Impossible trick question double integral

In summary: Glad I could help.In summary, the given double integral of sin(x-y)*e(x-y)^2-0y) 2--- dA over a disk of radius 2 centered at (1,1) cannot be solved through traditional integration methods. However, there is a trick involved, which is to notice the symmetry of the domain and integrate the top and bottom halves separately. This results in the integral being equal to 0.
  • #1
vandyboy73191
7
0

Homework Statement



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)

Homework Equations





The Attempt at a Solution



gee this problem stumped me. I've been working on it for over 3hrs. I've 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 I've 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. I am so frustrated.
 
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  • #2
Tex, I'm not sure what you're integrand is. (e times (x-y)?( e^(x-y))^2)?
 
  • #3
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
 
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  • #4
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?
 
  • #5
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 he's talking about.
 
  • #6
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 :)
 
  • #7
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.
 
  • #8
Hurray symmetry :)
 
  • #9
by the way. thank you so much. your a life saver
 
  • #10
No problem.
 

1. What is an impossible trick question double integral?

An impossible trick question double integral is a complex mathematical problem that is deliberately designed to deceive or trick the person attempting to solve it. It often involves a double integral, which is an advanced calculus concept that involves integrating a function of two variables over a defined region.

2. Why is it called an impossible trick question double integral?

It is called an impossible trick question double integral because it is designed to be impossible to solve using traditional methods. The person posing the question may have intentionally included errors or misleading information in order to make it unsolvable or extremely difficult to solve.

3. Is it possible to solve an impossible trick question double integral?

Yes, it is possible to solve an impossible trick question double integral. However, it requires advanced mathematical knowledge and creative problem-solving skills. It may also involve thinking outside of the box and using non-traditional methods to arrive at a solution.

4. How can I approach solving an impossible trick question double integral?

To approach solving an impossible trick question double integral, it is important to carefully read and analyze the question. Look for any potential errors or misleading information and try to simplify the problem by breaking it down into smaller parts. It may also be helpful to consult with other mathematicians or use online resources for guidance.

5. Why are impossible trick question double integrals used?

Impossible trick question double integrals are often used as a way to test and challenge the problem-solving abilities of students or to demonstrate the complexity and intricacy of higher-level mathematics. They can also serve as an entertaining and challenging brain teaser for those who enjoy solving difficult problems.

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