Moment of inertia about $x$-axis for solid of revolution

Question

Given a mass of density 1 in a region T of space, moment of inertia about the x-axis is given by

$I_x = \int\int\int_V (y^2+z^2)dV$

I need to show that, for a solid of revolution,

$I_x = \frac\pi2 \cdot \int_0^hr^4(x)dx$ where $0\leq x \leq h$

This problem is given in Advanced Engineering Calculus Erwin Keyring, 10ed, pg 458, Q25,

and I have no clue how to approach this. Help will be appreciated

Answer 1

You need to write the volume element in cylindrical coordinates. Due to the symmetry of the problem (revolution around $x$-axis and symmetry of the integrand with respect to the same axis), we can choose the volume element $dV$ to be a ring at some radius $R$, with thickness $dR$ and height $dx$. We know that $R^2=y^2+z^2$. $R$ varies between $0$ and $r(x)$. The volume of this ring is $$dV=2\pi R\ dR\ dx$$ so we write $$I_x=2\pi\int_0^h dx\int_0^{r(x)}R^3dR$$ The last step is trivial