<s>[INST] Joint To Marginal Density : Can't figure it out. Here goes the problem:
Problem:
Suppose $X$ and $Y$ have the joint density function:
$f(X,Y) = c \sqrt{1 - x^2 - y^2}, \,\,\,\,\, x^2 + y^2 \leq 1$


*

*Find $c$.

*Find the marginal densities of $X$ and $Y$


My Approach For Solution:
Part 1: 
To obtain $c$, we need to solve:
$\int\int_A f(x,y) dy dx = \int \int_A c \sqrt{1 - x^2 - y^2} = 1$
Now, let:
$x = r\,cos(\theta)\, ,\, \textrm{and let}\, y = r\,sin(\theta)$, then we have:
$dy dx = r dr d\theta$
So the original integral becomes:
$c\int_0^{2\pi}\int_0^1 \sqrt{1 - r^2}\,r dr d\theta$
Substituting $1 - r^2 = u\, \textrm{and}\, -2rdr = du$, we get (note the change of limits of the inner integral):
$-\frac{c}{2}\int_0^{2\pi}\int_1^0 \sqrt{u} du d\theta$
which is now solved:
$$-\frac{c}{2}\int_0^{2\pi}\int_1^0 \sqrt{u} du d\theta = -\frac{c}{2}\int_0^{2\pi}[\frac{2}{3}u^{3/2}]_1^0 d\theta = \frac{c}{3}\int_0^{2\pi}[u^{3/2}]_0^1 d\theta = \frac{c}{3}\int_0^{2\pi} d\theta = \frac{c}{3}\cdot 2\pi$$
Equating this result to 1 gives us $\color{red}{c = \frac{3}{2\pi}}$
Part 2: 
Part 2 is where I am stuck. Now, we have:
$$f_X(x) = \int_{-\sqrt{1 - x^2}}^{\sqrt{1 - x^2}} f(x,y) dy = \frac{3}{2\pi} \int_{-\sqrt{1 - x^2}}^{\sqrt{1 - x^2}}\sqrt{1 - x^2 - y^2} \color{red}{dy}$$
A similar expression will come for $f_Y(y)$, i.e.
$$f_Y(y) = \int_{-\sqrt{1 - y^2}}^{\sqrt{1 - y^2}} f(x,y) dy = \frac{3}{2\pi} \int_{-\sqrt{1 - y^2}}^{\sqrt{1 - y^2}}\sqrt{1 - x^2 - y^2} \color{red}{dx}$$
My question is: $\color{red}{\textrm{Now, how do I evaluate these integrals in Part 2}}$?
 [/INST] defint $y=\sqrt{1-x^2}\sin(t)$, then $dy=\sqrt{1-x^2}\cos(t)dt$.
$$
f_X(x)=\int_{-\sqrt{1-x^2}}^\sqrt{1-x^2}c\sqrt{1-x^2-y^2}dy = c(1-x^2)\int_{-\pi/2}^{\pi/2}\cos^2(t)dt=c\frac{\pi}{2} (1-x^2)=\frac{3}{4}(1-x^2)
$$
p.s. your integral limit for $y$ is wrong since $y$ must be smaller than $\sqrt{1-x^2}$.
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