Math 2310 Solution 2

MATH2310

1. Change of variables in double integrals 1.1. QUESTION. Calculate Z Z

1 dA R 4x − y where R is the region enclosed by the parallelogram with sides given by the lines 3x + y = −3,

3x + y = 4,

x − 2y = 6 and x − 2y = 36;

that is, R = {(x, y) ∈ R2 | − 3 ≤ 3x + y ≤ 4 and 6 ≤ x − 2y ≤ 36}.
1.2. ANSWER. 71 40 ln(40) + 3 ln(3) − 33 ln(33) − 10 ln(10) . 1.3. SOLUTION. We will solve this integral using a change of variables (x, y) → (u, v) (which amounts to a change in the coordinate system of R2 ) so that the bounds of integration become “uncoupled”; that is, so the rotated parallelogram region R in the xy coordinate system becomes an upright rectangular region S in the uv coordinate system. We change the coordinate system to simplify the region R by making the definitions u = 3x + y and v = x − 2y. Under this transformation of coordinate systems, the region R becomes S = {(u, v) ∈ R2 | − 3 ≤ u ≤ 4 and 6 ≤ v ≤ 36}. RR R 4 R 36 The integral . . . dA will then look like −3 6 . . . dvdu. R 1 , we will have In fact, defining f (x, y) = 4x−y Z Z Z 4 Z 36 ∂(x, y) dvdu, (1.1) f (x, y)dA = f (u(x, y), v(x, y)) ∂(u, v) R −3 6

where

∂(x,y) ∂(u,v)

is the Jacobian associated to the change in coordinate system, defined by !       ∂x ∂x ∂y ∂x ∂y ∂(x, y) ∂x ∂v = det ∂u = − . ∂y ∂y ∂(u, v) ∂u ∂v ∂v ∂u ∂u ∂v

To calculate the Jacobian, we first express x and y as functions of u and v. (At the moment we have u and v expressed as functions of x and y.) Recall that we have (1.2)

u = 3x + y

and (1.3)

v = x − 2y.

We aim to combine these two equations in two ways, each time eliminating one of x or y: To remove the x variable, we take the combination u − 3v: u − 3v = (3x + y) − 3(x − 2y) = 3x + y − 3x + 6y = 7y; solving for y gives 1 y = (u − 3v). 7

(1.4)

To eliminate the y variable, we take the combination 2u + v: 2u + v = 2(3x + y) + (x − 2y) = 7x, so 1 x = (2u + v). 7

(1.5)

(Alternatively, once we obtained the expression (1.4) for y, we could have substituted it into either of (1.2) or (1.3) to obtain (1.5).) Now we calculate the partial derivatives involved in the Jacobian: 2 ∂x = , ∂u 7 ∂y 1 = , ∂u 7

∂x 1 = , ∂v 7 ∂y 3 =− , ∂v 7

so ∂(x, y) = det ∂(u, v)

∂x ∂u ∂y ∂u

∂x ∂v ∂y ∂v

!

      2 3 1 1 = − − 7 7 7 7 6 1 − 49 49 1 =− . 7 =−

The last thing we need to do before calculating the integral is change f from a function of x and y to a function of u and v. We use the equations (1.4) and (1.5): 1 = 4x − y 4 =

1 (2u 7

1 . u+v

1
+ v) −

1 (u 7


− 3v)

Finally, we can calculate the integral. Using (1.1), we have: Z 4 Z 36 Z Z 1 1 1 dA = − dvdu u + v 7 −3 6 R 4x − y Z Z 1 4 36 1 dvdu = 7 −3 6 u + v Z 1 4 = [ln(u + v)]v=36 v=6 du 7 −3 Z 1 4 = ln(u + 36) − ln(u + 6)du 7 −3  Z 4 Z 4 1 (1.6) ln(u + 6)du ln(u + 36)du − = 7 −3 −3 At this point we have two integrals that look like

R4 −3

ln(u + a)du, so we’ll solve this

one and then substitute a = 36 and a = 6. We use integration by parts: defining R4 R u=4 functions s = s(u) and t = t(u), we want to express −3 ln(u + a)du as u=−3 s dt and use the relationship: Z

Z s dt = st −

t ds.

1 Defining s = ln(u + a) and dt = du, gives ds = u+a du and t = u. Therefore Z 4 Z 4  4 u du ln(u + a)du = u ln(u + a) −3 − −3 −3 u + a Z 4  4 u+a−a = u ln(u + a) −3 − du u+a −3 Z 4  4 a 1− = u ln(u + a) −3 − du u+a −3  4 (1.7) = u ln(u + a) − u + a ln(u + a) −3

= (4 + a) ln(4 + a) − (a − 3) ln(a − 3) − 7. (At this stage, it’s never a bad idea to check you’ve integrated correctly by differentiating the antiderivative inside the square brackets in (1.7) to make sure you get back the expression you were integrating.) Carrying on from (1.6), we have Z 4  Z 4 1 ln(u + 36)du − ln(u + 6)du 7 −3 −3 =

1 7

=

1 7

(40 ln(40) − 33 ln(33) − 7) − (10 ln(10) − 3 ln(3) − 7)
40 ln(40) + 3 ln(3) − 33 ln(33) − 10 ln(10) .




Without all the explanation, the solution should go something like this: (1) Find expressions for x and y in terms of u, v: x = 17 (2u + v) and y = 17 (u − 3v). (2) Calculate the Jacobian:

∂(x,y) ∂(u,v)

= − 71 .

(3) Change f from a function of x and y to a function of u and v: f (u, v) =

1 . u+v

(4) Use (1.1) to change the expression of the integral: Z Z Z 4 Z 36 1 1 1 dA = − dvdu. u + v 7 −3 6 R 4x − y
R 4 R 36 1 dvdu = 71 40 ln(40)+3 ln(3)−33 ln(33)−10 ln(10) . (5) Solve the integral: 17 −3 6 u+v The University of Newcastle, Australia