Issue with basic double integral

Question

I have $(X,Y)$ random variable in $\mathbb{R}^2$ with joint pdf defined as: $$f_{X,Y}(x,y)=\dfrac{c}{1+x^2+y^2}\mathbb{1}_{D}(x,y)$$ Where $c \in \mathbb{R}$ and $D=\{(x,y) \in \mathbb{R}^2 : x^2+y^2 \le 1\}$, and I have to calculate $c$. In other words, I have to find $c$ such that: $$\int_D f_{X,Y}(x,y)\text{d}x\text{d}y=1$$ I tried to solve the integral by using polar coordinates given by $\Psi(r,\theta)=(r \cos \theta,r\sin\theta)$, so that I obtain $$f_{X,Y}(r\cos\theta,r\sin\theta)=\dfrac{c}{1+r^2\cos ^2\theta+r^2\sin ^2\theta}=\dfrac{c}{1+r^2}\\ \Psi^{-1}(D)=\{(r,\theta) \in \mathbb{R}^2:r^2\cos ^2\theta+r^2\sin ^2 \theta \le 1\}=\{(r,\theta) \in \mathbb{R}^2:-1 \le r\le 1,\ 0\le \theta \le 2\pi\}$$ And the integral becomes $$c\int_{\Psi^{-1}(D)}\dfrac{r}{1+r^2}\text{d}r\text{d}\theta=c\int_{0}^{2\pi}\left(\int_{-1}^{1}\dfrac{r}{1+r^2}\text{d}r\right)\text{d}\theta$$ Where the additional $r$ is the determinant of the Jacobian of $\Psi$. But this doesn't make sense since the first integral it's equal to $0$. I haven't studied multidimensional calculus yet, so maybe there is some trivial mistakes in what I've done.

Answer 1

The radius is actually from $0$ to $1$, not $-1$ to $1$. You should have: $$c \int_0^{2\pi} \int_0^1 \frac{r}{1+r^2} \; dr d\theta=c \int_0^{2\pi} \frac{1}{2} \ln{2} \; d\theta=c\pi \ln{2}$$ Therefore, $$c=\frac{1}{\pi \ln{2}}$$

Answer 2

Your polar domain of integration $-1 \leq r \leq 1, 0 \leq \theta \leq 2\pi$ represents every point in $D$ other than the origin twice. For example, $\Phi(\frac{1}{2}, 0) = (\frac{1}{2}, 0) = \Phi(-\frac{1}{2}, \pi)$.

The new domain of integration after a change of variables should be chosen so that the map function is a bijection, not always the general function inverse of the original domain. So for polar coordinates, we usually restrict the domain to $r \geq 0$.