Expected distance of a point picked at disc A point $P$ is picked uniformly over the disc with radius $1$ and center in the origin, after this a new point $Q$ is picked uniformly over the disc with radius $|P|$ and center in the origin.

Im asked to compute $E|Q|$, and given the hint that a random point $(X,Y)$ uniformly distributed over the disc with radius $r&gt;0$ has density $f_{X,Y}(x,y)=\frac{1}{\pi r^{2}} $, for $x^{2}+y^{2}\leq r$.

Im not sure how to approach this problem, is it $E[Q|P=x ]$ that I should find? If thats the case, then how do I find the conditional distribution?
If we know $P$ then the probability of landing on a point in a region of area $d^2A=Q\,dQ\,d\theta$ is just the areal element over the total accessible area:
$$p(Q,\theta|P)dQ\,d\theta=\frac1{\pi P^2}Q\,dQ\,d\theta$$
In turn, the probability of landing on a point in a region of area $d^2A=P\,dP\,d\phi$ in the unit circle in the first place is the areal element of that region over the area of the unit circle:
$$p(P,\phi)dP\,d\phi=\frac1{\pi}P\,dP\,d\phi$$
So the mean value of $Q$ is
$$\int_0^{2\pi}\int_0^1\frac1{\pi}P\int_0^{2\pi}\int_0^P\frac1{\pi P^2}(Q)Q\,dQ\,d\theta\,dP\,d\phi=\frac1{\pi^2}(2\pi)^2\int_0^1\frac1P\frac13P^3\,dP=\frac49$$
EDIT: To be more step by step we can go back to the first equation and integrate over angles to get
$$f(Q|P)dQ=\int_0^{2\pi}\frac1{\pi P^2}Q\,dQ\,d\theta=\frac2{P^2}Q\,dQ$$
Then we can find
$$E[Q|P]=\int_0^PQf(Q|P)dQ=\int_0^PQ\frac2{P^2}Q\,dQ=\left.\frac2{3P^2}Q^3\right|_0^P=\frac{2P}3$$
With this formula in hand we can integrate over $\phi$ to find
$$f(P)=\int_0^{2\pi}\frac1{\pi}P\,dP\,d\phi=2P\,dP$$
Then we can find
$$E[Q]=\int_0^1E[Q|P]f(P)dP=\int_0^1\frac{2P}32P\,dP=\left.\frac49P^3\right|_0^1=\frac49$$
The notation where the probability density function is never separated from its differential goes back to statistical mechanics where it's always there to remind you of the extra steps (transformation of differential) that need to be undertaken when transforming coordinates in a complex situation.
Being the distribution of P uniform on the unit circle, the expected value of the radius of $P$ is $1/2$.
Therefore, the expected value of the radius of Q, uniformly distributed between $0$ and $|P|$ is $1/2\times1/2=1/4$