Moment Generating Function for distance from center in a circle, uniform PDF. So a Moment Generating Function $= \int_{-\infty}^{\infty} (pdf)(e^{tx}) dx$
Points are distributed inside a radius $r$ circle with uniform probability. What is the MGF for the distance from the center to a random point:
My pdf $= \frac{1}{\pi r^2}$ because pdf of uniform $= \frac{1}{b-a}$ where distributed between $a$ and $b$.
Am I correct so far?
Then the integral of this times $e^{tx}$. Integrating with two variables? Thanks for your time :).
Sorry I tried to format with the Latex code but it's not working?
 A: Let random variable $X$ be the distance from the centre. The probability that $X\le x$ is (for $0\le x\le r$) equal to the area of the circle of radius $x$, divided by the area of the full circle. 
So the cumulative distribution function of $X$ is $\frac{x^2}{r^2}$ (if $0\le x\le r$), and therefore the density function is $\frac{2x}{r^2}$ on $[0,r]$, and $0$ elsewhere.
Now use the formula you quoted to find the mgf. For the calculation, integration by parts will be useful. Note that effectively the integration is from $0$ to $r$.
Another way: You can also do the problem using a double integral over the region on or inside the circle. The function to be integrated is $\frac{1}{\pi}e^{t\sqrt{x^2+y^2}}$.  To carry out the integration, switch to polar coordinates. We end up with our previous integral.
This way of doing things saves us the trouble of finding the density function of $X$. So it is definitely more pleasant than the approach we first described. But it requires a little more calculus background.
