How to integrate this double integral? $$\iint \limits_D 2x^2e^{x^2+y^2}-2y^2e^{x^2+y^2} dydx $$ where D is the region $x^2+y^2=4$
I tried changing it to polar, but it didn't make any use. $\iint \limits_{D(r,\theta)}2r^3\cos2\theta e^{r^2} drd\theta$ 
This integral also seems difficult to integrate.
 A: Let
$$
f(x,y)= (2x^2-2y^2)e^{x^2+y^2}
$$
then the integral
$$
\iint_D f(x,y) dydx=0
$$
because $f(x,y)$ is an even function respect to both variables, while the domain is symmetric to both axes.
A: Noticing antisymmetry about the line $y=-x$ is the "best" solution, but here is another approach if you know Stoke's theorem:
$$\iint_D 2xe^{x^2+y^2} -2ye^{x^2+y^2} dydx  = \iint_D d(e^{x^2+y^2}dx +e^{x^2+y^2}dy) $$
$$=\int_{bD}e^{x^2+y^2}dx +e^{x^2+y^2}dy$$
But $x^2+y^2 = 4$ on $bD=$ circle of radius $4$, so 
$$=\int_{bD}e^4dx+e^4dy$$
Now compute directly that this equals $0$, or use Stoke's theorem again!
Ex. 
$$\int_{bD}dx = \int_{bbD} x = 0 \text{ because the circle has no boundary }$$
i.e. $dx$ is a conservative vector field so integrating around a closed loop gives $0$.
A: This simplifies to
$$\int_0^2 2r^3 \cdot e^{r^2} \, dr \cdot \int_0^{2\pi} \cos 2\theta \, d\theta$$
which is
$$\int_0^2 2r^3 \cdot e^{r^2} \, dr \cdot \underbrace{[\frac{1}{2} \sin 2\theta ]_0^{2\pi}}_{= 0} = 0.$$
A: By symmetry of your domain of integration, you have
$$\iint \limits_D 2x^2e^{x^2+y^2} \mathrm{d}y \,\mathrm{d}x = \iint \limits_D 2y^2e^{x^2+y^2} \mathrm{d}y\,\mathrm{d}x $$
