r r r Since on the disk z = 3, we get: rˆ E · n = 3k 3 and r = rˆ 2 + 9 r so E · n = 3k √ rˆ
rˆ 2 +9
3 .
D
E · dS =
2π 3 0
3 rˆ d r ˆ d θ = 6 π k 3/2 d rˆ 0 rˆ 2 + 9 0 rˆ 2 + 9 3/2 3k rˆ
Substituting u = rˆ 2 + 9 we get:
P
E · dS = 6π k
1 1 −√ 3 18
√ = 2 − 2 πk
23. Let v = x, 0, z be the velocity 2ft/s) y(in zof 2a fluid in R3 . Calculate the flow rate (in ft3 /s) through the upper x 2 field 2 2 2 Let S ellipsoid ≥ 0). = 1. Calculate the flux of F = z, 1, 0 over the portion of S hemisphere of be thethe sphere x + y + z+ = 1 (z + 4 3 2 where x, y, z ≤ 0 with upward-pointing normal. Hint: Parametrize S using a modified form of spherical coordinates SOLUTION We use the spherical coordinates: (θ , φ ). x = cos θ sin φ , y = sin θ sin φ , z = cos φ with the parameter domain 0 ≤ θ < 2π ,
0≤φ≤
π 2
The normal vector is (see Eq. (4) in Section 17.4): n = Tφ × Tθ = sin φ cos θ sin φ , sin θ sin φ , cos φ We express the function in terms of the parameters: v = x, 0, z = cos θ sin φ , 0, cos φ Hence, v · n = cos θ sin φ , 0, cos φ · sin φ cos θ sin φ , sin θ sin φ , cos φ = sin φ cos2 θ sin2 φ + cos2 φ The flow rate of the fluid through the upper hemisphere S is equal to the flux of the velocity vector through S. That is, S
v · dS =
π /2 2π cos2 θ sin3 φ + sin φ cos2 φ d θ d φ
=
0
0
0
2π
cos2 θ d θ
= π · 0 + 2π
π /2 0
sin3 φ d φ
+ 2π
π /2 0
sin φ cos2 φ d φ
2π 3 1 = ft /s 3 3
25. Calculate the flow rate of a fluid with velocity field v = x, y, x 2 y (in2 ft/s) 2through the portion of the ellipse x 2 Calculate y 2 the flow rate through the upper hemisphere of the sphere x + y + z 2 = R 2 (z ≥ 0) for v as in + Exercise 23.= 1 in the x y-plane, where x, y ≥ 0, oriented with the normal in the positive z-direction. 2 3 SOLUTION
We use the following parametrization for the surface (see remark at the end of the solution): : x = 2r cos θ , y = 3r sin θ , z = 0 π 0≤θ ≤ , 0≤r ≤1 2
(1)
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C H A P T E R 17
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(ET CHAPTER 16) z
n
y
3
2 x
Step 1. Compute the tangent and normal vectors. We have, ∂ ∂ = (2r cos θ , 3r sin θ , 0) = 2 cos θ , 3 sin θ , 0 ∂r ∂r ∂ ∂ = (2r cos θ , 3r sin θ , 0) = −2r sin θ , 3r cos θ , 0 Tθ = ∂θ ∂θ i j k 3 sin θ 0 = 6r cos2 θ + 6r sin2 θ k = 6r k Tr × Tθ = 2 cos θ −2r sin θ 3r cos θ 0 Tr =
Since the normal points to the positive z-direction, the normal vector is, n = 6r k = 0, 0, 6r Step 2. Compute the dot product v · n. We write the velocity vector in terms of the parameters: v = x, y, x 2 y = 2r cos θ , 3r sin θ , 4r 2 cos2 θ · 3r sin θ = 2r cos θ , 3r sin θ , 12r 3 cos2 θ sin θ Hence, v · n = 12r 3 cos2 θ sin θ · 6r = 72r 4 cos2 θ sin θ Step 3. Compute the flux. The flow rate of the fluid is the flux of the velocity vector through S. That is,
S
v · dS =
π /2
0
=
1
0
72r 4 cos2 θ sin θ dr d θ =
1
π /2
72r 4 dr
0
0
cos2 θ sin θ d θ
72 5 1 1 cos3 θ π /2 72 24 r · 0 + = 4.8 ft3 /s − = = 5 3 5 3 5 0 θ =0
Remark: We explain why (1) parametrizes the given portion of the ellipse. At any point (x, y) which satisfies (1) we have, x 2 y 3 + = r 2 cos2 θ + r 2 sin2 θ = r 2 ≤ 1 2 3 2 2 Therefore (x, y) is inside the ellipse x2 + 2y = 1. The limits of θ determine the part of the region inside the ellipse in the first quadrant. In Exercises 26–27, let T be the triangular region with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1) oriented with upwardpointing normal vector (Figure 15). Assume distances are in meters. 27. Calculate the flow rate through T if v = −j m/s. A fluid flows with constant velocity field v = 2k (m/s). Calculate: (a) The flow rate through T z (b) The flow rate through the projection of T onto the x y-plane [the triangle with vertices (0, 0, 0), (1, 0, 0), and (0, 0, 1) (0, 1, 0)] v = 2k (0, 1, 0) y (1, 0, 0) x
FIGURE 15
S E C T I O N 17.5
SOLUTION
Surface Integrals of Vector Fields
(ET Section 16.5)
619
We compute the flow rate through T . Since the unit normal vector is en = √1 , √1 , √1 we have, 3
3
3
1 1 1 −1 v · en = 0, −1, 0 · √ , √ , √ = √ 3 3 3 3 Therefore, the flow rate through T is the following flux:
S
v · dS =
S
(v · en ) d S =
√ √ −1 3 1 −1 √ d S = −Area(S)/ 3 = − √ = 2 2 3 S 3
The upward pointing normal to the projection D of T onto the x y-plane is n = 0, 0, 1. Since v = 0, −1, 0 is orthogonal to n, the flux of v through D is zero. In Exercises varying i(t) zflows through a long thex y-plane, x y-planethen as in Example 5. The Prove29–30, that if Sa is the partcurrent of a graph = g(x, y) lying over straight a domainwire D ininthe μ i wire is B = 0 T, where μ0 = current produces a magnetic fieldB whose magnitude at a distance r from the ∂g ∂g 2π r + F3 d x d y · dSthe = page at −F 1 P−inFthe 2 x y-plane. points 4π · 10−7 T-m/A. Furthermore, B pointsF into ∂x ∂y S D 29. Assume that i(t) = t (12 − t) A (t in seconds). Calculate the flux (t), at time t, of B through a rectangle of dimensions L × H = 3 × 2 m, whose top and bottom edges are parallel to the wire and whose bottom edge is located d = 0.5 m above the wire (similar to Figure 11). Then use Faraday’s Law to determine the voltage drop around the rectangular loop (the boundary of the rectangle) at time t. SOLUTION y L=3m
Loop C
Rectangle R H=2m
P = (x, y) 0.5 m
Wire
x
We choose the coordinate system as shown in the figure. Therefore the rectangle R is the region: R = {(x, y) : 0 ≤ x ≤ 3, 0.5 ≤ y ≤ 2.5} Since the magnetic field points into the page and R is oriented with normal vector pointing out of the page (as in Example 5) we have B = −Bk and n = en = k. Hence:
μ i B · n = B (−k) · k = −B = − 0 2π r μ i
The distance from P = (x, y) in R to the wire is r = y, hence, B · n = − 2π0 y . We now compute the flux (t) of B through the rectangle R, by evaluating the following double integral: (t) =
R
B · dS =
R
B · n dy dx =
3 2.5 0
μ i μ i 3 2.5 1 dy dx − 0 dy dx = − 0 2π y 2π 0 0.5 y 0.5
3μ i 2.5 d y 3μ i 3μ i 2.5 =− 0 = − 0 (ln 2.5 − ln 0.5) = − 0 ln 2π 0.5 y 2π 2π 0.5 =
−3 · 4π · 10−7 ln 5 t (12 − t) = −9.65 × 10−7 t (12 − t) T/m2 2π
We now use Faraday’s Law to determine the voltage drop around the boundary C of the rectangle. By Faraday’s Law, the voltage drop around C, when C is oriented according to the orientation of R and the Right Hand Rule (that is, counterclockwise) is, d d =− −9.65 · 10−7 t (12 − t) = 9.65 · 10−7 · 2(6 − t) = 1.93 · 10−6 (6 − t) volts E · dS = − dt dt C R
C
Assume that i = 10e−0.1t A (t in seconds). Calculate the flux (t), at time t, of B through the isosceles triangle of base 12 cm and height 6 cm, whose bottom edge is 3 cm from the wire, as in Figure 16. Assume the triangle is oriented with normal vector pointing out of the page. Use Faraday’s Law to determine the voltage drop around the
620
C H A P T E R 17
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(ET CHAPTER 16)
Further Insights and Challenges er 31. A point mass m is located at the origin. Let Q be the flux of the gravitational field F = −Gm 2 through the cylinder r x 2 + y 2 = R 2 for a ≤ z ≤ b, including the top and bottom (Figure 17). Show that Q = −4π Gm if a < 0 < b (m lies inside the cylinder) and Q = 0 if 0 < a < b (m lies outside the cylinder). z
b
R
m y a
x
FIGURE 17
Let the surface be oriented with normal vector pointing outward.
SOLUTION
n
z
b
R
m y
n a
x
n
We denote by S1 , S2 and S3 the cylinder, the top and the bottom respectively. These surfaces are parametrized by: • S1 :
1 (θ , z) = (R cos θ , R sin θ , z), 0 ≤ θ < 2π , a ≤ z ≤ b, n = R cos θ , sin θ , 0 • S2 :
2 (θ , r ) = (r cos θ , r sin θ , b), 0 ≤ θ < 2π , 0 ≤ r ≤ R, n = 0, 0, r • S3 :
3 (θ , r ) = (r cos θ , r sin θ , a), 0 ≤ θ < 2π , 0 ≤ r ≤ R, n = 0, 0, −r Using properties of integrals we have, F · dS = Q= S
S1
F · dS +
S2
F · dS +
S3
F · dS
(1)
Let us assume that a < 0. We compute the integrals over each part of the surface S separately. • S1 : On S1 , we have:
er Gm F (1 (θ , z)) = −Gm 2 = − 3/2 R cos θ , R sin θ , z 2 r R + z2 Hence, F (1 (θ , z)) · n(θ , z) = −
Gm R2 + z2
3/2 R cos θ , R sin θ , z · R cos θ , sin θ , 0 = −
Gm R 2 3/2 R2 + z2
We obtain the following integral: b 2π b dz Gm R 2 2 F · dS = − 3/2 dz d θ = −2π Gm R 3/2 2 2 2 S1 0 a a R +z R + z2
Surface Integrals of Vector Fields
S E C T I O N 17.5
(ET Section 16.5)
621
We compute the integral using the substitution z = R tan t. This gives: tan−1 b tan−1 b R R cos t dt = −2π Gm sin t 2 a −1 −1 R tan R t=tan Ra
b a = −2π Gm − b2 + R 2 a2 + R2
S1
F · dS = −2π Gm R 2
(2)
b2 + R2
b
a R
(
sin tan−1
)
b b = R b2 + R2
• S2 :
er Gm F (2 (θ , r )) = −Gm 2 = − 3/2 r cos θ , r sin θ , b r r 2 + b2 Hence, Gm Gmbr F (2 (θ , r )) · n(θ , r ) = − 3/2 r cos θ , r sin θ , b · 0, 0, r = − 3/2 2 2 2 r +b r + b2 We obtain the following integral: 2π R R r dr Gmbr F · dS = − 3/2 dr d θ = −2π Gmb 3/2 2 2 2 S2 0 0 0 r +b r + b2 We compute the integral using the substitution t = r 2 + b2 , dt = 2r dr , and we get:
2 2 R 2 +b2 1 dt 1 R +b 1 F · dS = −π Gmb = 2π Gmb √ = 2π Gmb − b t 3/2 t t=b2 S2 b2 b2 + R 2
(3)
• S3 :
er Gm F (3 (θ , r )) = −Gm 2 = − 3/2 r cos θ , r sin θ , a r r 2 + a2 Gm Gmar F (3 (θ , r )) · n(θ , r ) = − 3/2 r cos θ , r sin θ , a · 0, 0, −r = 3/2 2 2 2 r +a r + a2 √ Hence, by the same computation as for S2 we get (notice that since a < 0, we have a 2 = −a): S3
2 2 Gmar 1 R +a dr d θ = −2 π Gma √ 3/2 t t=a 2 0 0 r 2 + a2
1 1 1 1 = −2π Gma = −2π Gma −√ + a a2 R2 + a2 R2 + a2
F · dS =
2π R
Substituting (2), (3), and (4) in (1) we get:
b a 1 1 1 1 − 2π Gma + 2π Gmb − − + Q = −2π Gm b a b2 + R 2 a2 + R2 b2 + R 2 R2 + a2 = −2π Gm − 2π Gm = −4π Gm √ If 0 < a < b the only difference is in the integral in (4). In this case a = a therefore,
1 1 1 1 F · dS = −2π Gma −√ − = −2π Gma . a S3 a2 R2 + a2 R2 + a2
(4)
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C H A P T E R 17
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(ET CHAPTER 16)
Therefore, adding the integrals gives:
b 1 1 a 1 1 − − − Q = −2π Gm + 2π Gmb − 2π Gma b a b2 + R 2 a2 + R2 b2 + R 2 R2 + a2 = −2π Gm + 2π Gm = 0 In Exercises 32–33, let S be the surface with parametrization u u u cos u, 1 + v cos sin u, v sin (u, v) = 1 + v cos 2 2 2 for 0 ≤ u ≤ 2π , − 12 ≤ v ≤ 12 . 33. It is not possible to integrate a vector field over S because S is not orientable. However, it is possible to Use a computer algebra system. integrate functions over S. Using a computer algebra system: (a) Plot S and confirm visually that S is a M¨obius strip. (a) Verify that (b) The intersection of S with the x y-plane is the unit circle (u, 0) = (cos u, sin u, 0). Verify that the normal 1 3 u vector along this circle is n(u, v)2 = 1 + v 2 + 2v cos + v 2 cos u 2 2 4 u u u n(u, 0) = places. cos u sin , sin u sin , − cos (b) Compute the surface area of S to four decimal 2 2 2 2 2 2 ) d point S to four decimal places. (c) Compute (c) As u varies (x from+0yto + 2πz, the (u, 0) moves once around the unit circle, beginning and ending at (0, 0) = S 0, 0). Verify that n(u, 0) is a unit vector which varies continuously but that n(2π , 0) = −n(0, 0). (2π , 0) = (1, SOLUTION This shows that S is not orientable, that is, it is not possible to choose a nonzero normal vector at each point on S in a continuously (a) Using a CAS, we varying discovermanner that (if it were possible, the unit normal vector returns to itself rather than its negative when carried around the circle). u u 1 3u ∂n ∂n × = + 2 cos u + v cos , n(u, v) = −v cos sin ∂u ∂v 2 2 2 2 1 3u v + 2 cos(u/2) + 2v cos(u) − 2 cos − v cos(2u) , 4 2 u u 1 + v cos − cos 2 2 and after taking the norm of this, we find that 1 3 u n(u, v)2 = 1 + v 2 + 2v cos + v 2 cos u 4 2 2 (b) We calculate the area of S as follows: A(S) =
1/2 2π 1 3 u 1 + v 2 + 2v cos + v 2 cos u du dv ≈ 6.3533 n(u, v) du dv = 4 2 2 −1/2 0
(c) We proceed as follows. Since x 2 + y2 + z2 =
2 2 u u u 2 cos u + 1 + v cos sin u + v sin 1 + v cos 2 2 2
and
1 3 u 1 + v 2 + 2v cos + v 2 cos u 4 2 2 2 2 2 then, substituting these expressions into the double integral S (x + y + z ) d S 1 1 2 z )n(u, v) dudv, and integrating over 0 ≤ u ≤ 2π , − 2 ≤ v ≤ 2 , we find that (x 2 + y 2 + z 2 ) d S ≈ 7.4003 n(u, v) =
S
CHAPTER REVIEW EXERCISES 1. Compute the vector assigned to the point P = (−3, 5) by the vector field: (a) F = x y, y − x (b) F = 4, 8
=
S (x
2 + y2 +
Chapter Review Exercises
623
(c) F = 3x+y , log2 (x + y) SOLUTION
(a) Substituting x = −3, y = 5 in F = x y, y − x we obtain: F = −3 · 5, 5 − (−3) = −15, 8 (b) The constant vector field F = 4, 8 assigns the vector 4, 8 to all the vectors. Thus: F(−3, 5) = 4, 8 x+y , log2 (x + y) we obtain (c) Substituting x = −3, y = 5 in F = 3 F = 3−3+5 , log2 (−3 + 5) = 32 , log2 (2) = 9, 1 In Exercises sketch theFvector Find 3–6, a vector field in the field. plane such that F(x, y) = 1 and F(x, y) is orthogonal to G(x, y) = x, y for all x, y. 3. F(x, y) = y, 1 SOLUTION
Notice that the vector field is constant along horizontal lines. y
x
F = 〈y, 1〉
2 , where ϕ (x, 5. ∇ ϕF(x, 4,y) y) = 1 = x − y SOLUTION The gradient of ϕ (x, y) = x 2 − y is the following vector: ∂ϕ ∂ϕ , F(x, y) = = 2x, −1 ∂ x ∂y
This vector is sketched in the following figure: y
x
∇j = 〈2x, −1〉
In Exercises 7–14,determine whether or not thevector field is conservative and, if so, find a potential function. 4y −x F(x, y) = y , 2 2 2 , z 7. F(x, y, z) = sin x, e x + 16y x + 16y 2 SOLUTION Hint:
2= We examine crossvector partials of tangent F. SincetoFthe sin x, F = e y , Fx32=+z4y we have: c2 . Show that F isthe a unit field of2 ellipses 1 =family
∂F1 =0 ∂y
∂F2 =0 ∂z
∂F3 =0 ∂x
∂F2 =0 ∂x
∂F3 =0 ∂y
∂F1 =0 ∂z
⇒
∂F1 ∂F2 = , ∂y ∂x
∂F2 ∂F3 = , ∂z ∂y
∂F3 ∂F1 = ∂x ∂z
Since the cross partials are equal, F is conservative. We denote the potential field by ϕ (x, y, z). So we have:
ϕx = sin x
ϕy = e y
ϕz = z
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By integrating we get:
ϕ (x, y, z) = ϕy = C y = e y
sin x d x = − cos x + C(y, z) C(y, z) = e y + D(z)
⇒
ϕ (x, y, z) = − cos x + e y + D(z) ϕz = Dz = z
⇒
D(z) =
z2 2
2
We conclude that ϕ (x, y, z) = − cos x + e y + z2 . Indeed: ∂ϕ ∂ϕ ∂ϕ , , ∇ϕ = = sin x, e y , z = F ∂ x ∂y ∂z 1 x 2 z, 2z 2 y 9. F(x, y, z) = x yz, F(x, y, z) = 2,24, e z SOLUTION No. We show that the cross partials for x and z are not equal. Since the equality of the cross partials is a necessary condition for a field to be a gradient vector field, we conclude that F is not a gradient field. We have: ∂ ∂ F1 = (x yz) = x y ∂z ∂z ∂ F3 ∂ = (2z 2 y) = 0 ∂x ∂x
⇒
∂ F1 ∂ F3 = ∂z ∂x
Therefore the cross partials condition is not satisfied, hence F is not a gradient vector field. y −1 1 2z 2 y2 11. F(x, y, z)y,= x 2 yz,x, x F(x, z) =1 +x yx22z,, tan 2 SOLUTION
We examine the cross partials of F. Since F1 =
y , F2 = tan−1 x, F3 = 2z we have: 1+x 2
1 ∂F1 = ∂y 1 + x2 1 ∂F2 = ∂x 1 + x2
⇒
∂F2 ∂F1 = ∂y ∂x
∂F2 =0 ∂z ∂F3 =0 ∂y
⇒
∂F3 ∂F2 = ∂z ∂y
∂F3 =0 ∂x ∂F1 =0 ∂z
⇒
∂F3 ∂F1 = ∂x ∂z
Since the cross partials are equal, F is conservative. We denote the potential field by ϕ (x, y, z). We have:
ϕx =
y , 1 + x2
ϕ y = tan−1 (x),
ϕz = 2z
By integrating we get:
ϕ (x, y, z) =
y d x = y tan−1 (x) + c(y, z) 1 + x2
ϕ y = tan−1 (x) + c y (y, z) = tan−1 (x)
⇒
c y (y, z) = 0
⇒
c(y, z) = c(z)
Hence ϕ (x, y, z) = y tan−1 (x) + c(z). ϕz = c (z) = 2z ⇒ c(z) = z 2 . We conclude that ϕ (x, y, z) = y tan−1 (x) + z 2 . Indeed: y ∂ϕ ∂ϕ ∂ϕ −1 x, 2z = F , , = , tan ∇ϕ = ∂ x ∂y ∂z 1 + x2 2x 13. F(x, y, z) = xe , ye2z, ze2y F(x, y) = x 2 y, y 2 x
Chapter Review Exercises SOLUTION
625
We have: ∂ 2y ∂ F3 = = 2ze2y ze ∂y ∂y ∂ F2 ∂ 2z = ye = 2ye2y ∂z ∂z
F2 Since ∂∂Fy3 = ∂∂z , the cross-partials condition is not satisfied , hence F is not conservative. 4· ds, 4 y 3 ϕ (x, y, z) = x 4 y 3 z 2 and c(t) = (t 2 , 1 + t, t −1 ) for 1 ≤ t ≤ 3. 15. Calculate F(x, y) =∇ ϕ y x 3 , xwhere c
SOLUTION
The initial point P and the terminal point Q are the following points: P = c(1) = 12 , 1 + 1, 1−1 = (1, 2, 1) 1 Q = c(3) = 32 , 1 + 3, 3−1 = 9, 4, 3
Using the Fundamental Theorem for Gradient Vector Fields we obtain: c
∇ϕ
· ds = ϕ (Q) − ϕ (P) = 94 · 43 ·
2 1 − 14 · 23 · 12 = 46,648 3
1, 1, Find 17–20, a gradient vectorthe field the form Ff (x, = g(y), h(x) F(0, 0) =path where g(y) and h(x) are In Exercises compute lineofintegral y) ds for thesuch giventhat function and or curve. differentiable functions. Determine all suchCvector fields. 17. f (x, y) = x y, the path c(t) = (t, 2t − 1) for 0 ≤ t ≤ 1 SOLUTION
Step 1. Compute ds = c (t) dt. We differentiate c(t) = (t, 2t − 1) and compute the length of the derivative vector: √ c (t) = 1, 2 ⇒ c (t) = 12 + 22 = 5 Hence, ds = c (t) dt =
√
5 dt
Step 2. Write out f (c(t)) and evaluate the line integral. We have: f (c(t)) = x y = t (2t − 1) = 2t 2 − t Using the Theorem on Scalar Line Integral we have: f (x, y) ds = C
1 0
f (c(t)) c (t) dt =
√ 1 √ √ 2 3 1 2 1 √ 2 1 5 2 t − t = 5 − 5 dt = 5 2t − t = 3 2 3 2 6 0 0
√ y 19. f (x, y, z) = e x − √ , the path c(t) = 2 ln t,2 2t, 12 t 2 for 1 ≤ t ≤ 2 + y = 1, y ≥ 0 f (x, y) = x − y, the unit semicircle x 2 2z SOLUTION
Step 1. Compute ds = c (t) dt. We have: √ 1 d 1 √ c (t) = ln t, 2t, t 2 = , 2, t dt 2 t 2 2 √ 2 1 1 1 1
2= c (t) = + t + + 2 + t = +t 2 + t2 = t t t t2 Hence:
1 ds = c (t) dt = t + dt t
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Step 2. Write out f (c(t)) and evaluate the integral. √ 2t y 1 ln t √ =e − √ 1 =t− 2 t 2 2z 2 2 · 2t
f (c(t)) = e x −
We use the Theorem on Scalar Line Integrals to compute the line integral: 2 2 1 1 f (x, y) ds = f (c(t)) c (t) dt = t− t+ dt t t 1 1 C
=
2 1
1 t2 − 2 t
2 8 1 1 1 11 t3 + = + +1 = dt = − 3 t 3 2 3 6 1
21. Find the total mass of an L-shaped rod consisting of the segments (2t, 2) and (2, 2 − 2t) for 0 ≤ t ≤ 1 (length in f (x, y, z) = x + 2y + z, the helix c(t) = (cos t, sin t, t) for −1 ≤ t ≤ 3 centimeters) with mass density ρ(x, y) = x 2 y g/cm. SOLUTION y
A = (0, 2)
B = (2, 2)
x
C = (2, 0)
The total mass of the rod is the following sum:
M=
AB
x 2 y ds +
x 2 y ds
(1)
BC
The segment AB is parametrized by c1 (t) = (2t, 2), 0 ≤ t ≤ 1. Hence c 1 (t) = 2, 0 , c 1 (t) = 2 and f (c1 (t)) = x 2 y = (2t)2 · 2 = 8t 2 . The segment BC is parametrized by c2 (t) = (2, 2 − 2t), 0 ≤ t ≤ 1. Hence c 2 (t) = 0, −2 , c 2 (t) = 2 and f (c2 (t)) = x 2 y = 22 (2 − 2t) = 8 − 8t. Using these values, the Theorem on Scalar Line Integrals and (1) we get: 1 1 1 16t 3 1 2 = 40 = 13 1 M= 8t 2 · 2 dt + (8 − 8t) · 2 dt = + 16t − 8t 3 3 3 0 0 0 0 23. Calculate y 3 d x + x 2 y d y, where C1 is the zoriented curve in Figure 1(A). F · ds, where Calculate C1F = ∇ ϕ , where ϕ (x, y, z) = x ye , and compute C
(a) C is any curve from (1, 1, 0) to (3, y y e, −1). (b) C is the boundary of the square−30 ≤ x, y ≤ 1 oriented counterclockwise. −3 C2 C1
3
x
3
x
(B)
(A)
FIGURE 1 SOLUTION
arc BA.
We compute the line integral as the sum of the line integrals over the segments AO, O B and the circular
Chapter Review Exercises
627
y 3
A
B 0
The vector field is F = y 3 , x 2 y . We have: C1
3
F · ds =
x
AO
F · ds +
OB
F · ds +
arc B A
F · ds
(1)
We compute each integral separately. • The line integral over AO. The segment AO is parametrized by c(t) = (0, −t),
−3 ≤ t ≤ 0 . Hence:
F (c(t)) = y 3 , x 2 y = −t 3 , 0 c (t) = 0, −1 F (c(t)) · c (t) = −t 3 , 0 · 0, −1 = 0 Therefore: AO
F · ds =
0 −3
F (c(t)) · c (t) dt = 0
• The line integral over O B. We parametrize the segment O B by c(t) = (t, 0),
(2) 0 ≤ t ≤ 3 . Hence:
F (c(t)) = y 3 , x 2 y = 0, 0 c (t) = 1, 0 F (c(t)) · c (t) = 0 Therefore: OB
F · ds =
3 0
F (c(t)) · c (t) dt = 0
(3)
• The line integral over the circular arc B A. We parametrize the circular arc by c(t) = (3 cos t, 3 sin t), 0 ≤ t ≤ π . 2 Then c (t) = −3 sin t, 3 cos t and F (c(t)) = y 3 , x 2 y = 27 sin3 t, 27 cos2 t sin t . We compute the dot product:
F (c(t)) · c (t) = 27 sin3 t, cos2 t sin t · −3 sin t, 3 cos t = 81 − sin−4 t + cos3 t sin t We obtain the integral: arc B A
F · ds =
π /2 0
81 − sin4 t + cos3 t sin t dt
π /2 3 t sin 2t sin3 t cos t cos4 t − − = 81 − 4 4 2 4 4 t=0 1 3π 1 243 + = 81 − π + 20 =− 4·4 4 16 4
(4)
Combining (1), (2), (3), and (4) gives: 243 F · ds = 0 + 0 − π + 20.25 ≈ −27.463 16 C1 √ y (x 3 , eline 2 − 3x) and In Exercises 25–28, compute integral F · ds. thethe given vectorcurve fieldin and path.1(B). let Cfor oriented Figure Let F(x, y) = 9y − ythe 2 be c (a) Show that F is not conservative. 2y x , explicitly computing the integral. Hint: Consider the direction of F along the 25. (b) F(x,Show y) = that2 F2· ,ds 2= 0 without x +C24y x + 4y 2 edges of the square. the path c(t) = cos t, 12 sin t for 0 ≤ t ≤ 2π
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SOLUTION
Step 1. Calculate the integral F (c(t)) · c (t). 1 c(t) = cos t, sin t 2 2 · 12 · sin t 2y x cos t F (c(t)) = 2 , , = x + 4y 2 x 2 + 4y 2 cos2 t + 4 · 14 sin2 t cos2 t + 4 · 14 sin2 t sin t cos t = , = sin t, cos t cos2 t + sin2 t cos2 t + sin2 t 1 c (t) = − sin t, cos t 2 The integral is the dot product: 1 1 1 1 F (c(t)) · c (t) = sin t, cos t · − sin t, cos t = − sin2 t + cos2 t = cos 2t − sin2 t 2 2 2 2 Step 2. Evaluate the line integral. F · ds = C
2π 0
F (c(t)) · c (t) dt =
2π 1 0
1 cos 2t − sin2 t 2 2
t sin 2t 2π π sin 2t − + dt = =− 4 4 8 0 2
27. F(x, y) = x 2 y, y 2 z, 2z 2 x , 2 the path c(t) = e−t , e−2t , e−3t for 0 ≤ t < ∞ F(x, y) = 2x y, x + y , the part of the unit circle in the first quadrant oriented counterclockwise. SOLUTION
Step 1. Calculate the integrand F (c(t)) · c (t). c(t) = e−t , e−2t , e−3t c (t) = e−t , −2e−2t , −3e−3t F (c(t)) = x 2 y, y 2 z, z 2 x = e−2t · e−2t , e−4t · e−3t , e−6t · e−t = e−4t , e−7t , e−7t The integrand is the dot product: F (c(t)) · c (t) = e−4t , e−7t , e−7t · e−t , −2e−2t , −3e−3t = −e−5t − 2e−9t − 3e−10t Step 2. Evaluate the line integral. ∞ ∞ F · ds = F (c(t)) · c (t) dt = −e−5t − 2e−9t − 3e−10t dt C
0
0
1 −5R 2 −9R 1 2 3 13 3 13 e + + =− + e + e−10R − =0− 9 10 5 9 10 18 18 R→∞ 5
= lim
3 2 ), the 2 ), et two 29. Consider the line integrals ds2for F and 2. tWhich ln(1the +vector y 4 + zfields pathpaths c(t) c=int Figure , ln(1 + for 0of≤the t ≤line 1 integrals F = ∇ ϕ , where ϕ (x, y, z) F=· 4x c
appear to have a value of zero? Which of the other two is negative?
Chapter Review Exercises y
629
y
C x
x
(A)
(B)
y
y Q P
Q
P x
x
(C)
(D)
FIGURE 2 SOLUTION In (A), the line integral around the ellipse appears to be positive, because the negative tangential components from the lower part of the curve appears to be smaller than the positive contribution of the tangential components from the upper part. In (B), the line integral around the ellipse appears to be zero, since F is orthogonal to the ellipse at all points except for two points where the tangential components of F cancel each other. In (C), F is orthogonal to the path, hence the tangential component is zero at all points on the curve. Therefore the line integral C F · ds is zero. In (D), the direction of F is opposite to the direction of the curve. Therefore the dot product F · T is negative at each point along the curve, resulting in a negative line integral.
31. Find constants a, b, c such that Calculate the work required to move an object from P = (1, 1, 1) to Q = (3, −4, −2) against the force field inv)meters, in + Newtons), where r = x 2 + y 2 + z 2 . Hint: Find a F(x, y, z) = −12r −4 x, y, z (distance (u, = (u +force av, bu v, 2u − c) potential function for F. parametrizes the plane 3x − 4y + z = 5. Calculate Tu , Tv , and n(u, v). SOLUTION
We substitute x = u + av, y = bu + v and z = 2u − c in the equation of the plane 3x − 4y + z = 5, to
obtain: 5 = 3x − 4y + z = 3(u + av) − 4(bu + v) + 2u − c = (5 − 4b)u + (3a − 4)v − c or (5 − 4b)u + (3a − 4)v − (5 + c) = 0 This equation must be satisfied for all u and v, therefore the following must hold: 5 4 4 3a − 4 = 0 ⇒ a = 3 5 − 4b = 0
5+c =0
b=
c = −5
We obtain the following parametrization for the plane 3x − 4y + z = 5: 4 5 φ (u, v) = u + v, u + v, 2u + 5 3 4 We compute the tangent vectors Tu and Tv : Tu = The normal vector is their cross product: i n = Tu × Tv = 1 4 3
4 5 ∂φ ∂φ = 1, , 2 ; Tv = = , 1, 0 ∂u 4 ∂v 3
j 5 4
1
k 2 0
5 = 4 1
1 2 4 i − 0 3
1 2 j + 4 0 3
5 2 8 8 k = −2, , − = −2i + j + 1 − 3 3 3 3
5 4 k
1
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33. Let S be the surface parametrized by Calculate the integral of f (x, y, z) = e z−y over of the plane the portion 6x + 4y + z = 24, where x, y, z ≥ 0. v v (u, v) = 2u sin , 2u cos , 3v 2 2 for 0 ≤ u ≤ 1 and 0 ≤ v ≤ 2π . (a) Calculate the tangent vectors Tu and Tv , and normal vector n(u, v) at P = (1, π3 ). (b) Find the equation of the tangent plane at P. (c) Compute the surface area of S. SOLUTION
(a) The tangent vectors are the partial derivatives: ∂ v v v ∂ϕ v = Tu = 2u sin , 2u cos , 3v = 2 sin , 2 cos , 0 ∂u ∂u 2 2 2 2 ∂ v v v v ∂ϕ = 2u sin , 2u cos , 3v = u cos , −u sin , 3 Tv = ∂v ∂v 2 2 2 2 The normal vector is their cross-product: i j k v 2 sin v 0 0 2 cos v2 2 j + 2 sin 2v 2 cos v2 0 = n = Tu × Tv = 2 sin v2 i − v v u cos u cos −u sin 2 3 u cos v −u sin v 3 2 3 2 2 2 v v v v i − 6 sin j + −2u sin2 − 2u cos2 k = 6 cos 2 2 2 2 v v v v = 6 cos i − 6 sin j − 2uk = 6 cos , −6 sin , −2u 2 2 2 2 π At the point P = 1, 3 , u = 1 and v = π3 . The tangents and the normal vector at this point are,
2 cos v2 k v −u sin 2
π π π √ Tu 1, = 2 sin , 2 cos , 0 = 1, 3, 0 3 6 6 √ π 1 π π 3 = 1 · cos , −1 · sin , 3 = ,− ,3 Tv 1, 3 6 6 2 2 π √ π π n 1, = 6 cos , −6 sin , −2 · 1 = 3 3, −3, −2 3 6 6 π √ (b) A normal to the plane is n 1, 3 = 3 3, −3, −2 found in part (a). We find the tangency point: π π π π √ = 2 · 1 sin , 2 · 1 cos , 3 · = 1, 3, π P = φ 1, 3 6 6 3
The equation of the tangent plane is, thus, √ √ x − 1, y − 3, z − π · 3 3, −3, −2 = 0 or
√ √ 3 3(x − 1) − 3 y − 3 − 2(z − π ) = 0 √ 3 3x − 3y − 2z + 2π = 0
(c) In part (a) we found the normal vector: v v n = 6 cos , −6 sin , −2u 2 2 We compute the length of n: n =
36 cos2
v v + 36 sin2 + 4u 2 = 36 + 4u 2 = 2 9 + u 2 2 2
Using the Integral for the Surface Area we get: 1 2π 1 n(u, v) du dv = 2 9 + u 2 du dv = 4π 9 + u 2 du Area(S) = D
= 4π
0
0
1 u 2 9 u + 9 + ln u + 9 + u 2 2 2 u=0
= 4π
0
√ 9 9 1√ 10 + ln 1 + 10 − ln 3 2 2 2
Chapter Review Exercises
631
√ √ √ √ 1 + 10 ≈ 38.4 = 2 10π + 18π ln 1 + 10 − 18π ln 3 = 2 10π + 18π ln 3 35. Express the surface area of the surface z = 10 − x 2 − y 2 , −1 ≤ x ≤ 1, −3 ≤ y ≤ 3 as a double integral. Plot the surface with parametrization Evaluate the integral numerically using a CAS. (u, v) =Let (u + 4v,y)2u=−10 v,− 5uv) SOLUTION We use the Surface Integral over a graph. g(x, x 2 − y 2 . Then gx = −2x, g y = −2y hence 2 . The area at the surface is the following integral which we compute using a CAS: 2 2 1+ 1 +≤ 4xu2 ≤ + 4y forgx−1+≤g yv ≤=1, −1 1. Express the surface area as a double integral and use a computer algebra system to compute the area numerically. 3 1 1 + gx 2 + g y 2 d x d y = 1 + 4x 2 + 4y 2 d x d y ≈ 41.8525 Area(S) = D
−3 −1
2 √ with equation x 2 + y 2 = 9 for 0 ≤ z ≤ 10. 37. Calculate x 2+ y 2 e−z d S, where S is the cylinder Evaluate S x y d S, where S is the surface z = 3x + y 2 , −1 ≤ x ≤ 1, 0 ≤ y ≤ 1. SOLUTION
S
We parametrize the cylinder S by, (θ , z) = (3 cos θ , 3 sin θ , z)
with the parameter domain: 0 ≤ θ ≤ 2π ,
0 ≤ z ≤ 10.
We compute the tangent and normal vectors: ∂φ ∂ 3 cos θ , 3 sin θ , z = −3 sin θ , 3 cos θ , 0 = ∂θ ∂θ ∂φ ∂ 3 cos θ , 3 sin θ , z = 0, 0, 1 = Tz = ∂θ ∂θ
Tθ =
The normal vector is their cross product: i j n = Tθ × Tz = −3 sin θ 3 cos θ 0 0
k 0 1
3 cos θ = 0
−3 sin θ 0 i − 1 0
−3 sin θ 0 j + 1 0
3 cos θ k 0
= (3 cos θ )i + (3 sin θ )j = 3 cos θ , sin θ , 0 We compute the length of the normal vector: n = 3 cos2 θ + sin2 θ + 0 = 3 We now express the function f (x, y, z) = x 2 + y 2 e−z in terms of the parameters: f φ (θ , z) = x 2 + y 2 e−z = 9 cos2 θ + 9 sin2 θ e−z = 9e−z Using the Theorem on Surface Integrals, we obtain:
S
x 2 + y 2 e−z d S =
10 2π 0
0
9e−z 3 d θ dz = 27 · 2π
= 54π −e−10 + 1 ≈ 54π
−z 10 −z e dz = 54π −e z=0
10 0
39. Let S be a small patch of surface with a parametrization (u, v) for 0 ≤ u, v ≤ 0.1 such that the normal vector S be upper x 24. + Use y 2 +Eq. z 2 (1) = in 1, Section z ≥ 0. For of the functions (a)–(d), determine whether n(u,v) Let for (u, v) the = (0, 0) ishemisphere n = 2, −2, 17.4each to estimate the surface area of S. f d S is positive, zero, or negative (without evaluating the integral). Explain your reasoning. SOLUTION S
(a) f (x, y, z) = y 3
(b)y f (x, y, z) = z 3
(c) f (x, y, z) = x yz
(d) f (x, y, z) = z 2 − 2 n(0, 0)
Φ 0.1
P = Φ(0,
R (0, 0)
S
0.1
u
x
0)
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We use Eq. (1) in section 17.4 with u i j , vi j = (0, 0), Ri j = R = [0, 0.1] × [0, 0.1] in the (u, v)-plane and Si j = S = (R), in the (x, y)-plane to obtain the following estimation for the area of S: Area(S) ≈ n(0, 0)Area(R) That is: Area(S) ≈ 2, −2, 4 0.12 =
√ 22 + (−2)2 + 42 · (0.1)2 = 0.02 6 ≈ 0.049
In Exercises F · dS given + y 2 + surface z 2 = 9 or andparametrized let (r, θ ) =surface. (r cos θ , r sin θ , 9 − r 2 ) be Let S41–46, be the compute upper hemisphere of for the the sphere x 2oriented S its parametrization by cylindrical coordinates (Figure 3). (a) Calculate thex 2normal n= × −3 Tθ at y, e x z vector y 2 T=r 9, ≤ the z ≤point 3, (2, π3 ). , x2 + 41. F(x, y, z) = y, outward-pointing normal (b) Use Eq. (1) in Section 17.4 to estimate the surface area of (R), where R is the small domain defined by SOLUTION
The part of the cylinder is parametrized by: π π 2 ≤ r ≤ 2.1, ≤ θ ≤ + 0.05 3 3 (θ , z) = (3 cos θ , 3 sin θ , z), 0 ≤ θ ≤ 2π , −3 ≤ z ≤ 3 z
3
3
y x
−3
Step 1. Compute the tangent and normal vectors. ∂ ∂ 3 cos θ , 3 sin θ , z = −3 sin θ , 3 cos θ , 0 = ∂θ ∂θ ∂ ∂ 3 cos θ , 3 sin θ , z = 0, 0, 1 = Tz = ∂z ∂z
Tθ =
We compute the cross product: Tθ × Tz = ((−3 sin θ )i + (3 cos θ )j) × k = (3 sin θ )j + (3 cos θ )i = 3 cos θ , 3 sin θ , 0 The outward pointing normal is (when θ = 0, the x-component must be positive): n = 3 cos θ , 3 sin θ , 0 Step 2. Evaluate the dot product F · n. We write F(x, y, z) = y, x 2 y, e x z in terms of the parameters by substituting x = 3 cos θ , y = 3 sin θ . We get: F ((θ , z)) = 3 sin θ , 9 cos2 θ · 3 sin θ , e3z cos θ = 3 sin θ , 27 cos2 θ sin θ , e3z cos θ Hence:
F ((θ , z)) · n = 3 sin θ , 27 cos2 θ sin θ , e3z cos θ · 3 cos θ , 3 sin θ , 0 = 9 sin θ cos θ + 81 cos2 θ sin2 θ
Step 3. Evaluate the surface integral. The surface integral is equal to the following double integral (we use the trigonometric identities sin θ cos θ = sin22θ and sin2 2θ = 12 (1 − cos 4θ )): S
F · dS =
2π 3
=6 =
−3
0
F ((θ , z)) · n(θ , z) dz d θ =
2π 9 2
0
54 − 2
sin 2θ + 81 ·
cos 2θ 2
sin 2θ 2
2
2π 3 0
dθ =
−3
9 sin θ cos θ + 81 cos2 θ sin2 θ d θ
6 · 9 2π 6 · 81 2π sin 2θ d θ + (1 − cos 4θ ) d θ 2 0 8 0
2π 2π + 243 θ − sin 4θ = 243 · 2π = 243 π ≈ 381.7 4 4 4 2 0 0
Chapter Review Exercises
2 2 2 2 + z 2 = 4, z ≥ 0, outward-pointing normal 43. F(x, y, z)y,= x ,−y, y , xz, −x, + y 2 , (u, x 2 +v)y= F(x, z) = (u + 3v, v − 2u, 2v + 5), 0 ≤ u, v ≤ 1, upward-pointing normal SOLUTION The upper hemisphere is parametrized by: (θ , φ ) = (2 cos θ sin φ , 2 sin θ sin φ , 2 cos φ ),
0 ≤ θ ≤ 2π ,
0≤φ≤
π 2
As seen in section 17.4, since 0 ≤ φ ≤ π2 then the outward-pointing normal is: n = 4 sin φ cos θ sin φ , sin θ sin φ , cos φ We express F in terms of the parameters: F (θ , φ ) = x 2 , y 2 , x 2 + y 2 = 4 cos2 θ sin2 φ , 4 sin2 θ sin2 φ , 4 sin2 φ cos2 θ + sin2 θ = 4 cos2 θ sin2 φ , 4 sin2 θ sin2 φ , 4 sin2 φ The dot product F · n is thus F (θ , φ ) · n(θ , φ ) = 4 sin 4 cos3 θ sin3 φ + 4 sin3 θ sin3 φ + 4 sin2 φ cos φ = 16 cos3 θ sin4 φ + sin3 θ sin4 φ + sin3 φ cos φ We obtain the following integral: F · ds = S
F (θ , φ ) · n(θ , φ ) d θ d φ
D
π /2 2π = 16 cos3 θ sin4 φ + sin3 θ sin4 φ + sin3 φ cos φ d θ d φ 0
= 16
0
2π
0
+ 16
cos3 θ d θ 2π
0
π /2 0
sin4 φ d φ
sin3 θ d θ
π /2 0
sin4 φ d φ
+ 32π
π /2 0
sin3 φ cos v d φ
Since 02π cos3 θ d θ = 02π sin3 θ d θ = 0, we get:
π /2 sin4 φ π /2 1−0 3 = 32 F · ds = 32π sin φ cos φ d φ = 32π π = 8π 4 φ =0 4 S 0 45. F(x, y, z) = 0, 0, x ze x2y , z = x y, 0 ≤ x, y ≤ 1, F(x, y, z) = normal z, 0, z , (u, v) = (v cosh u, v sinh u, v), 0 ≤ u, v ≤ 1, upward-pointing SOLUTION
upward-pointing normal
We parametrize the surface by: (x, y) = (x, y, x y)
Where the parameter domain is the square: D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} Step 1. Compute the tangent and normal vectors. ∂ ∂ x, y, x y = 1, 0, y = ∂x ∂x ∂ ∂ x, y, x y = 0, 1, x = Ty = ∂y ∂y i j k 0 y i − 1 Tx × T y = 1 0 y = 0 1 x 0 1 x Tx =
1 y j + 0 x
0 k = −yi − xj + k = −y, −x, 1 1
Since the normal points upwards, the z-coordinate is positive. Therefore the normal vector is: n = −y, −x, 1
633
634
C H A P T E R 17
L I N E A N D S U R FA C E I N T E G R A L S
(ET CHAPTER 16)
Step 2. Evaluate the dot product F · n. We express F in terms of x and y: F ((x, y)) = 0, 0, x ze x y = 0, 0, x(x y)e x y = 0, 0, x 2 ye x y Hence:
F ((x, y)) · n(x, y) = 0, 0, x 2 ye x y · −y, −x, 1 = x 2 ye x y
Step 3. Evaluate the surface integral. The surface integral is equal to the following double integral: F · ds = F ((x, y)) · n(x, y) d x d y D
S
=
1 1 0
0
x 2 ye x y d y d x =
1
x2
1
0
ye x y d y
dx
0
We evaluate the inner integral using integration by parts: 1 0
1 1 1 xy 1 1 y x y 1 ex ex e y=0 − e dy = − 2 e x y − 2 ex − 1 = x x x x x 0 x y=0
ye x y d y =
Substituting this integral in (1) gives: F · ds = S
1 0
xe x − e x − 1 d x =
1 0
xe x d x −
1 0
ex − 1 d x
1 x 1 x xe d x − e − x = xe x d x − (e − 2) = 0 0 0 1
Using integration by parts we have: S
1 F · d S = xe x − e x − (e − 2) = 1 − (e − 2) = 3 − e 0
47. Calculate the total charge on the cylinder F(x, y, z) = 0, 0, z, 3x 2 + 2y 2 + z 2 = 1, z ≥ 0, x 2 + y 2 = R2 , upward-pointing normal
0≤z≤H
if the charge density in cylindrical coordinates is ρ(θ , z) = K z 2 cos2 θ , where K is a constant. SOLUTION The total change on the surface S is S ρ d S. We parametrize the surface by, (θ , z) = (R cos θ , R sin θ , H z) with the parameter domain, 0 ≤ θ ≤ 2π , 0 ≤ z ≤ 1. We compute the tangent and normal vectors: ∂ ∂ R cos θ , R sin θ , H z = −R sin θ , R cos θ , 0 = ∂θ ∂θ ∂ ∂ R cos θ , R sin θ , H z = 0, 0, H = Tz = ∂z ∂z
Tθ =
The normal vector is their cross product: i j k n = Tθ × Tz = −R sin θ R cos θ 0 0 0 H R cos θ 0 i − −R sin θ = 0 H 0
−R sin θ 0 j + H 0
= (R H cos θ )i + (R H sin θ )j = R H cos θ , sin θ , 0 We find the length of n: n = R H cos2 θ + sin2 θ = R H
R cos θ k 0
(1)
Chapter Review Exercises
635
We compute ρ ((θ , z)):
ρ ((θ , z)) = K (H z)2 cos2 θ = K H 2 z 2 cos2 θ Using the Theorem on Surface Integrals we obtain: 2π 1 ρ · dS = ρ ((θ , z)) · n(θ , z) dz d θ = K H 2 z 2 cos2 θ · H R dz d θ S
D
= =
1
K H 3 Rz 2 dz 0
0
2π
0
cos2 θ d θ
=
0
K H 3 Rz 3 1 sin 2θ 2π θ + 3 2 4 0 0
π K H3R · π = K H3R 3 3
x2 49. With v asthe in flow Exercise flow rate across part elliptic cylinder = 1 where = 0,9 Find rate 48, of acalculate fluid withthe velocity field v = the 2x, y, xofythe m/s across the part of + they 2cylinder x 2 +x,yy2 ≥ 4 x,4.y ≥ 0, and 0 ≤ z ≤ 4 (distance in meters). and 0where ≤z≤ SOLUTION
The flow rate of a fluid with velocity field v = 2x, y, x y through the elliptic cylinder S is the surface
integral:
S
v · dS
(1)
z
1
2
y
x
To compute this integral, we parametrize S by, (θ , z) = (2 cos θ , sin θ , z),
0≤θ≤
π , 2
0≤z≤4
y
0 ≤ θ ≤ π/2
x
Step 1. Compute the tangent and normal vectors. ∂ ∂ 2 cos θ , sin θ , z = −2 sin θ , cos θ , 0 = ∂θ ∂θ ∂ ∂ 2 cos θ , sin θ , z = 0, 0, 1 = Tz = ∂z ∂z i j k n = Tθ × Tz = −2 sin θ cos θ 0 = (cos θ )i + (2 sin θ )j = cos θ , 2 sin θ , 0 0 0 1 Tθ =
Step 2. Compute the dot product v · n v ((θ , z)) · n = 4 cos θ , sin θ , 2 cos θ sin θ · cos θ , 2 sin θ , 0 = 4 cos2 θ + 2 sin2 θ = 2 cos2 θ + 2 cos2 θ + sin2 θ = 2 cos2 θ + 2 Step 3. Evaluate the flux of v. The flux of v in (1) is equal to the following double integral (we use the equality 2 cos2 θ = 1 + cos 2θ in our calculation): 4 π /2 v · dS = v ((θ , z)) · n d θ dz = 2 cos2 θ + 2 d θ dz S
D
0
0
π /2 π /2 sin 2θ π /2 2 =4 (3 + cos 2θ ) d θ = 4 3θ + 2 cos θ + 2 d θ = 4 = 6π 2 θ =0 0 0