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A point $(x,y,z)$ is picked uniformly at random inside the unit ball. Find the joint density function of $Z$ and $R$, where $R^2=X^2+Y^2+Z^2$.

First, I find the joint density of $Y$ and $Z$, that is $f_{Y,Z}(y,z)=\frac{3}{2\pi}\sqrt{1-y^2-z^2}$ with $Y^2+Z^2\leq R^2$. Let $Y=\pm\sqrt{R^2-Z^2}$ and $W=Z$. The Jacobin determinant is $$J=\begin{vmatrix}1&0\\\frac{R}{\sqrt{R^2-Z^2}}&\frac{Z}{\sqrt{R^2-Z^2}}\end{vmatrix}=R/\sqrt{R^2-Z^2}$$ Then, $$f_{R,Z}(r,z)=f_{Y,Z}(\sqrt{1-z^2},z)|J|=\frac{3\sqrt{1-(1-z^2)-z^2}}{4\pi}\left(\frac{r}{\sqrt{r^2-z^2}}\right)$$ I think I miss something. Can someone give me a hint or suggestion to work on this problem? Thanks

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    1. Is $r=1$ in your first formula? 2. There's no way to come from $Y,Z$ to $R$. I suggest you to first find the density of $Y,Z,R$ and then extract the density of $(Z,R)$ from there. 3. There's no way to get $0$ from non-negative densities. I suspect that you forgot that there is the absolute value of Jacobian in the formula for density. – 2017-02-03
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    Alternatively, instead of using any formulas, you can try to compute the probability that $R\in dr, Z\in dz$. This is fairly easy. – 2017-02-03
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    @zhoraster it is a unit ball, radius should be $1$. I don't understand your second suggestion. I don't see how to compute $R\in\,dr,Z\in\,dz$ – 2017-02-03
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    I asked to write the formula *with* Jacobian, not that *for* Jacobian (which is wrong as written btw). That is, how do you use Jacobian to find the density? – 2017-02-03
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    How did the inequality $Y^2 + Z^2\le R^2$ become an equality $Y=...$? Once you turned to the variables $Y$ and $Z$, there's no way to recover $R$. So do the following. Transform the variables: $V= Y$, $W=Z$, $R = \sqrt{X^2 + Y^2 +Z^2}$, find two inverses: $X = \pm\sqrt{...}$, find the Jacobian (be careful here, you have problems with that), substitute into formula (which should have two summands - as there are two inverse functions). I'll write alternative method only after you make these steps (otherwise it will not be very helpful). – 2017-02-03
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    @zhoraster I still don't understand your suggestion. I used spherical coordinate then got $3r/2$ with $ |z| – 2017-02-03
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    Then I can't help, sorry. – 2017-02-03
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    @zhoraster I finally understand your suggestion. Transform x,y,z to r,y,z. The determinant is easy to compute because the Jacobi an matrix is simple. Then integrate about y, and it is simple. (I am using my phone) – 2017-02-03
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    You can try to post the final answer, at least. – 2017-02-03
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    @zhoraster the answer is $3r/2$ with $|z| – 2017-02-03
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    Sounds very realistic! I'll write alternative approach tomorrow. – 2017-02-03

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Let $|z|\le r$. The probability that $R\in[r,r+dr]$, $Z\in [z,z+dz]$ is $\frac{3}{4\pi}$ (total volume$^{-1}$) times the volume of the corresponding set. The latter is a belt cut from the sphere of radius $r$ and thickness $dr$ by the plane $Z=z$ of thickness $dz$. The volume is equal to the area of the corresponding spherical belt, i.e. $2\pi \sqrt{r^2-z^2} \cdot dz$ times the "height" of this belt, which, by simple geometry, is equal to $\frac{r\, dr}{\sqrt{r^2-z^2}}$ (I hope the picture explains it well); so the volume is $2\pi r\cdot dr\,dz$. Thus, the probability is $\frac{3r}{2}dr\,dz$, which confirms your formula for density.

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