Double integration with various curves

Question

$$\iint_D e^{2(x^2+y^2)}\, dxdy$$ where $D\colon\, x^2+y^2 \geq 4, x^2+y^2\leq 25, y=x, x\geq 0, y\geq0$.

I cannot understand how to solve this; please help.

Answer 1

First of all, are you sure that $x=y$ is part of the region $D$? This renders the integral equal to zero, since it is a line with $\theta=\pi/4$ and $\int_{a}^{a}dx=0$.

If $x=y$ condition is gone:

The region $D$ lies in the first quadrant of the $xy$ plane between the circles with radii 2 and 5 ($x \geq 0$ and $y \geq 0$). After converting your problem into polar coordinates (like said by Olivier Oloa), we have

$$\int_{0}^{\pi/2}\int_{2}^{5}e^{2r^2}r\,dr\,d\theta.$$

We can solve this by first integrating by $r$ and then $\theta$:

$$\int_{0}^{\pi/2}\int_{2}^{5}e^{2r^2}r\,dr\,d\theta=\int_{0}^{\pi/2}\, \frac{e^{8}}{4}\left( e^{42}-1\right)d\theta=\frac{\pi e^{8}}{8}\left( e^{42}-1\right).$$

Note that the first integration by $r$ was made by substitution $u=r^2$ and $du=2r\, dr$.

Answer 2

Hint. One would be tempted to make a polar change of variables : $$ x=r\cos\theta, \\\\ y=r\sin\theta, \\\\ \sqrt{x^2+y^2}=\sqrt{r^2}=r\ge0 $$ with $0\le\theta \le \frac \pi2$ then you have $dxdy=r dr d\theta$.

Hope you can take it from here.