Impossible trick question double integral

[vandyboy73191]

1. The problem statement, all variables and given/known data

Evaluate the double integral sin(x-y)*e(x-y)^2�-0y) 2--- dA where D is a disk of radius 2 whose center is (1; 1)

2. Relevant equations



3. The attempt at a solution

gee this problem stumped me. Ive been working on it for over 3hrs. Ive tried changing into polar form and integrating that. That just takes me to an even messier integral. My professor says their is a trick, but I can't find it. I have a hunch somehow the integral will come down to finding the area of the disk, but I'm not sure how to get to that point.

Just to prove Ive tried something: sin(r(cos(theta)-sin(theta)))*e^(r^2(cos(theta)-sin(theta)))*r

that is what my integrand would be if I change it to polar

Please help me. Im so frustrated.

[Quantumpencil]

Tex, I'm not sure what you're integrand is. (e times (x-y)?( e^(x-y))^2)?

[vandyboy73191]

http://i.imagehost.org/t/0153/mathproblem3.jpg (http://i.imagehost.org/view/0153/mathproblem3)

that is the original integrand. I basically just substituted x=rcos(theta) and y=rsin(theta) and multiplied by r to convert to polar. But I don't know what to do after that

[Quantumpencil]

That isn't integrable over this domain in closed form, because integral (sin u)(e^[u^2]) has no closed form integral, at least not that I know of.

Are you sure you have the statement entirely correct?

[vandyboy73191]

the statement is correct. my professor says its a trick question that can not be solved by straight integrating. Just to reiterate it is over the domain of a disk of radius 2 and center (1,1).

I reached the conclusion that its not integrable about 1.5 hrs ago, but I can't find this trick hes talking about.

[Quantumpencil]

The center is (1,1). So it's centered at x=y. Think about what that means.

btw, the integral is zero. You tell me why :)

[vandyboy73191]

holy sh*t. lol wow that was eye opening. so above y=x within the circle y is greater than x. Below y=x x is greater than y. This means that the integral of the top half is the opposite of the integral of the bottom half and thus they sum to 0. Wow.

That was cool.

[Quantumpencil]

Hurray symmetry :)

[vandyboy73191]

by the way. thank you so much. your a life saver

[Quantumpencil]

No problem.