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$$\iint \limits_D 2x^2e^{x^2+y^2}-2y^2e^{x^2+y^2} dydx $$ 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.

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4 Answers 4

up vote 4 down vote accepted

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.

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2  
Well, $(2x^2+2y^2)e^{x^2+y^2}$ is also even in both variables. –  Jean-Claude Arbaut Dec 11 '13 at 14:25
    
Indeed. The correct symmetry to observe is pointed out in my answer and (implicitly) in arbaujic's answer. Namely that the integrand is antisymmetric about the line $y=x$, i.e. $f(x,y) = -f(y,x)$. Mark's answer is actually incorrect, but had the right idea of looking for a symmetry. –  Steven Gubkin Dec 11 '13 at 18:21

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 $0$.

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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 $$\int_0^2 2r^3 \cdot e^{r^2} \, dr \cdot \underbrace{[\frac{1}{2} \sin 2\theta ]_0^{2\pi}}_{= 0} = 0.$$

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By symmetry of your domain of integration, you have

$$\iint \limits_D 2x^2e^{x^2+y^2} \mathrm{d}y \,\mathrm{d}x = \iint \limits_D 2y^2e^{x^2+y^2} \mathrm{d}y\,\mathrm{d}x $$

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