Probability the $X \leq Y$ when $(X,Y)$ is uniformly distributed on the unit disk

Question

Given the probability density function $f_{XY}(x,y)=\frac1\pi,$ within the region $\{(x,y):x^2+y^2\leq1\}$, what is the probability $\mathbb{P}(X\leq Y)?$

My solution:

$$\int_0^1\int _{\frac\pi4}^{\frac{5\pi}4}\frac1\pi\,\textrm{d}y\,\textrm{d}x=1.$$

However, I know that my answer is wrong. The correct answer is $\displaystyle\frac12$. I can't detect my mistake.

Answer 1

Easy realize that your joint pdf is constant, thus it is uniform on the unit disk... thus whitout further calculations:

$$\mathbb{P}(X\leq Y)=0.5$$

If you want to integrate you have to do

$$\mathbb{P}(X\leq Y)=\int_0^1\rho\left[ \int_{\pi/4}^{5\pi/4}\frac{1}{\pi }d\theta \right]d\rho=0.5$$

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Of course the best way to proceed is to look at the following drawing.

The purple area is the favourable one; it is exactly half of the total area thus, being the joint pdf constant everywhere, the requested probability is exactly 50%

Answer 2

you are integrating over rectangle $[0,1] \times[\pi/4, 5\pi/4]$, first you have to apply changing of coordinates to your integral https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Calculus_(OpenStax)/15%3A_Multiple_Integration/15.3%3A_Double_Integrals_in_Polar_Coordinates

since your task looks like university assignment I will leave it there :)