Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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.

share|cite|improve this answer
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


Now compute directly that this equals $0$, or use Stoke's theorem again!


$$\int_{bD}dx = \int_{bbD} x = 0 \text{ because the circle has no boundary }$$

i.e. $dx$ is a conservative vector field so integrating around a closed loop gives $0$.

share|cite|improve this answer

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

share|cite|improve this answer

By symmetry of your domain of integration, you have

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.