# Multiple Integrals: Moment of inertia of a cylinder

Question: Calculate the moment of inertia of the cylinder deﬁned below when the cylinder is rotated around the $x$-axis. The cylinder’s axis lies along the z-axis and is deﬁned by $x^2+y^2=1$, $z ≥ 0$ and $z ≤ 2$ and has constant mass density $ρ$. State your answer in terms of the mass of the cylinder, $M$. (End of question)

My issue is setting up the integral for this. I understand we will need to use cyclindrical polar co-ordinates. Also, the moment of inertia is calculated by squaring the axis which is perpendicular to the rotation axis (Is this correct?).

So, how would I go about setting up the integral? This is what I have so far:

$\int_{z=0}^2\int_{\theta=0}^{2\pi}\int_{r=0}^1$[unkown integrand]$r drd\theta dz$

I want to say that the integrand is $r^2$, but I think that give the moment of inertia about the rotation axis. I would appreciate an explanation of what the integrand should be and why. I will have no trouble computing the integral after this stage.

The moment of inertia is calculated by using: $$I = \int _V \rho(x,y,z)\bar{r}^2 dv,$$ where $\bar{r}$ is the distance from the rotation axis. When you change to cylindrical Coordinates you need to take into account that: $$dv = rdrd\theta dz$$ And the distance from the rotation axis is: $$\bar{r}=\sqrt{y^2+z^2}= \sqrt{r^2\sin^2(\theta)+z^2}$$ With $\rho(x,y,z)=\rho$ you get: $$I = \int_{z=0}^2 \int_{\theta=0}^{2\pi} \int_{r=0}^1 \rho (r^2\sin^2(\theta)+z^2) rdrd\theta dz$$
• Wait, are you sure it's $r^3$? We're rotating about the $x$-axis, not the axis of symmetry. – Mr Croutini Jun 5 '14 at 10:57
• How did you get the distance from the rotation axis? Also, if we were rotating about the $y$-axis instead, would the distance from the rotation axis be $\sqrt{x^2+z^2}$? – Mr Croutini Jun 5 '14 at 14:37
• Also, why have you put $\sin^2(\rho)$? should it not be $\sin^2(\theta)$? – Mr Croutini Jun 5 '14 at 14:53
• It is the shortest distance from the axis to that point. So that is correct, and for z-axis it's $\sqrt{x^2+y^2}$. I think you have understood. It should of course be $\theta$. – oholmer Jun 5 '14 at 15:02