# Doubts whether my integral is right

1. Dec 5, 2005

### elle

Hi, can someone please quickly help me check if I've made a correct attempt at the following question. I'm having doubts whether my integral is right or not because I tried integrating it and applying the limits, but I didn't get the given answer 15/8. Can someone help? thank you!

http://i26.photobucket.com/albums/c109/mathsnerd/84c1d9ff.jpg

I've not included my diagram but I found to limits to be 1 < u < 2 and 1 < v < 2 .

2. Dec 5, 2005

### BerkMath

What is R? What is the power of y in the original integrand?

3. Dec 5, 2005

### elle

oops sorry I forgot to mention that part of the question.

R is the region in the first quadrant bounded by xy=1, xy=2, y=x and y=2x.
The power of y is 2 (same for x)

4. Dec 5, 2005

### BerkMath

Why did you change variables? There exist a nice symmetry in the Region R which would make the integral easily solved after breaking the region into two.

5. Dec 5, 2005

### BerkMath

Just looking at your Jacobian: You calculated J(u,v)/J(x,y) correctly, but unfortunately you want/need J(x,y)/J(u,v). You have int(int(f(x,y))dA. The change of coordiantes given by x=g_1(u,v) and y=g_2(u,v) is a mapping from the domain, U<R2, of f(x,y), to a new domain, V<R2. But you were given the change of coordiantes u=xy, v=y/x or u=h_1(x,y) v=h_2(x,y), which is a mapping from the domain V of f(x,y) to U. The maps are of course inverses, however in order to calculate your integral using the change of coordiantes given implicity, you must first solve for x and y to find the transformation from U to V. You did this to calculate what x^2+2y^2 was, but then you used the Jacobian for the other C.O.C. So you transformed x and y to u and v, but then used the Jacobian for the transformation from u and v to x and y. Find J(x,y)/J(u,v) and then carry out the integration.

6. Dec 5, 2005

### BerkMath

It was a surprise to me that your substitution at the end of the determinant ameliorated any confusion whether to use J(u,v)/J(x,y) or J(x,y)/J(u,v). Sorry! Your integral is good and your limits of integration in terms of u and v are good. You must have made a mistake in calculation 'cuz I get 15/8 also.

7. Dec 5, 2005

### elle

haha sorry for confusing you!! Thanks very much for the advice and yes I think I did make a mess of the integration which was why i couldn't get 15/8 but I've got it nows thnks again!