$P(X^2 + Y^2 \leq 1/2)$ geometric intuition I have the following density function $c\sqrt{(1 - X^2 - Y^2)}$ where $x^2 + y^2 \leq 1$ and I want to think about it geometrically. For example, one question asked me what the value of c is and I realized that it should be $\frac{1}{area}$ of the hemisphere and thus is 3/(2pi). 
Now, I am asked to find $P(X^2 + Y^2 \leq 1/2)$ and I also want to think about this geometrically. I know the density is defined on $x^2 + y^2 \leq 1$ so my guess is that $P(X^2 + Y^2 \leq 1/2)$ is just a section of the area of the region but I'm not sure... any help is appreciated! 
 A: 
I have the following density function $c\sqrt{(1 - X^2 - Y^2)}$ where $x^2 + y^2 \leq 1$ and I want to think about it geometrically. For example, one question asked me what the value of c is and I realized that it should be $\frac{1}{area}$ of the hemisphere and thus is $3/(2\pi)$. 

Close, $c$ is the inverse of the volume of the unit hemisphere.   Which is $3/(2\pi)$

Now, I am asked to find $P(X^2 + Y^2 \leq 1/2)$ and I also want to think about this geometrically. I know the density is defined on $x^2 + y^2 \leq 1$ so my guess is that $P(X^2 + Y^2 \leq 1/2)$ is just a section of the area of the region but I'm not sure... any help is appreciated! 

It's the volume of the intersection of the unit hemisphere and the (infinite height) cylinder of radius $1/2$.   Imagine: Sculpt playdough into a hemisphere and cut its centre out with a cookie cutter.
In polar coordinates: 

 $$\begin{align}\mathsf P(R\leq r)~=~&3\int_0^r \rho\sqrt{1-\rho^2~}~\mathbf 1_{0< \rho\leq 1}\operatorname d \rho \\ =~& (1-(1-r^2)^{2/3})\mathbf 1_{0<r<1}+\mathbf 1_{1\leq r}\end{align}$$

