Compute $$\iint_D e^{(x+y)^2} \ dxdy,$$

where $D=\{ y\leq3x, \ x\leq 3y, \ 0 \leq x+y \leq 2 \}.$

The area $D$ is easily drawn:

enter image description here

From this image, it's easy to seet that we can split the integral into two parts:

$$\iint_D e^{(x+y)^2} \ dxdy = \int_0^{1/2}\int_{x/3}^{3x}e^{(x+y)^2} \ dydx+\int_{1/2}^{3/2}\int_{x/3}^{2-x}e^{(x+y)^2} \ dydx = \frac{e^4-1}{4}.$$

This is the correct answer. The problem here is that I actually did not compute those two integrals myself, I entered them in Maple and for each one of them I got the answers in terms of the error function, however they canceled out luckily and I was left with the correct answer.

Can anyone help me evaluating this integral by a clever substitution so that I don't have to deal with non-elementary functions?

  • $\begingroup$ Looks like you should use polar coordinates. $\endgroup$ – Surb Feb 18 '18 at 17:55
  • $\begingroup$ I thought so too. But the exponent would have been nicer if it was $x^2+y^2$. I will try polar for this too then! $\endgroup$ – Parseval Feb 18 '18 at 18:00

HINT...You can use the following change of variables: let $$u=x+y$$ and $$v=\frac yx$$

The Jacobian is $$\frac{u}{(1+v)^2}$$ and the integral works out quite easily to get your answer. I shall leave this to do yourself.

I hope this helps.

  • $\begingroup$ Thank you very much. However, once I know the proper substitutions, things are straight forward. How did you know that making these substitutions, will work out so well? Experience or is there anything specieal to look for? $\endgroup$ – Parseval Feb 18 '18 at 18:20
  • $\begingroup$ Look for a substitution which makes the limits simple, and see if the resulting Jacobian will make the integrand integrable $\endgroup$ – David Quinn Feb 18 '18 at 19:07

Let use polar coordinates

  • $x=r \cos \theta$
  • $y=r \sin \theta$


  • $\theta_1=\arctan \frac13$
  • $\theta_2=\arctan 3$
  • $0\le r \le \frac{2}{\sin \theta + \cos \theta}$ (since $y+x=2 \iff r\sin \theta + r \cos \theta=2$)


$$\iint_D e^{(x+y)^2} \ dxdy =\int_{\theta_1}^{\theta_2}d\theta\int_0^{\frac2{\sin \theta+\cos \theta}}re^{r^2(1+\sin 2\theta)}dr$$

we obtain

$$\int_{\theta_1}^{\theta_2}d\theta\int_0^{\frac1{\sin \theta+\cos \theta}}re^{r^2(1+\sin 2\theta)}dr=\int_{\theta_1}^{\theta_2} \left[\frac{e^{r^2(1+\sin 2\theta)}}{2(1+\sin 2\theta)}\right]_{0}^{\frac2{\sin \theta+\cos \theta}}d\theta=\\\frac{e^4-1}2\int_{\theta_1}^{\theta_2} \frac{1}{1+\sin 2\theta} d\theta=\frac{e^4-1}2\left[ \frac{\tan \theta}{1+\tan \theta}\right]_{\theta_1}^{\theta_2}=\frac{e^4-1}2\left(\frac34-\frac14\right)=\frac{e^4-1}4$$


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