Calculate the following integral in the following region:
$$Q_r=\lbrace (x,y) \in \mathbb{R}^2: -r \le x,y\leq r\rbrace$$
$$\iint_{Q_r} e^{-x^2-y^2} \mathrm{d}x \mathrm{d}y$$
Since the region where the integration occurs is infinite, I first figured out that it would be easier to convert the cartesian coordinates to polar. I transformed the function and put the central point from which I want to integrate in the origin
So I would have to integrate $f(x,y)=e^{-(x-r)^2-(y+r)^2}$ ,with $x\ge 0$ and $y \le 0$ thus the angle $\theta$ will vary between $-\frac{\pi}{2}$ and $0$
How I proceed from here to get the volume of the region in terms of $r$?
Denote $O = (0,0), A = (1,0), B = (1,1), C = (0,1)$, then the region $Q_r$ is made up of $4$ such congruent squares. Thus you only need to integrate over one of them and multiply the result by $4$ to obtain the main result. Thus call the square $\square OABC$ $S_r$, then $S_r$ again is made up of a quarter of a circle centered at the origin that we call $T_r$, and a tiny wedge between this quarter circle and the square. Call this wedge $W_r$, then $S_r = T_r \cup W_r\implies \int \int_{Q_r}= 4\int \int_{S_r}= 4\int \int_{T_r\cup W_r} = 4\left(\displaystyle \int_{0}^{\frac{\pi}{2}} \displaystyle \int_{0}^{r} \delta e^{-\delta^2} d\delta d\theta+\displaystyle \displaystyle \int_{0}^{\frac{\pi}{4}} \int_{r}^{\frac{r}{\cos \theta}}\delta e^{-\delta^2}d\delta d\theta+ \displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \displaystyle \int_{r}^{\frac{r}{\sin \theta}}\delta e^{-\delta^2} d\delta d\theta\right)$. You can work it out to simplify it.