density functions if $f X,Y (x,y) = c(x^2 + y^2)$ for $x^2 + y^2 \leq 16$ then find (use polar coordinates to set up your integrals):
a) the value of c to make this a density function
b) $P(Y<2X)$ where y is in the first quadrant
c) $P(X + Y > 1, X > 0, Y > 0)$ (Set up only)
I know how to do these problems when there is only one variable, but I'm not sure how to do them with two. Any help is appreciated. 
 A: (a) We want the integral of $c(x^2+y^2)\,dy\,dx$ over the disk to be $1$. Go to polar coordinates. Then $dy\,dx=r\,dr\,d\theta$. So we want
$$\int_0^{2\pi} \left(\int_0^4 cr^3\,dr\right)\,d\theta=1.$$
The integral is $(2\pi)(64c)$. Set this equal to $1$ and solve for $c$.
(b) I am not sure what is being asked. Will interpret it as the probability that $Y\lt 2X$ and $Y\gt 0$. Draw the circle, and the line $y=2x$. We want the probability that $(X,Y)$ ends up in the sector of the circle which is below the line and in the first quadrant. The angle of this sector is $\arctan(2)$.  By the circular symmetry of the density function, the required probability is $\frac{\arctan(2)}{2\pi}$.
(c) Draw the circle, and the line $x+y=1$. We want the probability of being in the first quadrant and above the line $x+y=1$.
By symmetry, the probability of being in the first quadrant is $\frac{1}{4}$. From this we subtract the probability $p$ of being in the first quadrant and below the line $x+y=1$. 
This region is a little triangle. We can use polar coordinates or rectangular. Rectangular is probably more familiar.  We get
$$p=\int_{x=0}^1 \left(\int_{y=0}^{1-x}c(x^2+y^2)\,dy\right)\,dx.$$
If we really want to use  polar, we can set up more directly, integrate $cr^3 \,dr\,d\theta$. The line $x+y=1$ becomes $r(\cos\theta+\sin\theta)=1$, so $r$ goes from $\frac{1}{\cos\theta+\sin\theta}$ to $4$, then $\theta$ from $0$ to $\frac{\pi}{2}$. Looks horrible, but it is not quite as bad as it looks. Anyway, you are not asked to calculate.
