Thread: Double integral change of variables

1. Double integral change of variables

I have to integrate x+y with 0<=x<=1 and 0<=y<=x by making the change of variables x=u+v and y=u-v.

I got the Jacobian to be 2, and I solved for u and v and got u=(x+y)/2 and v=(x-y)/2, but I can't seem to find the right limits of integration. Any hints?

2. Re: Double integral change of variables

Your region of integration is bound by the triangle whose vertices are at (0,0), (1,1), and (1,0).

If x=0 & y=0,the (u, v) = (0, 0). If x=1 & y=1, then (u, v) = (1, 0). If x=1 & y=0, then u = 1/2 & v = 1/2 .

Integrate 0 ≤ u ≤ 1, 0 ≤ v ≤ (1/2) - |x - (1/2)|

3. Re: Double integral change of variables

Thanks! I'm also having trouble with this one.

Let R denote the region inside x^2 + y^2 = 1 but outside x^2 + y^2 = 2y with x=>0 and y=>0. Let u=x^2 + y^2 and v=x^2 + y^2 -2y. Compute the integral of x*e^y over the region D in the uv-plane which corresponds to R under the specified change of coordinates.

My first attempt at figuring out the new limits of integration yielded 0<=u<=1 and 0<=v<=u, which seems wrong to me. I'm also not sure how to change the integrand to make it a function of u and v.

4. Re: Double integral change of variables

Originally Posted by SammyS
Your region of integration is bound by the triangle whose vertices are at (0,0), (1,1), and (1,0).

If x=0 & y=0,the (u, v) = (0, 0). If x=1 & y=1, then (u, v) = (1, 0). If x=1 & y=0, then u = 1/2 & v = 1/2 .

Integrate 0 ≤ u ≤ 1, 0 ≤ v ≤ (1/2) - |u - (1/2)|

Switch order of integration. It makes more sense.

Integrate: 0 ≤ v ≤ 1/2 , v ≤ u ≤ 1-v

5. Re: Double integral change of variables

What about the second problem in post #3?

6. Re: Double integral change of variables

Yup. That's a lot tougher.

Have you posted that elsewhere?

7. Re: Double integral change of variables

yes, but still no solution.