# Evaluating integral of Green's theorem

Applying Green's theorem, I've obtained a double integral of $$\iint_c 4ye^{-x^2 - y^2} \cos (2xy) dx dy = 0$$ along the curve $x^2 + y^2 \le R^2$.

Why is it equal to $0$?

The explanation I got was because "the integral in anti-symmetric (odd) in $y$ and the area of integration is symmetric in $y$."

Will anyone please tell me what does the above sentence means exactly? Thanks.

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"Antisymmetric in $y$" means that $f(x,-y)=-f(x,y)$. Imagine cutting up the region of integration into a "left" and "right" portion. Note that the integral on the "left" is precisely the negative of the integral on the "right" due to the oddness. – J. M. Nov 6 '11 at 4:38
You might be interested in this thread. – J. M. Nov 6 '11 at 4:43

## 1 Answer

Maybe a few pics might be of use. Here's the surface whose volume you're trying to get:

Here are the "left" and "half" bits I was talking about in the comments:

This is the antisymmetry that was being referred to. The integrand on the left is precisely the negative of the integrand of the right. Due to this, the integral for the left section would be the negative of the integral for the right section, and adding those two integrals will yield the value of zero.

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Hey! Thanks for the great graphics! really helped in my understanding. – adsisco Nov 6 '11 at 11:05