pdf without using polar coordinates

Question

If we have a joint pdf that is $f_{X,Y}(x,y) = \frac{1}{\pi}$ when $x^2+y^2\le 4$ and $x,y>0$ then it can easily be shown (using a change of coordinates that) the PDF for $Z=X^2+Y^2$ is $g(z) = \frac{1}{4}$ when $0<z<4$. However, my friend is unfamiliar with the change of coordinates (very unfortunately) so the integral $G(z) = \int_0^{\pi/2}\int_0^{z^2} \frac{1}{\pi} r drd\theta$ doesn't make sense to her. She's wanting the integral in cartesian coordinates only but I am unfamiliar with how to approach the problem like this.

Answer 1

She can use Fubini: if $F$ is the cdf of $Z$ and $R=\left\{\ (x,y)\ \middle|\ x^2+y^2\le r,\ x\ge 0, y\ge 0\ \right\}$, then

$$F(r)=P(Z\le r)=\iint_{R}f_{X,Y}(x,y)\ dA = \int_{0}^{\sqrt{r}}\int_{0}^{\sqrt{r-x^2}}\frac{1}{\pi}\ dy\ dx= \frac{1}{\pi}\int_{0}^{\sqrt{r}}\sqrt{r-x^2}\ dx$$

The last integral can be computed by a trigonometric substitution, or by remarking that this is a quarter of the area of a disk of radius $\sqrt{r}$.

Therefore, $F(r)=\frac{1}{\pi}\frac{\pi r}{4}=\frac{r}{4}$. This implies that the pdf is $F'(z=r)=\frac{1}{4}$.

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Edit: actually, since $f_{X,Y}$ is constant equal to $\frac{1}{\pi}$, then $F(r)$ is equal to $F(r)=\frac{area(R)}{\pi}=\frac{r}{4}$ directly, since the area of $R$ is $area(R)=\iint_R 1\ dA$