Calculus Of Multivariable - Solved Assignments - Semester Spring 2008

Solution Assignment No: 1 1.

In the form of interval, real line can be denoted by (-∞ , +∞)

2.

To each point on a co-ordinate line or real line, there is associated i. A real number ii. An integer iii. A natural number iv. A rational number.

3.

How many real numbers lie in the interval [1, 5] i. 5 ii. 3 iii. 10 iv. Infinite

4.

All the three axes are positive in i. Third octant ii. First octant iii. Second octant iv. Eigth octant

5.

Let w = f(x, y, z) such that w =

( x − 1)2 ( y − 2) 2 ( z − 4) 2 where a, + + a2 b2 c2

b and c are real numbers, not equal to zero. Is the function defined at origin? If yes, what is its value and if not, give the reason.

Solution: Origin means (x, y, z) = (0, 0, 0). So putting these values in given equation

( x − 1) 2 ( y − 2) 2 ( z − 4) 2 + + a2 b2 c2 (0 − 1) 2 (0 − 2) 2 (0 − 4) 2 = + + a2 b2 c2 1 4 16 = 2 + 2 + 2 a b c

w=

Thus the given function is defined at origin and its value is 1 4 16 + 2 + 2. 2 a b c 6.

Find the domain of the following functions. i.

f ( x , y) =

4 − x2 − 2 y2

Domain of f consists of region in xy-plane where 4 − x 2 − 2 y 2 ≥ 0

4 ≥ x2 + 2 y2 x2 + 2 y2 ≤ 4 Now x 2 + 2 y 2 = 4 is an equation of surface of elliptic cylinder in three dimensional space which has infinite length along z-axis. Thus domain of f is inside and on the surface of elliptic cylinder.

ii.

f ( x, y ) =

1 ( x + y 2 )2 2

Domain of f is entire xy-plane except the origin because at origin function is not defined. iii.

f ( x, y ) = x 2 + y 3 − xy + cos ( xy ) Domain of f is entire xy-plane.

7.

Which of the following statement is true? Correct the false statement.

i. A straight line is a special kind of curve. TRUE Generally, a curve is considered to be any one-dimensional collection of points. For your convenience, think a curve as a thread which we use in our daily life. The straight line is a special kind of curve.

ii. Plane is an example of surface. TRUE Surface is a two-dimensional geometric figure (a collection of points) in three-dimensional space. The simplest example is a plane—a flat surface. Some other common surfaces are spheres, cylinders, and cones.

iii. The intersection of two surfaces in three dimensional space gives a surface. FALSE Correct statement is “Intersection of two surfaces in three dimensional space gives a curve”.

8.

Identify the curve or surface in three dimensional space for each of the following equation. i. y2 + z2 = 100 This is the equation of surface of circular cylinder which has infinite length along x-axis.

ii. x = 0 This is the equation of yz-plane.

iii. y 2 + 4 z 2 = 1 , x = 0 Since intersection of two surfaces is a curve in three dimensional space. So a curve in three dimensional space is represented by two equations representing the intersecting surfaces. y 2 + 4 z 2 = 1 is the equation of surface of an elliptic cylinder which has infinite length along x-axis and x = 0 is the equation of yzplane. The intersection of these two surfaces is an elliptic curve. Thus these two equations shows elliptic curve in yz-plane.

iv.

y = x2 This is the equation of surface of half cylinder in three dimensional space.

v. z = z0 , if (r , θ , z ) represent spherical co-ordinates, and z0 is any

real number. z = z0 is a plane at point z0 on z-axis. It is parallel to xy-plane.

9.

Following are the co-ordinates of point in different co-ordinate systems. Complete the table.

Rectangular Co-ordinate

Spherical Co-ordinate

Cylindrical Co-ordinate

( x, y , z )

(ρ , θ , φ )

(r , θ , z )

π π   3, ,  3 2 

r = ρ sin φ r = 3 sin

π 2

r = 3(1) r= 3 θ =θ

θ=

π

3 z = ρ cos φ z = 3 cos

π 2

z = 3(0) z=0

π   ⇒  3, , 0  3  

x = ρ sin φ cos θ x = 3 sin

π

cos

2 1 x = 3(1)    2

π 3

3 2 y = ρ sin φ sin θ

x=

y = 3 sin

π 2

sin

π 3

 3 y = 3(1)    2  3 y= 2 z = ρ cos φ z = 3 cos

π 2

z = 3(0) z=0  3 3  , , 0  ⇒  2 2   x = r cos θ

ρ = r2 + z2

x = (0) cos π

ρ = (0) 2 + (− 2) 2

x = (0)(−1)

y = (0)sin π

ρ= 2 θ =θ θ =π

y = (0)(0)

tan φ =

x=0 y = r sin θ

y=0

tan φ =

z=z z=− 2

(

⇒ 0, 0, − 2

)

r z 0

− 2 tan φ = 0

φ = tan −1 0 φ =0 ⇒

(

2, π , 0

)

( 0, π , − 2 )

10.

Show that whether the limit of the following function exist or not at the origin. x2 y2 . x4 + 3 y 4

f ( x, y ) = Solution:

Let us recall Rule for Non-Existence of a Limit If in lim

( x , y ) → ( a ,b )

f ( x, y )

we get two or more different values as we approach (a, b) along different paths, then

lim

( x , y ) → ( a ,b )

f ( x, y )

does not exist. The paths along which (a, b) is approached may be straight lines or plane curves .

Two of the more common paths to check are the x and y-axis so let’s try those.

If we approach (0, 0) along the x-axis, their y = 0 . This means that along the x-axis, we will plug-in y = 0 into the function and then take the limit as x approaches zero. lim

( x , y ) → (0,0)

f ( x, y ) =

x2 y2 ( x , y ) → (0,0) x 4 + 3 y 4 lim

x 2 (0) 2 ( x , y ) →(0,0) x 4 + 3(0) 4 = lim 0 =

lim

( x , y ) →(0,0)

=0 So, along the x-axis the function will approach zero as we move towards the origin. Now, let’s try the y-axis. Along this axis we have x = 0 and so the limit becomes, x2 y 2 lim f ( x, y ) = lim ( x , y ) → (0,0) ( x , y ) →(0,0) x 4 + 3 y 4

(0) 2 y 2 = ( x , y ) →(0,0) (0) 4 + 3 y 4

=

lim

lim

( x , y ) →(0,0)

0

=0

The same limit along two paths Let’s take another path. Move towards the origin along the path y = x. To do this we will replace all the y’s with x’s and then let x approach zero. lim

( x , y ) → (0,0)

x2 y 2 ( x , y ) →(0,0) x 4 + 3 y 4

f ( x, y ) =

lim

( x )( x ) 2

= = = =

lim

( x , x ) →(0,0)

lim

( x , y ) →(0,0)

lim

( x , y ) →(0,0)

2

x 4 + 3x 4 x4 4 x4 1 4

1 4

A different value from the previous two paths. This means that the limit DOES NOT exist.

11.

In which region the following function is continuous? f (r , s ) = ln (r 2 + s 2 − 1) Solution: The function is continuous in the region where r 2 + s 2 − 1 > 0 because the natural logarithm (ln) is defined for non-zero positive real numbers only. This implies r 2 + s 2 > 1 Since r 2 + s 2 = 1 is the equation of a circle, center at origin and unit radius. Thus f is continuous in the region outside a unit circle center at origin.

12.

Let r is a function and s, t, u, v are variables. Partial derivative is a derivative of a function r with respect to u

i. If r is a function of s and s is a function of u.

FALSE

Say r = f ( s ) , s = g (u ) then we have dr dr ds = du ds du ii. If r is a function of s and s is a function of two variables u and v. TRUE Say r = f ( s ) , s = g (u , v) then we have ∂r dr ∂s = ∂u ds ∂u iii. If r is a function of s and t and each of s and t is a function of u. FALSE

Say r = f ( s, t ) , s = g (u ) , t = h(u ) then we have dr ∂r ds ∂r dt = + du ∂s du ∂t du iv. If r is a function of s and t and each of s and t is a function of both u and v.

TRUE

Say r = f ( s, t ) , s = g (u , v) , t = h(u , v) then we have ∂r ∂r ∂s ∂r ∂t = + ∂u ∂s ∂u ∂t ∂u 13.

Let f ( x, y ) =

x 3 y − xy 3 and f (0, 0) = 0 x2 + y 2

Find f x ( x, y ) and f y ( x, y ) . Solution: x3 y − xy 3 f ( x, y ) = 2 x + y2 f x ( x, y ) =

(3 x 2 y − y 3 )( x 2 + y 2 ) − ( x3 y − xy 3 )(2 x) ( x 2 + y 2 )2

=

3x 4 y + 3 x 2 y 3 − x 2 y 3 − y 5 − 2 x 4 y + 2 x 2 y 3 ( x 2 + y 2 )2

=

x4 y + 4 x2 y3 − y5 ( x 2 + y 2 )2

f y ( x, y ) =

14.

( x3 − 3 xy 2 )( x 2 + y 2 ) − ( x3 y − xy 3 )(2 y ) ( x 2 + y 2 )2

=

x5 + x 3 y 2 − 3x3 y 2 − 3xy 4 − 2 x3 y 2 + 2 xy 4 ( x 2 + y 2 )2

=

x5 − 4 x3 y 2 − xy 4 ( x 2 + y 2 )2

x  Show whether the function z ( x, y ) = sin  y −  satisfy Laplace’s 2  Equation or not? Solution:

As we know for a function z(x, y) ,Laplace Equation is ∂2 z ∂2 z + =0 ∂x 2 ∂y 2 x  So for z ( x, y ) = sin  y −  2  ∂z x  = cos  y −  ∂x 2 

 −1     2  ∂2 z x 1  = − sin  y −    2 2 4 ∂x  ∂z x  = cos  y −  ∂y 2  ∂2 z x  = − sin  y −  2 2 ∂y  ∂2 z ∂2 z x x 1   + 2 = −   sin  y −  − sin  y −  2 2 2 ∂x ∂y 4   5  x = − sin  y −  ≠ 0 4  2 Since for given function z,

∂2 z ∂2 z + ≠ 0 so the given function does not ∂x 2 ∂y 2

satisfy Laplace Equation. 15.

For a function f ( x, y ) = y 2 x 4 e x + 2 ,

∂5 f =0 ∂y 3 ∂x 2 Show the calculation steps. Solution: f ( x, y ) = y 2 x 4 e x + 2 ∂5 f means differentiating f with respect to x, two times and then with ∂y 3 ∂x 2 respect to y three times. If we differentiate this function in this order it will be difficult because due to the presence of x 4 e x , number of terms increase after each differentiation. But if we first differentiate this function three times with respect to y and then two times with respect to x, this fifth derivative can be calculated in few steps. As we know that this function and all of its partial derivatives are defined and continuous everywhere, so we use Euler’s Theorem for mixed derivatives by which ∂5 f ∂5 f = ∂y 3 ∂x 2 ∂x 2 ∂y 3 So f ( x, y ) = y 2 x 4 e x + 2

∂f = 2 yx 4 e x ∂y ∂2 f = 2 x4ex 2 ∂y ∂3 f =0 ∂y 3 ∂4 f =0 ∂x∂y 3 ∂5 f =0 ∂x 2 ∂y 3

Thus ∂5 f =0 ∂y 3 ∂x 2

16.

Let w =

x where x = et , y = ln t . Find the derivative of w with respect y

to t. Solution:

Since w is a function of two variables x and y , and each x and y are function of one variable t. So by chain rule dw ∂w dx ∂w dy = + − − − − − − − − − −(1) dt ∂x dt ∂y dt Since w =

x , x = et , y = ln t so y

dx dy 1 = et and = dt dt t

∂w 1 ∂w − x = and = 2 ∂x y ∂y y Put values of x and y ∂w 1 ∂w − et and = = ∂x ln t ∂y (ln t ) 2 Put values in equation (1) −et  1  dw 1 t = e )+ (   dt ln t (ln t ) 2  t  =

tet ln t − et t (ln t ) 2

et (t ln t − 1) t (ln t ) 2 dw Observe that is a function of t only. dt =

Solution Assignment No: 2 Q 1.

Consider a point P(1, 2, 2) in rectangular co-ordinate system. a. What is the position vector of this point? b. Find the unit vector in the direction of this position vector?

Solution: →

a)

The position vector r of the point (1, 2, 2) is →

r = i + 2 j + 2k

b)

Since any vector can be expressed as product of its magnitude and unit vector so we can write →

^

r= r r → ^

r=

r r

r = (1)2 + (2) 2 + (2) 2 = 9 =3 ^

So,

i + 2 j + 2k 3 1 2 2 = i+ j+ k 3 3 3

r=

→

Q 2.

Find whether the vectors a = i − 3 j + 7 k ,

→

b = 8i − 2 j − 2k are

orthogonal to each other or not?

Solution: As we know, if two vectors are orthogonal their dot product is zero. So let us find the dot product of given vectors.

→

→

a . b = (i − 3 j + 7k ) . (8i − 2 j − 2k ) = 8(i . i ) + 6( j. j ) − 14(k .k ) = 8 + 6 − 14 =0

Thus, the two vectors are orthogonal or perpendicular to each other.

Q 3.

Find the area of the parallelogram determined by the vectors

→

→

a = i − j + 2k and b = 3 j + k

Solution: →

→

→

→

Area of a parallelogram = magnitude of a × b = a × b →

→

a × b = (i − j + 2k ) × (0i + 3 j + k ) i j k = 1 −1 2 0 3 1 = (−1 − 6)i + (0 − 1) j + (3 − 0)k = − 7i − j + 3k →

→

a × b = (−7) 2 + (−1) 2 + (3)2 = 49 + 1 + 9 = 59 = 7.68

Thus area of parallelogram is 7.68 square unit

Q 4.

Find the directional derivative of f ( x, y ) = y 2 ln x →

at the point (1, 4) in the direction of the vector u = − 3i + 3 j Solution:

f ( x, y ) = y 2 ln x y2 , f y ( x, y ) = 2 y ln x x f x (1, 4) = 16 , f y (1, 4) = 0 f x ( x, y ) =

→

u = − 3i + 3 j →

^

u= uu → ^

u=

u u

u = (−3) 2 + (3)2 =

9+9

= 18 = 3 2 ^

u=

−3i + 3 j

3 2 1 1 i+ j = − 2 2

^ ^     Du f (4, 1) = f x (1, 4)  x − component of u  + f y (1, 4)  y − component of u     

 −1   1  = 16   + 0   2  2 −16 = 2

Q 5.

Find a unit vector in the direction in which the function f ( x, y ) = 4 x 3 y 2 increases most rapidly at the point P(-1, 1).

Solution:

f ( x, y ) = 4 x 3 y 2 f x ( x, y ) = 12 x 2 y 2 ,

f y ( x, y ) = 8 x 3 y

∇f ( x, y ) = f x ( x, y )i + f y ( x, y ) j = 12 x 2 y 2 i + 8 x 3 y j f x (−1, 1) =12 , f y (−1, 1) = −8

∇f (−1, 1) = f x (−1, 1)i + f y (−1, 1) j = 12 i − 8 j ^

A unit vector u in the direction of ∇f (−1, 1) is ^

u=

∇f (−1, 1) ∇f (−1, 1)

∇f (−1, 1) = (12) 2 + (−8) 2 = 144 + 64 = 4 13 ^

u= =

Q 6.

∇f (−1, 1) 12 i − 8 j = ∇f (−1, 1) 4 13 3 2 i− j 13 13

Write down the general equation in parametric form for the following. a. Straight line in two dimensional space. Let (x0, y0) is any fixed point on the line and the line is parallel to the vector ai + bj then parametric form of straight line in two dimensional space in terms of parameter t is x = x0 +at , y = y0 + bt b. Straight line in three dimensional space. Parametric equation of a line in three dimensional space passing through the point (x0, y0, z0) and parallel to the vector ai + bj + ck is given by x = x0 +at , y = y0 + bt

,

z = z0 +ct

c. Parametric form of curve in two dimensional space is x = f(t) , y = g(t) , where t is parameter

Q 7.

Find the equation of a. Tangent plane b. Normal line of the surface f ( x, y, z ) = x 2 y − 4 z 2 + 7 at the point (-3, 1, -2)

Solution: f ( x, y , z ) = x 2 y − 4 z 2 + 7 f x ( x, y, z ) = 2 xy f y ( x, y , z ) = x 2 f z ( x, y, z ) = −8 z f x (−3, 1, − 2) = 2(−3)(1) = − 6 f y (−3, 1, − 2) = (−3) 2 = 9 f z (−3, 1, − 2) = − 8(−2) = 16

Equation of the tangent plane to the surface at (x0, y0, z0) = (-3, 1, -2) is f x ( x0 , y0 , z0 ) ( x − x0 ) + f y ( x0 , y0 , z0 ) ( y − y0 ) + f z ( x0 , y0 , z0 ) ( z − z0 ) = 0 −6( x + 3) + 9( y − 1) − 2( z + 2) = 0 −6 x − 18 + 9 y − 9 − 2 z − 4 = 0 −6 x + 9 y − 2 z − 31 = 0 Equation of the normal line of the surface through (x0, y0, z0) = (-3, 1, -2) in SYMMETRIC FORM is

x − x0 y − y0 z − z0 = = f x ( x0 , y0 , z0 ) f y ( x0 , y0 , z0 ) f z ( x0 , y0 , z0 ) x + 3 y −1 z + 2 = = −6 9 16 Equation of the normal line of the surface through (x0, y0, z0) = (-3, 1, -2) in PARAMETRIC FORM is x = x0 + f x ( x0 , y0 , z0 ) y = y0 + f y ( x0 , y0 , z0 ) z = z0 + f z ( x0 , y0 , z0 ) Thus x = − 3 − 6t y = 1 + 9t z = −2 + 16t

Q 8.

Show that the sphere x 2 + y 2 + z 2 = a 2 and the cone z 2 = x 2 + y 2 are

orthogonal at every point of intersection. Solution:

f ( x, y, z ) = x 2 + y 2 + z 2 − a 2 = 0 − − − − − − − − − − − − − −(1) g ( x, y, z ) = x 2 + y 2 − z 2 = 0 − − − − − − − − − − − − − − − −(2) f x ( x, y , z ) = 2 x f y ( x, y , z ) = 2 y f z ( x, y , z ) = 2 z g x ( x, y , z ) = 2 x g y ( x, y , z ) = 2 y g z ( x, y , z ) = − 2 z

f x g x + f y g y + f z g z = (2 x)(2 x) + (2 y )(2 y ) + (2 z )(−2 z ) = 4x2 + 4 y 2 − 4 z 2 = 4( x 2 + y 2 − z 2 ) Since from equation (2) , x 2 + y 2 − z 2 = 0 So f x g x + f y g y + f z g z = 4( x 2 + y 2 − z 2 ) = 0 Thus, the sphere x 2 + y 2 + z 2 = a 2 and the cone z 2 = x 2 + y 2 are orthogonal at every point of intersection.

Q 9.

For evaluating RELATIVE EXTREME VALUES of a function at some

point in the domain, what we consider as a neighborhood of a point if it lies in a. Two dimensional space b. Three dimensional space Solution:

a. For evaluating RELATIVE EXTREME VALUES of a function at some point (x0, y0) in the domain D, neighborhood of that point is an open disc. Open disc K ∈ D, centered at point (x0, y0) and of radius r is defined as K = {( x, y ) ∈ R 2 : ( x − x0 ) 2 + ( y − y0 ) 2 < r 2 } b. For evaluating RELATIVE EXTREME VALUES of a function at some point (x0, y0, z0) in the domain D, neighborhood of that point is an open

sphere. Open sphere S ∈ D, centered at point (x0, y0, Z0) and of radius r is defined as S = {( x, y, z ) ∈ R 3 : ( x − x0 )2 + ( y − y0 ) 2 + ( z − z0 ) 2 < r 2 }

Q 10.

Locate all relative maxima, relative minima and saddle point of

f ( x, y ) = x 2 + 2 y 2 − x 2 y Solution: f ( x, y ) = x 2 + 2 y 2 − x 2 y f x ( x, y ) = 2 x − 2 xy , f y ( x, y ) = 4 y − x 2 The critical points of f satisfy the equations 2 x − 2 xy = 0 − − − − − − − − − −(1) 4 y − x 2 = 0 − − − − − − − − − −(2)

From equation (1) 2 x(1 − y ) = 0 x(1 − y ) = 0 x = 0 , 1− y = 0 ⇒ y = 1 Use this value of x and y in equation (2), If x = 0 then 4 y − x2 = 0 ⇒ 4 y = 0 ⇒y=0 If y = 1 then 4 y − x2 = 0 ⇒ 4 − x2 = 0 x2 = 4 x = ±2 So (0, 0), (2, 1) and (-2, 1) are the critical points Now find second-order partial derivatives f xx ( x, y ) = 2 − 2 y , f yy ( x, y ) = 4 , f xy ( x, y ) = − 2 x For critical point (0, 0) f xx (0, 0) = 2 , f yy (0, 0) = 4 , f xy (0, 0) = 0 D = f xx (0, 0) f yy (0, 0) − f xy 2 (0, 0) = (2)(4) − 0 = 8

D > 0 and f xx (0, 0) > 0 So f has relative minimum at (0, 0) For critical point (2, 1)

f xx (2, 1) = − 2 , f yy (2, 1) = 4 , f xy (2, 1) = − 4 D = f xx (2, 1) f yy (2, 1) − f xy 2 (2, 1) = (−2)(4) − (−4) 2 = −8 − 16 = −24 D < 0 so f has a saddle point at (2, 1)

For critical point (-2, 1) f xx ( x, y ) = 2 − 2 y , f yy ( x, y ) = 4 , f xy ( x, y ) = − 2 x f xx (−2, 1) = − 2 , f yy (−2, 1) = 4 , f xy (−2, 1) = 4 D = f xx (−2, 1) f yy (−2, 1) − f xy 2 (−2, 1) = (−2)(4) − (4) 2 = −8 − 16 = −24 D < 0 so f has a saddle point at (-2, 1)

Solution Assignment No: 3 Q 1.

Find the absolute extrema of the function f ( x, y ) = x 2 − 3 y 2 − 2 x + 6 y on the

square region R with vertices (0, 0), (0 , 2), (2, 2) and (2, 0) Solution: There are three steps of finding the Absolute extrema of a continuous function of two variables on a closed and bounded set R. STEP 1: Find the critical points of f that lie in the interior of R. STEP 2: Find all boundary points at which the absolute extrema can occur. STEP 3: Evaluate f(x, y) at the points obtained in the proceeding steps. The largest of these values is the absolute maximum and the smallest the absolute minimum.

Above figure shows a yellow-colored square region R with vertices (0, 0), (0 , 2), (2, 2) and (2, 0). STEP 1:

f ( x, y ) = x 2 − 3 y 2 − 2 x + 6 y f x ( x, y ) = 2 x − 2

f y ( x, y ) = −6 y + 6

So all the critical points occur where 2x − 2 = 0

−6y + 6 = 0

That is: x − 1 = 0 and − y + 1 = 0 x = 1 and

y =1

(1, 1) is the only critical point inside the given region R. You can easily see in the above figure that point (1, 1) lies inside the square R. STEP 2: i. Line segment between (0, 0) and (0, 2): On this line segment we have x = 0. So given function simplifies to a function of the single variable,

f (0, y ) = −3 y 2 + 6 y Say f (0, y ) = u ( y ) then

0≤ y≤2

u ( y ) = −3 y 2 + 6 y u′( y ) = −6 y + 6

0≤ y≤2

For critical point u′( y ) = −6 y + 6 = 0 − y +1 = 0 y =1 which corresponds to a point (0, 1). (As x = 0 on this line segment) The extreme values of u(y) may occur at the endpoints, that is, y = 0 and y = 2 which corresponds to the points (0, 0) and (0, 2) respectively. ii. Line segment between (0, 2) and (2, 2): On this line segment we have y = 2. So given function simplifies to a function of the single variable,

f ( x, 2) = x 2 − 12 − 2 x + 12

0≤ x≤2

= x − 2x Say f ( x, 2) = v( x) then 2

v( x) = x 2 − 2 x v′( x) = 2 x − 2

0≤ x≤2

For critical point v′( x) = 2 x − 2 = 0 x −1 = 0 x =1 which corresponds to a point (1, 2). (As y = 2 on this line segment) The extreme values of v(x) may occur at the endpoints, that is, x = 0 and x = 2 which corresponds to the points (0, 2) and (2, 2) respectively. iii. Line segment between (2, 2) and (2, 0): On this line segment we have x = 2. So given function simplifies to a function of the single variable,

f (2, y ) = 4 − 3 y 2 − 4 + 6 y

0≤ y≤2

= −3 y + 6 y Say f (2, y ) = t ( y ) then 2

t ( y) = − 3 y 2 + 6 y t ′( y ) = −6 y + 6

0≤ y≤2

For critical point

t ′( y ) = −6 y + 6 = 0 − y +1 = 0 y =1 which corresponds to a point (2, 1). (As x = 2 on this line segment) The extreme values of t(y) may occur at the endpoints, that is, y = 0 and y = 2 which corresponds to the points (2, 0) and (2, 2) respectively. iv. Line segment between (2, 0) and (0, 0): On this line segment we have y = 0. So given function simplifies to a function of the single variable, f ( x, 0) = x 2 − 2 x

0≤ x≤2

Say f ( x, 0) = w( x) then w( x) = x 2 − 2 x w′( x) = 2 x − 2 For critical point

0≤ x≤2

w′( x) = 2 x − 2 = 0 x −1 = 0 x =1

which corresponds to a point (1, 0). (As y = 0 on this line segment) The extreme values of w(x) may occur at the endpoints, that is, x = 0 and x = 2 which corresponds to the points (0, 0) and (2, 0) respectively. STEP 3: (x, y)

(0,1)

(0, 0)

(0, 2)

(2, 2)

(2, 0)

(1, 0)

(1, 2)

(2, 1)

f(x, y)

3

0

0

0

0

-1

-1

3

Thus Absolute maximum value of f is f (0, 1) = f (2, 1) = 3 (highest value among others) Absolute minimum value of f is f (1, 0) = f (1, 2) = -1 (lowest value among others)

Remarks: My advice to all the students is to first draw the figure of region R according to the given points (or equations). Seeing visually makes the solution easier. Also, revise Article 4 of first chapter in the book ‘Calculus with Analytic Geometry’ by Howard Anton, for equations of line.

Q 2.

Prove that 1

∫ ∫

2

−1 0

( x 2 − 2 y )dx dy =

Solution:

2

1

0

−1

∫∫

( x 2 − 2 y )dy dx

2

1

∫ ∫

2

−1 0

( x − 2 y )dx dy = 2

∫

x3 − 2 xy dy 3 0

1

−1

8 − 4 y dy 3

1

=∫

−1

1

8 = y − 2 y2 3 −1 8 8 = −2+ +2 3 3 16 = 3 2

∫∫ 0

1

−1

2

∫ =∫ =∫

( x 2 − 2 y )dy dx =

0 2

0

2

0

x2 y − y2

1 −1

dx

x 2 − 1 − (− x 2 − 1) dx 2 x 2 dx

2 3 = x 3 0

2

2 = (2)3 3 16 = 3 Thus Q 3.

1

∫ ∫

2

−1 0

( x 2 − 2 y )dx dy =

Evaluate the iterated integral.

∫ ∫ (4x 1

2y

0 0

Solution:

)

y + y dx dy

2

1

0

−1

∫∫

( x 2 − 2 y )dy dx =

16 3

∫∫ ( 1

2y

0 0

)

1

4 x y + y dx dy = ∫ 2 x 2 y + xy 0

(8 y

1

=∫ 0

2

2y

dy 0

)

y + 2 y 2 dy

1  5  = ∫  8 y 2 + 2 y 2  dy  0  1

=8

7 2

y y3 +2 7 3 2

= 16

0 1

7 2

3

y y +2 7 3

0

=

Q 4.

16 2 + 7 3

=

48 + 14 62 = 21 21

Evaluate the iterated integral by first changing the order of integration. 1 1

∫∫ 0

y

2

e x dx dy

(Hint: Yellow colored region in following figure is the region of integration)

Solution: First observe that the leftmost integral has limits 0 to 1 while at rightmost we have dy. So this means that variable y varies from zero to one. Next integral has limit y to 1 and dx is before dy, so this means that variable x varies from y to 1. In short we can say that for each fixed y on the interval [0, 1], x ranges from y to 1. You can also see that in the yellow-colored region of above figure. Now, to reverse the order of integration, see in the yellow-colored region of the figure that x varies from 0 to 1. To express y in terms of x we can say that y varies from 0 to x. So, this region is also said to be enclosed by 0 ≤ x ≤1

y=0, y=x,

Thus we can write, 1 1

∫∫ 0

y

e x dx dy = ∫ 2

1

∫

x

0 0

2

e x dy dx

1

= ∫ ye x 0

( xe

1

=∫ 0

x

2

x2

dx 0

1

= ∫ xe x dx 2

0 1

2 1 = ∫ 2 xe x dx 20

1 x2 1 e 0 2 1 = ( e1 − e0 ) 2 1 = ( e − 1) 2 =

)

− (0)e x dx 2

Solution Assignment No.4 Q 1.

Use double integral in rectangular co-ordinates to compute area of the

region inside the circle x 2 + y 2 = 4 and above x-axis. Confirm your result by first converting double integral from rectangular to polar co-ordinates and then evaluating it. a  π a2  2 2 Hint: a − x dx =   ∫ 2  −a  Solution: The given region is as follows

2

4 − x2

−2

0

∫ ∫

dy dx

2

=

∫

−2 2

=

∫

−2 2

=

∫

−2

y0

4− x2

dx

)

(

4 − x 2 − 0 dx

(

(2) 2 − x 2 dx

)

According to the HINT of this question, here a = 2. Thus 2 π (2)2 2 2 (2) − x dx = = 2π ∫ 2 −2 Now let us confirm our result by changing it to polar integral.

(

)

π

2

0

0

r dr dθ

∫ ∫ π

r2 2

=∫ 0

π

=∫

2

dθ 0

( 2 ) dθ

0

=2θ

Q 2.

π 0

= 2π

Find all possible polar co-ordinates representations of the point P whose

rectangular co-ordinates are ( 2, − 2 ) . (Hint: Consider negative value of r also) Solution:

As we know r = x 2 + y 2 and

tan θ =

y x

So r = (2) 2 + (−2) 2 = tan θ =

−2 = −1 2

4+4 = 8= 2 2

−π 7π = 4 4 −π 7π     + 2nπ  ,  2 2 , + 2nπ  , 2 2 , 4 4     3π −5π     + 2nπ  ,  −2 2 , + 2nπ   −2 2 , 4 4     n = 0, 1, 2, − − − − − − −

θ = tan −1 (−1) =

Q 3.

What is the geometrical representation of the following equations?

Here ( r , θ ) are polar co-ordinates of an arbitrary point.

a. r sin θ = c , where and c is an arbitrary constant. Solution:

It’s a line parallel to x-axis that meets y-axis at point (0, c) b. θ = α where α is an arbitrary constant angle. Solution:

It’s a straight line passing through origin making an angle α with x-axis or initial line or polar axis. c. r = c where c is an arbitrary constant. Solution:

It’s a circle center at origin and radius c d. r = 2a cos θ where a is arbitrary constant. Solution:

It’s a circle with center on x-axis, passing through origin and lie at right side of yaxis. Its center in polar co-ordintes is (a, 00 ) and in rectangular co-ordinates is

(a, 0) . Its radius is a. e. r = a(1 − sin θ ) where a is arbitrary constant Solution:

It’s a Cardioid, symmetric about y-axis

f.

r 2 = 4 cos 2θ

Solution:

It’s a lemniscate, symmetric about x-axis and y-axis

Q 4.

Use double integral to find the area enclosed by four-petaled rose

Solution:

1

0.5

-1

-0.5

0.5

1

-0.5

-1

Graph of r = cos 2θ is symmetric about x-axis and y-axis. Let us find the area enclosed by one of the petal and multiply it by 4 to get the total area enclosed by the four-petaled rose. To find the upper and lower limit of θ , put r = 0 r = cos 2θ 0 = cos 2θ cos −1 0 = 2θ 2θ =

θ=

π

π

or

2

−π 2 −π θ= 4

2θ =

or

4

So area A enclosed by one of the petal is π

A=

4

cos 2θ

−π 4

0

∫ ∫

π

=

4

∫

−π 4

π

=

4

∫

−π 4

r2 2

r dr dθ

cos 2θ

dθ 0

cos 2 2θ dθ 2

As we know cos 2 2θ =

1 + cos 4θ , so putting this value 2

π

1 4 1 + cos 4θ = ∫ dθ 2 −π 2 4

π

π

4

1 1 4 = ∫ dθ + ∫ cos 4θ dθ 4 −π 4 −π 4

=

1 θ 4

4

π −π 4

4

+

1 sin 4θ 4 2

π 4 −π 4

1π π  1 =  +  + ( sin π − sin(−π ) ) 4 4 4  8 1π  =  +0 4 2  =

π 8

Thus area enclosed by four-petaled rose = 4 ×

π 8

=

π 2

Solution Assignment # 5 Note: Text in blue color is the information related to the given question and text in black color gives the solution of the question. Q 1.

Let C be the curve with parametric equations

x = t , y = t2 , z = t3

; t ≥0

Find parametric equations for the tangent line to C at the point corresponding to t = 2.

Solution: DEFINITION: →

→

If a vector-valued function r (t ) has a tangent vector r ′(t0 ) at any point t0 on its →

graph, then the line parallel to r ′(t0 ) and passing through the tip of the radius →

→

→

vector r (t0 ) is called the tangent line of the graph of r (t ) at r (t0 ) . Vector equation of the tangent line at t0 is →

→

→

r = r (t0 ) + t r ′(t0 ) − − − − − − − − − −( A)

We will use the above definition for the solution of this question. →

The vector equation r (t ) of the given curve C is →

^

^

^

^

^

^

r (t ) = x i + y j + z k = t i + t 2 j + t 3 k

for t ≥ 0

The tangent vector to C at the point corresponding to t is →

^

^

^

r ′(t ) = i + 2t j + 3t 2 k

In particular, radius vector of the point corresponding to t = 2 is →

^

^

^

r (2) = 2 i + 4 j + 8 k

And a tangent vector at the point whose radius vector is as above is →

^

^

^

r ′(2) = i + 4 j + 12 k

For vector equation of tangent line at the point where t = 2, put values in Equation (A) →

^

^

^

→

→

r = x i + y j + z k = r (2) + t r ′(2)

^ ^ ^ ^ ^ ^ ^  ^  ^  x i + y j + z k =  2 i + 4 j + 8 k  + t  i + 4 j + 12 k      ^

^

^

^

^

^

= 2 i + 4 j + 8 k + t i + 4t j + 12t k ^

^

^

= ( 2 + t ) i + ( 4 + 4t ) j + ( 8 + 12t ) k

Equating the corresponding components gives the parametric equation of tangent line at the point where t = 2 x = 2 + t , y = 4 + 4t , z = 8 + 12t

→

Q 2.

^

^

Find parametric equations for the curve r (t ) = (3t − 2) i + (4t + 3) j using arc

length s as a parameter. Use the point on the curve where t = 0 as the reference point.

Solution: →

^

^

^

^

Since r (t ) = x i + y j = (3t − 2) i + (4t + 3) j So in parametric form we can write x = 3t − 2   − − − − − − − − − −(1) y = 4t + 3 Now rewrite these parametric equations with u in place of t x = 3u − 2 y = 4u + 3

dx =3 du dy =4 du

Theorem: Let C be a curve in 2D-space given parametrically by x = x(t) , y = y (t) /

/

where x (t) and y (t) are continuous functions. If an arc-length parameter s is introduced with its reference point at (x(t0), y (t0)), then the parameters s and t are related by 2

t

2

 dx   dy    +   du − − − − − − − − − − − − − − − ( A)  du   du 

s=∫ t0

Use this theorem for further procedure. Since it is stated in the question that use the point on the curve where t = 0 as the reference point, so here t0 = 0 . Put values in equation (A) 2

t

2

 dx   dy    +   du  du   du 

s=∫ t0 t

( 3) + ( 4 )

s =∫

2

2

du

0 t

=∫

9 + 16 du

0 t

=

∫

5 du

0

=5u0

t

= 5 (t − 0) = 5t So, s = 5t ⇒ t =

s 5

Substituting the value of t in the parametric equations (1).

s x = 3  − 2 5 s y = 4  + 3 5

Or 3 x = s−2 5 4 y = s +3 5

Q 3.

Determine whether the following differential is exact? If so, find z by

integration of exact differentials.

dz = ( y sec2 x + sec x tan x) dx + (tan x + 2 y ) dy Solution: As we know that if z = f(x, y) then dz =

∂z ∂z dx + dy ∂x ∂y

Let ∂z ∂x ∂z Q= ∂y P=

   − − − − − − − − − − − −( A)  

Then above differential can be written as

dz = P dx + Q dy

Test for exact differential: Any differential

dz = P dx + Q dy − − − − − − − −( B) where P and Q are functions of x and y, is an exact differential if

∂P ∂Q = ∂y ∂x

Equate the given differential dz = ( y sec2 x + sec x tan x) dx + (tan x + 2 y ) dy with equation (B) we get P = y sec2 x + sec x tan x and Q = tan x + 2 y ∂P = sec 2 x ∂y ∂Q = sec2 x ∂x Since

∂P ∂Q = , so the given differential is exact. ∂y ∂x

Integration of Exact differential to determine z: For an exact differential dz = P dx + Q dy where

∂z ∂x ∂z Q= ∂y P=

   − − − − − − − − − − − − As in equation ( A)  

z = ∫ P dx + f ( y ) − − − − − − − − − − − − − (i ) where f ( y ) is an arbitrary function of y only and is same as cons tan t of int egration in a normal int egration and also z = ∫ Q dy + f ( x) − − − − − − − − − − − − − − − − − −(ii ) where f ( x) is an arbitrary function of x only and is same as cons tan t of int egration in a normal int egration

Since expressions (i) and (ii) has same left hand sides so their right hand sides are also equal. Compare both expressions and find the value of f(y) and f(x), and show z.

Since it is proved that the given differential is exact, so we will do the above stated procedure to determine z. P = y sec2 x + sec x tan x and Q = tan x + 2 y z = ∫ P dx = ∫

( y sec

2

x + sec x tan x ) dx

= y tan x + sec x + f ( y ) − − − − − − − (1) z = ∫ Q dy = ∫

( tan x + 2 y ) dy

= y tan x + y 2 + f ( x) − − − − − − − − − − − −(2)

Compare right hand side of both the expressions (1) and (2) we get

f ( y) = y 2 and f ( x) = sec x Thus z = y tan x + sec x + y 2

Q 4.

Evaluate

∫

xy dx + x 2 dy if

C

i. C consists of line segments from (2, 1) to (4, 1) and from (4, 1) to (4, 5) . ii. C is the line segment from (2, 1) to (4, 5) .

Solution: (a)

Let us sub-divide C into two parts c1 (line segment joining the points (2, 1) to (4, 1) ) and c2 (line segment joining the points (2, 1) to (4, 5) ) , as shown

in the figure.

Line integral along C may be expressed as a sum of two line integrals, the first along c1 and the second along c2. That is,

∫ C

xy dx + x 2 dy = ∫ xy dx + x 2 dy + ∫ xy dx + x 2 dy − − − − − − − ( A) c1

c2

c1 is parallel to line y = 1. So dy = 0. To find the limits of integral we can see that along c1 , x varies from 2 to 4. Put values in given line integral

∫ c1

4

xy dx + x 2 dy = ∫ x (1) dx + x 2 (0) 2 4

= ∫ x dx 2

x2 = 2

4

= 8 − 2 = 6 − − − − − − − − − − − (1) 2

c2 is parallel to line x = 4. So dx = 0. To find the limits of integral we can see that along c2 , y varies from 1 to 5. Put values in given line integral

∫ c2

5

xy dx + x 2 dy = ∫ xy (0) + (4)2 dy 1 5

= 16 ∫ dy 1

= 16 y 1

5

=16 (5 − 1) = 16(4) = 64 − − − − − − − − − (2) Put values from equations (1) and (2) in (A)

∫ C

xy dx + x 2 dy = ∫ xy dx + x 2 dy + ∫ xy dx + x 2 dy − − − − − − − ( A) c1

= 6 + 64 = 70

Thus line integral along C is 70.

c2

(b) Here, since C is the line segment from (2, 1) to (4, 5) . So first we will find the equation of C. Let ( x1 , y1 ) = (2, 1) and

( x2 , y2 ) = (4, 5) then Slope = m = m=

y2 − y1 x2 − x1

5 −1 4 = =2 4−2 2

Use Point-Slope form of the line, to find equation of C.

y − y1 = m( x − x1 ) y − 1 = 2( x − 2) y = 2x − 4 +1 = 2x − 3 Now

dy = 2 dx As we are converting given line integral to one variable x, so for limits of integral we see how x varies along C, which is from 2 to 4. Put values in given line integral

∫ C

4

xy dx + x 2 dy =

∫

x(2 x − 3) dx + x 2 ( 2 dx )

2 4

=∫

( 2x

2

− 3x + 2 x 2 ) dx

2

− 3 x ) dx

2 4

=∫

( 4x

2

x3 x2 = 4 −3 3 2

4

2

 256 48   32  = −  −  − 6 2   3  3  512 − 144 − 64 + 36 = 6 340 170 = = 6 3

Attention please: In the above question we get different values of the integral while considering two separate paths joining the same two end points. Can anyone of you give me the reason? Send me through mail.

Solution Assignment No: 6 Show that the line integral ∫ ( y 2 + 2 xy ) dx + ( x 2 + 2 xy ) dy is independent

Q 1.

C

of path. Then find its value if C is the curve joining the points (−1, 2) to (3, 1) .

Solution: THEOREM: If P(x, y) and Q(x, y) have continuous first partial derivatives on a simply connected region D, then the line integral

∫ P( x, y) dx + Q( x, y) dy C

is independent of path in D if and only if ∂P ∂Q = ∂y ∂x

(That is, integrand is exact differential.)

Use this theorem to show that the given line integral is independent of path. Given line integral is

∫(y

2

+ 2 xy ) dx + ( x 2 + 2 xy ) dy

C

Here P( x, y ) = y 2 + 2 xy Q( x, y ) = x 2 + 2 xy ∂P = 2 y + 2x ∂y ∂Q = 2x + 2 y ∂x Since

∂P ∂Q , that is, integrand is exact differential, so given line integral is = ∂y ∂x

independent of path.

Now, let dz = ( y 2 + 2 xy ) dx + ( x 2 + 2 xy ) dy Find z from the exact differential dz (As we did in last assignment also).

P( x, y ) = y 2 + 2 xy and Q( x, y ) = x 2 + 2 xy

z = ∫ P dx = ∫

(y

2

+ 2 xy ) dx

= xy 2 + x 2 y + f ( y ) − − − − − − − (1)

z = ∫ Q dy = ∫

(x

2

+ 2 xy ) dy

= x 2 y + xy 2 + f ( x) − − − − − − − − − − − −(2)

As there is no dissimilar term at the right side of equations (1) and (2) so f ( y) = 0 and f ( x) = 0 Thus z = xy 2 + x 2 y So for dz = ( y 2 + 2 xy ) dx + ( x 2 + 2 xy ) dy , we have z = xy 2 + x 2 y Or d ( xy 2 + x 2 y ) = ( y 2 + 2 xy ) dx + ( x 2 + 2 xy ) dy Put this value in given integral, with limits (−1, 2) to (3, 1) (3, 1) 2 2 ∫ ( y + 2 xy) dx + ( x + 2 xy) dy = C

∫

d ( xy 2 + x 2 y )

( −1, 2)

= xy 2 + x 2 y

(3, 1) ( −1, 2)

= 3(1) + (3) 2 (1)  − (−1)(2) 2 + (−1) 2 (2)  2

= 3+9+ 4− 2 = 14

Q 2.

Use Green’s Theorem to evaluate the integral


∫

ln(1 + y ) dx −

C

xy dy 1+ y

where C is the triangle with vertices (0, 0) , (2, 0) and (0, 4) . Solution:


∫

ln(1 + y ) dx −

C

xy dy 1+ y

If we compare the given integral with this:


∫

P dx + Q dy

C

We get P = ln(1 + y ) xy Q=− 1+ y

∂P 1 = ∂y 1 + y ∂Q − y = ∂x 1 + y By Green’s Theorem


∫ C

 ∂P ∂Q  P dx + Q dy = − ∫ ∫  −  dy dx  ∂y ∂x  R  ∂Q ∂P  = ∫∫  −  dy dx  ∂x ∂y  R

So


∫ C

ln(1 + y ) dx −

 −y xy 1  dy = ∫ ∫  −  dy dx 1+ y  1+ y 1+ y  R  − y −1  = ∫∫   dy dx 1 + y   R  −( y + 1)  = ∫∫   dy dx  1+ y  R = ∫∫

( −1) dy dx

R

Now there are three ways to proceed further. Each will give the same answer.

First method: For the limits of region R, we use the previously discussed concept “Double integration over a nonrectangular region”. First draw the figure of R, using given vertices.

Find the equation of line joining the points (2, 0) and (0, 4) . Let ( x1 , y1 ) = ( 2, 0 ) and ( x2 , y2 ) = ( 0, 4 )

Then Slope = m =

y2 − y1 4 − 0 = = −2 x2 − x1 0 − 2

Use Point-Slope form of the line, to find equation

y − y1 = m( x − x1 ) y − 0 = −2( x − 2) y = −2 x + 4 I)

For integration with respect to y first and then x, proceed as follows:

From the figure we can easily say that

x varies from 0 to 2 and y varies from 0 to −2 x + 4


∫ C

ln(1 + y ) dx −

2  −y xy 1  dy = ∫ ∫  − dy dx =  ∫0 1+ y  1+ y 1+ y  R

−2 x + 4

∫ ( −1) dy dx 0

2

= ∫ −y 0

−2 x + 4

dx

0 2

=∫

( 2 x − 4 ) dx

0

= x2 − 4 x

2 0

= 4 −8 = − 4

II)

For integration with respect to x first and then y, proceed as follows:

From the figure we can easily say that

y varies from 0 to 4 and

x varies from 0 to 2 −


∫ C

y 2

4  −y xy 1  ln(1 + y ) dx − dy = ∫ ∫  − dy dx =  ∫0 1+ y  1+ y 1+ y  R 4

2−

y 2

∫ ( −1) dx dy 0

= ∫ −x 0

2−

y 2

dy

0 4

y  = ∫  −2 +  dy 2 0  y2 = −2 y + 4

4

0

= −8 + 4 = −4

Second Method: As we studied in lecture 22,

∫∫

dy dx gives the area of the region R. Here R is a

R

triangle and Area of a triangle =

( Base)(Vertical height ) 2

Here, base = 2, and Vertical Height = 4 So Area of a triangle =

(2)(4) =4 2

Thus


∫

ln(1 + y ) dx −

C

→

Q 3.

xy dy = (−1) ∫ ∫ dy dx = (−1) ( Area of triangle ) = −4 1+ y R ^

^

Let F = 3 xz i + 2 y j and the surface S

is the upper half of the sphere x 2 + y 2 + z 2 = 4

(as shown in the figure). Find

∫

→

^

^

F i n dS if n is a unit upper normal to S.

S

Solution: First, I would like to share with you that given surface can also be stated as: “Surface S is defined by x 2 + y 2 + z 2 = 4 which is bounded by z = 0” x 2 + y 2 + z 2 = 4 is equation of a sphere center at origin and radius 2. z = 0 is the equation of xy-plane (As we studied in lecture 3).

Lets move toward the solution of this question. Here S : x 2 + y 2 + z 2 − 4 = 0 The normal vector, often simply called the "normal" to a surface is a vector perpendicular to it.

Given a three-dimensional surface →

^

n=

∇S

→

where ∇S denotes the gradient calculated by following formula.

→

∇S →

∇S =

∂S ^ ∂S ^ ∂S ^ i+ j+ k ∂x ∂y ∂z ^

Find value of n . →

∇S =

∂ ( x2 + y 2 + z 2 − 4) ∂x ^

^

^

= 2x i + 2 y j+ 2z k

^

i+

∂ ( x2 + y 2 + z 2 − 4) ∂y

^

j+

∂ ( x2 + y 2 + z 2 − 4) ∂z

^

k

→

∇S = (2 x) 2 + (2 y ) 2 + (2 z )2 = 4( x 2 + y 2 + z 2 ) As x 2 + y 2 + z 2 = 4. So →

∇S = 16 = 4

→

^

n=

^

∇S

^

^

2x i + 2 y j+ 2z k x ^ y ^ z ^ = = i + j+ k 4 2 2 2

→

∇S

As it is given that →

^

^

F = 3 xz i + 2 y j →

^

Put these values of F and n in given integral

∫ S

→ ^ ^ ^ x ^ y ^ z ^ F i n dS = ∫ (3 xz i + 2 y j ) i  i + j + k  dS 2 2  2 S

3  = ∫  x 2 z + y 2  dS 2  S  Before integrating over the given surface, we convert it to spherical polar coordinates. x = 2sin φ cos θ y = 2sin φ sin θ z = 2 cos φ Limits are:

φ ranges from 0 to π θ ranges from 0 to 2π For selecting proper limits for φ and θ , please see the following link http://www.math.montana.edu/frankw/ccp/multiworld/multipleIVP/spherical/bod y.htm dS is the element of surface area of given surface.

And

∫

dS gives the total surface area of the given surface.

S

Now, surface area of a sphere in spherical co-ordinates is 2π

π

0

0

∫ ∫

r 2 sin φ dφ dθ

But in this question, the surface is upper half of the sphere so we change the limits of θ and φ accordingly. Put these values

∫ S

→ ^ 3  F i n dS = ∫  x 2 z + y 2  dS 2  S 

=

π

2π

∫ ∫ 0

=

0

π

2π

∫ ∫ ( 48sin 2

0

=

3 2 2  (2sin φ cos θ ) (2 cos φ ) + (2sin φ sin θ )  4sin φ dφ dθ 2 

2

3

φ cos 2 θ cos φ + 16sin 3 φ sin 2 θ ) dφ dθ

0

π

2π

∫ ∫ ( 48sin 2

0

3

φ cos θ cos φ ) dφ dθ + 2

0 2π

= 48 ∫ 0

2π

∫ ∫ (16sin 0

π 2

2π

sin n x dx =

0

0

0

π

2 ∫ ( sin φ ) dφ = 3 2

3

0

So

2

0

U sin g this formula, , here n = 3

φ sin 2 θ ) dφ dθ

π

∫

2. 4. 6. 8 − − − − − −(n − 5)(n − 3)(n − 1) 1. 3. 5. 7 − − − − − −(n − 4)(n − 2)n

3

0

∫

By Wallis sine formula, when n is odd

∫

2

cos 2 θ sin 3 φ cos φ dφ dθ + 16 ∫

π 2

π

sin 2 θ sin 3 φ dφ dθ

∫

→

2π

^

sin 4 φ cos θ 4

F i n dS = 48 ∫

S

π

0

2π

2 dθ + 16 ∫ sin 2 θ   dθ 3 0

2

2

0

2π

32 1 = 48 ∫ cos 2 θ   dθ + 3 4 0 2π

= 12 ∫ cos 2 θ dθ + 0 2π

0

=

∫

6(1 + cos 2θ ) dθ +

∫ 0

=

= =

34 3

sin 2 θ dθ

0

sin 2 θ dθ

∫

2π

∫ 0

1 − cos 2θ dθ 2

16 (1 − cos 2θ ) dθ 3

16 16    6 + 6 cos 2θ + − cos 2θ  dθ 3 3  

∫ 0

=

2π

0

2π

∫

0

1 + cos 2θ 32 dθ + 2 3

= 12 ∫ 2π

32 3

2π

2π

2π

∫ 0

34 θ 3 68π 3

dθ + 2π 0

+

2 3

2π

∫

cos 2θ dθ

0

2 sin 2θ 3 2

2π

0

Solution Assignment # 7 Let V be the region bounded by the surface x 2 + y 2 = 4 and the planes

Q 1.

z = 0 and z = 3 in three dimensional space and let S denote the surface →

^

^

^

of V . If F = x 3 i + y 3 j + z 3 k , use Divergence Theorem to find

→

^

∫ F • n dS S

^

where n denote the unit outer normal to S. Solution: Divergence Theorem or Gauss’s Theorem states that Let V be a region in three dimensional space which is bounded by a closed surface S (In three dimensional space such region V is a volume of the solid ^

whose surface is S). Also suppose that n denote the unit outer normal to S at any →

point (x, y, z). If F is a vector function that has continuous partial derivatives on →

V then the volume integral of the divergence of F

i-e

→    ∇ • F  over V and the  

→

surface integral of F over the surface S of V are related by →

^

∫ F • n dS = ∫ S

→

∇ • F dV

V

Or →

→

^

∫ F • n dS = ∫ div F S

dV

V

Now come up to the solution of this question. According to the given statement, V is a cylinder of radius 2 and height 3 (as shown in figure)

→

^

^

^

F = x3 i + y 3 j + z 3 k →  ∂ ^ ∂ ^ ∂ ^  3^ 3 ^ 3 ^ ∇ • F =  i+ j+ k  •  x i+ y j+ z k  ∂z     ∂x ∂y 3 3 3 ∂x ∂y ∂z = + + ∂x ∂y ∂z

= 3x 2 + 3 y 2 + 3 z 2 = 3( x 2 + y 2 + z 2 )

So by Divergence Theorem →

→

^

∫ F • n dS = ∫ ∇ • F dV S

V

= 3∫ ( x 2 + y 2 + z 2 ) dV V

Use cylindrical co-ordinates to evaluate the volume integral at right hand side.

x = r cos θ y = r sin θ z=z x 2 + y 2 + z 2 = r 2 cos 2 θ + r 2 sin 2 θ + z 2 = r2 + z2 dV = r dz dr dθ

Limits for z is 0 to 3 r is 0 to 2

θ is 0 to 2π To have better understanding of cylindrical co-ordinates, see the following link.

http://www.math.montana.edu/frankw/ccp/multiworld/multipleIVP/cylindrical/bo dy.htm

So, putting values

→

→

^

2 2 2 ∫ F • n dS = ∫ ∇ • F dV = 3∫ ( x + y + z ) dV S

V

V 2π

2

= 3∫

3

(r 2 cos 2 θ + r 2 sin 2 θ + z 2 ) r dz dr dθ

∫ ∫

0

0

0

2π

2

3

∫ ∫ (r

= 3∫ 0

0

0

2π

2

3

0

0

0

2π

2

= 3∫

∫

0

0

2π

2

rz 3 r z+ 3

0

3

dr dθ

3

∫ ( 3r

= 3∫

(cos 2 θ + sin 2 θ ) + z 2 ) r dz dr dθ

(r 3 + rz 2 ) dz dr dθ

∫ ∫

= 3∫

2

3

0

+ 9r ) dr dθ

0

2π

3r 4 9r 2 + 4 2

= 3∫ 0

2π

2

dθ 0

dθ

= 90 ∫ 0

= 90 θ

2π 0

= 180π

Thus →

^

∫ F • n dS = 180π S

Q 2.

Determine the Fourier Series of the periodic function f ( x) shown in the figure.

Solution:

First we will find how the periodic function in the figure be defined. π f ( x) =  x f ( x) = f ( x + 2π )

−π ≤ x ≤ 0

0≤ x ≤π

As we know, Fourier Series is of the form a0 ∞ + ∑ {an cos nx + bn sin nx} 2 n =1 where n is a positive int eger f ( x) =

Before doing calculation for Fourier co-efficients, its better to find whether the given function is even or odd. As then, you know we can apply some known results and reduce our work. If the function is odd, all the Fourier co-effiients an for n = 0, 1, 2… are zero. If the function is even, all the Fourier co-efficients bn for n = 0, 1, 2… are zero.

Here, even by looking at the figure we can easily say that the given function is neither even nor odd as it is not symmetric about y-axis or origin.

Fourier co-efficients a0 =

1

π

π

∫

f ( x) dx

−π

Put values

a0 =

1

π

π

∫

−π

0 π  1  ∫ f ( x) dx + ∫ f ( x) dx  π  −π 0  0 π  1 =  ∫ π dx + ∫ x dx  π  −π 0 

f ( x) dx =

=

1

π

π x −π

=π + an = an =

1

π 1

π

π 2

=

0

1 x2 + π 2

π

0

3π 2

π

∫

f ( x) cos nx dx

−π

π

∫

−π

π 0  1 f ( x) cos nx dx =  ∫ f ( x) cos nx dx + ∫ f ( x) cos nx dx  π  −π 0 

0 π  1 =  ∫ π cos nx dx + ∫ x cos nx dx  π  −π 0 

sin nx 1 sin nx ( − cos nx ) = π + x − n −π π n n2 π 0 1

0

π

 (−1) n 1  (0 − 0) +  0 + 2 − 0 + 2  n n  π  n (−1) + 1 = π n2

=

1

If n is odd (−1) n + 1 =0 π n2 and if n is even an =

(−1) n + 1 2 = 2 2 πn πn Or we can say that an =

an =

2 if n is any positive int eger π (2n)2

bn =

1

bn =

π 1

π

π

∫

f ( x)sin nx dx

−π

π

∫

−π

0 π  1 f x nx dx + f ( x) sin nx dx  ( ) sin ∫ ∫ π  −π  0 0 π  1 =  ∫ π sin nx dx + ∫ x sin nx dx  π  −π 0 

f ( x)sin nx dx =

(− cos nx) 1 (− cos nx) ( − sin nx ) = π + x − n n n2 π π −π 0 1

0

 1  −π −π (−1) n  1  π (−1) n +1 =  − − 0 + 0 − 0 +  n n π n  π  1  −π + π (−1) n  1  π (−1) n +1   +   n n π  π  −1 + (−1)n (−1) n+1 −1 + (−1) n + (−1) n (−1) = + = n n n n n −1 + (−1) − (−1) −1 = = n n =

Hence

π

a0 ∞ + ∑ {an cos nx + bn sin nx} 2 n =1 ∞ ∞ a f ( x) = 0 + ∑ an cos nx + ∑ bn sin nx 2 n =1 n =1

f ( x) =

f ( x) =

3π ∞ 2 cos 2nx ∞ sin nx +∑ −∑ 4 n =1 π ( 2n )2 n=1 n

=

3π 2 ∞ cos 2nx ∞ sin nx + ∑ −∑ 4 π n =1 ( 2n ) 2 n n =1