Given integral $\iint_D (e^{x^2 + y^2}) \,dx \,dy$ in the domain $D = \{(x, y) : x^2 + y^2 \le 2, 0 \le y \le x\}.$ Move to polar coordinates.

Question

Given integral $\iint_D (e^{x^2 + y^2}) \,dx \,dy$ in the domain $D = \{(x, y) : x^2 + y^2 \le 2, 0 \le y \le x\}.$ Move to polar coordinates.

First of all I tried to find the domain of $x$ and $y$:

$0 \le y \le x$ (given)

$0 \le x^2 + y^2 \le 2 \implies -y^2 \le x^2 \le 2 - y^2 \implies y \le x \le \sqrt{2 - y^2}$.

If I rewrite the integral:

$$\int^0_x\int_y^{\sqrt{2-y^2}} e^{x^2 + y^2} \,dx\,dy,$$

I'm getting that $x$ dependent with $y$, and $y$ dependent with $x$.

I guess this is why I need to move to polar coordinates.

But How can I do it?

Thanks in advance.

Answer 1

Draw a picture of $D$: The first condition means that $D$ lies within the disk around $0$ of radius $2$, and the second condition means that $D$ lies below the line $y = x$, but above $0$. That is, it's a slice of pie lying in the first quadrant. This leads us to the bounds

$$0 \le r \le 2$$ and $$0 \le \theta \le \frac{\pi}{4}$$

Now make the change of variables.

Answer 2

Using polar coordinates gives the answer

$$ \int_{0}^{pi/4}\int_{0}^{\sqrt{2}}e^{r^2}r\,dr\,d\theta $$

Plot the region to see what's going on.