DefinitionsLet X and Y be random variables defined on the same random experiment. We define: 
It follows from this definition that, 
It follows from this definition that, 
> f is obtainable from F by taking derivatives
2 d f(s, t) =  F(s, t) dt ds> F is obtainable from f by integration
t s / /   F(s, t) =   f(x, y) dx dy   / / infinity infinity
ProblemLet (X,Y) be uniformly distributed over the unit circle centered at the origin. Compute the joint pdf, joint cdf, marginal density of X (i.e. fx(s)) and the marginal cdf of Y (i.e. Fy(t)). 
The joint pdf is constant over the circle and since the total area of the circle is Pi we get that for (x,y) inside the circle, 
> f is
1 f := (x, y) >  Pi
The joint cdf is computed from f by integration. When (s,t) is a point inside the circle the answer will depend on which cuadrant (s,t) is. 
> when (s,t) is in the III or IV cuadrant we have,
s t / /   1 F(s, t) =    dy dx   Pi / / 1 2 1/2 (1  x )> integrating over y,
s / 2 1/2  t + (1  x ) F(s, t) =   dx  Pi / 1
and computing this integral (e.g. with MAPLE) 
> F(s,t) = int(int(f(x,y),y=sqrt(1x^2)..t),x=1..s);
2 1/2 4 t s + 2 s (1  s ) + 2 arcsin(s) + 4 t + Pi F(s, t) = 1/4  Pi
The other two simmilar cases: i.e. when (s,t) is in the II or in the I
quadrant, are left as exercises. HINT: draw the pictures of the
regions.
Now, is this all? 
This is given by FY(t) where, 
> FY(t) = 2*int(int(1/Pi,x=0..sqrt(1y^2)),y=1..t);
2 1/2 2 t (1  t ) + 2 arcsin(t) + Pi FY(t) = 1/2  Pi
where t varies from 1 to 1.

By symmetry FY = FX. Thus, the marginal pdf of X is just the derivative w.r.t. t of FY(t) above. This is 
> given by
2 2 1/2 t 2 2 (1  t )  2  +  2 1/2 2 1/2 (1  t ) (1  t ) 1/2  Pi
Done! 