Double Integrals

Question

How do we solve the question marked in red colour?

Question: drive(DOT)google(DOT)com/file/d/1A6f8Y9ZxKl2d26c4uYyWXi-qMVQ5uhxt/view?usp=share_link

Answer 1

Draw the parallelogram. Find the equations of the lines for each side and put those in general form (Ax + By = C). You should get:

x - y = -1

x - y = 1

x + y = 1

x + y = 3

If we let u = x - y and let v = x + y, we get new boundaries***:

-1 ≤ u ≤ 1

1 ≤ v ≤ 3

Also, if we use linear combinations, we can find x and y in terms of u and v:

First, add these:

u = x - y

+ (v = x + y)

____________

u + v = 2x

So x = 1/2 (u + v)

Then subtract them:

u = x - y

- (v = x + y)

____________

u - v = - 2y

So y = -1/2 (u - v)

We then find the partial derivatives ∂x/∂u, ∂x/dv, ∂y/∂u, and ∂y/dv and form a matrix. These answers are placed in left to right order, top to bottom, as shown here:

∂/∂u (1/2u + 1/2v) = 1/2 ∂/∂v (1/2u + 1/2v) = 1/2

∂/∂u ( - 1/2u + 1/2v) = -1/2 ∂/∂v ( - 1/2u + 1/2v) = 1/2

Now we find the determinant of this matrix:

(1/2)(1/2) - (1/2)(-1/2) = 1/4 + 1/4 = 1/2

Next, we find the new integrand:

(x + y)³ cos( x - y)

becomes

[1/2 (u + v) + -1/2 (u - v)]³ cos[(1/2 (u + v)) - (-1/2 (u + v))] = v³cos(u)

Don't forget to multiply by the determinant we found earlier, which was 1/2.

So our new integrand is:

1/2 v³cos(u)

This is much easier to integrate with our new boundaries***!