Volume integral over a cone

Question

I'm trying to integrate $I = \rho\int_V(y^2+z^2)dv$ over a cone of base radius $R$ and height $H$, where $\rho$ is a constant. The coordinates $y$ and $z$ are coordinates of the volume element. In cylindrical coordinates, we have $$dv = rdrd\theta dz$$

and $$y = r\sin \theta$$

Now the integral $I$ should be $$I = \int_0^R\int_0^{2\pi}\int_0^H(r^3\sin^2\theta+z^2r)drd\theta dz$$

Is this conversion to cylindrical coordinates correct? The background of the question is the moment of inertia tensor of said cone.

Answer 1

Judging from the limits, the order in which you want to perform the integration is $dz~d\theta~dr$. You are almost correct, except for the upper limit on $z$: it should be $H(1-r/R)$.

Here is a cross-section of the cone. $\triangle AHF\sim\triangle ADC$ gives $\frac{H-z}r=\frac HR$ which yields the expression above.

Answer 2

No. That would be the integral over the cylinder with the same base and the same height. It should be$$\rho\int_0^{2\pi}\int_0^R\int_0^{H(1-r/R)}(r^3\sin^2(\theta)-z^2r)\,\mathrm dz\,\mathrm dr\,\mathrm d\theta,$$which is equal to$$\frac{\rho\pi HR^2}{60}(3R^2-2H^2).$$