We have a homogenous rotational cylinder height $h$ that is deformed on the axis into a rotational body:
$$\Omega = \{(x,y,z): -h/2 \leq z \leq h/2 \,\, \text{and} \,\, x^2+y^2 \leq r_0(1 + \epsilon \sin(\frac{\pi z}{h}))\}$$
We need to find its density if the mass stays the same.
My try:
We find the volume of the new body: $\iiint_\Omega dV$
If we use cylindrical coordinates:
$x^2 + y^2 = r^2$
$0 \leq r^2 \leq r_0(1+ \epsilon \sin(\pi z/h))$
So bounds of the integral are: $0 \leq r \leq \sqrt{r_0(1+ \epsilon \sin(\pi z/h))}$
$-h/2 \leq z \leq h/2$
However, I don't know on what interval is $\phi$
What I would do then is: $\frac{V_1}{\rho_1}= \frac{V_2}{\rho_2} \implies: \rho_2 = \frac{V_2 \rho_1}{V_1}$
Then after I would do the first part, the question also is to calculate the momentum about the symmetric axis. But first I need to know how to do the upper.
There’s no bound on $\phi$, since the body is rotationally symmetric about the $z$ axis. You can calculate the volume as the integral of the areas of the circular discs it consist of:
$$ V=\int_{-h/2}^{h/2}\pi r_0\left(1+\epsilon\sin\left(\frac{\pi z}h\right)\right)\mathrm dz=\pi r_0h\;. $$