This is a statistics problem, but it requires integration. And I'm getting a little lost in the problem because there are multiple variables involved when you switch over to polar coordinates, and I feel like I'm making many mistakes (it's been about 1-2 years since I last took calculus).
Question:
I first need to find the marginal probability of X, which is:
$f_X(x) = \int^{\infty }_{-\infty}f(x,y)dy$
I know the following:
$x = acos(\theta), y=asin(\theta), r = \sqrt{x^2 + y^2} = a$
Okay, so we have:
$f_X(x) = \int^{2\pi}_{0}\frac{1}{\pi a^2}acos(\theta)d\theta = \int^{2\pi}_{0}\frac{1}{\pi a}cos(\theta)d\theta = \frac{1}{\pi a^2}sin(\theta) = 0$
So then the expected value of $X$ is:
$E(X) = \int^{-\infty}_{\infty} x \times f_X(x) dx$
Which would be zero? The book says zero is the correct answer, but I'm not sure if the steps I did here are correct or if I arrived at that answer coincidentally.
Also, if $f_X(x)$ were nonzero and I actually had to proceed with the integration, what would the limits have been? From $0$ to $2\pi$ again?
Your answer cannot be correct as the marginal density you found does not integrate to $1$. You need to be careful with the bounds of integration. The set $\{(x,y)\mid x^2+y^2\leq a^2\}$ is the closed disk of radius $a$ about $(0,0)$. Hence $$ f_X(x)=\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\frac{1}{\pi a^2}\,dy =\frac{1}{\pi a^2}2\sqrt{a^2-x^2};\quad -a\leq x\leq a. $$ Hence $$ EX=\int_{-a}^axf_X(x)\,dx=\int_{-a}^a \frac{1}{\pi a^2}2x\sqrt{a^2-x^2}\,dx =0 $$ since the integrand is odd.