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

Question

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:

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?

Answer 1

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.

Answer 2

Let use polar coordinates

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

with

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

thus

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

we obtain

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