A point is uniformly distributed in a unit sphere in three dimensions.
Find the marginal densities of the $x$, $y$, and $z$ coordinates.
My attempt: Since the volume of a sphere is $\frac{4}{3} \pi r^3$ and $r = 1$ for a unit sphere, the joint probability $f(x, y, z) = \frac{3}{4\pi}$.
In order to find $f_X(x)$, we must integrate over all of $y$ and $z$.
$$ x^2 + y^2 + z^2 \leq 1 \implies z^2 \leq 1 - x^2 - y^2 \implies - \sqrt{1 - x^2 - y^2} \leq z \leq \sqrt{1 - x^2 - y^2}$$
This inequality defines the limits of integration for the $z$ coordinate. For the $y$ coordinate, we simply integrate from $-1$ to $1$ (this is the first part I'm not sure about. I would insert different limits of integration defined by $x$ but there doesn't seem to be any suitable substitution for it).
$$ \int_{-1}^{1} \int_{-\sqrt{1 - x^2 - y^2}}^{\sqrt{1 - x^2 - y^2}} \frac{3}{4\pi} dz dy = \frac{3}{2\pi} \int_{-1}^{1} \sqrt{1 - x^2 - y^2} dy$$
This integral is the main problem. Should I use substituion, change to polar coordinates or something else? I'm a little lost on polar coordinates in 3-dimensional space too, if converting is the way to go.
If you are concerned only aboutevaluating this integral,substitute $y=\sqrt{1-x^2}\cos\theta$