N-dimensional broken stick problem -- find joint density

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Homework Statement
A point ##P## is chosen uniformly in an ##n##-dimensional ball of radius ##1##. Next, a point ##Q## is chosen uniformly within the concentric sphere, cantered at the origin, going through ##P##. Let ##X## and ##Y## be the distances of ##P## and ##Q##, respectively, to the common center. Find the joint density function of ##X## and ##Y## and the conditional expectations ##E(Y\mid X=x)## and ##E(X\mid Y=y)##.
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I am not sure.
There are also two hints, which I will share with you now. The first hint says to start with the case ##n=2##. I've drawn a unit disc and a circle inside this unit disc, but I do not know how to reason further.

The second hint says that the volume of an ##n##-dimensional ball of radius ##r## is equal to ##c_nr^n##, where ##c_n## is some constant, and that this is of no interest to the problem. Somewhere this makes sense as we are only concerned with distances.

Then there's also a remark to the problem, namely that for ##n=1##, we rediscover the broken stick problem.

I'd be grateful for any help on this problem. The answer for the joint distribution should be ##f_{X,Y}(x,y)=n^2\frac{y^{n-1}}{x^n}## for ##0<y<x<1## (I also have the answer for the conditional expectations, if anyone's interested). But how to obtain these answers I have yet to understand.
 
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Start with finding the radial distribution for a uniformly sampled sphere and go from there as you would in the broken stick problem. The only difference is that you are not breaking the stick uniformly.
 
Ok. There's actually an example in the book where the author determines the radial distribution of a dart board for a beginning dart player (i.e. the darts are assumed to land uniformly on the dart board). The author gets that the radial distribution of ##R##, the distance from the origin, has density ##f_R(r)=2r## for ##0<r<1##. So in this example I believe we simply have ##X=R##. I don't see yet how this can help me determine the joint distribution. Moreover, I think the density of ##Y## is a bit trickier and I'm not sure how to obtain it.
 
You have the conditional probability ##f_{Y|X=x}## and you have ##f_X##. From this you shoukd be able to obtain the joint pdf.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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