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A machine produces cylindrical containers with the radii and the heights varying according to a joint pdf: \[ f_{R,H} = \begin{cases} 2r(r+2)h^{r+1}, & \text{if 0 < r, h < 1} \\ 0, & \text{otherwise} \\ \end{cases}\] What is the joint density for the volume and surface area of the containers?

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    I actually don't know how to start solving this question. Do you have any suggestions I can start with?2012-10-22

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Volume and surface area depend on radius and height as

$ V=\pi r^2h\;,\quad A=2\pi rh+2\pi r^2\;. $

Solving the first equation for $h$ and substituting into the second leads to a cubic equation for $r$. Since that would lead to multiple solutions and horribly complicated formulas, I'll assume that the problem is badly posed and only the curved surface area of the cylinders was meant to be included. In that case we have

$ V=\pi r^2h\;,\quad A=2\pi rh\;, $

and substituting $\pi rh=A/2$ into the first equation yields $V=rA/2$, or $r=2V/A$. Then substituting that into the second equation and solving for $h$ yields $h=A^2/(4\pi V)$.

Now you can calculate the Jacobian of the transformation.

To find the boundaries of the distribution in the transformed representation, transform each of the given four boundaries. Both boundaries $h=0$ and $r=0$ map to the single point $A=V=0$. The boundary $r=1$ maps to $V=1\cdot A/2=A/2$, and the boundary $h=1$ maps to $A^2=4\pi V$, or $V=A^2/4\pi$, so the support of the distribution is enclosed by the two curves $V=A/2$ and $V=A^2/4\pi$, which intersect at the points $A=V=0$ (as they must) and $A=2\pi$, $V=\pi$ (corresponding to $r=h=1$).

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    @woaini: I think you're misinterpreting $0\lt r,h\lt1$ (which shows that this is not a good way to write inequalities). Under your interpretation the given distribution would be unnormalizable. I think the only interpretation that yields a well-defined problem is $0\lt h\lt1$ and $0\lt r\lt1$. (Note also that the given distribution is normalized under this interpretation, which is unlikely to be a coincidence.)2012-10-22