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I have a cone (vertex at $(x,y,z)=0$) with height $H$ and radius $a$ (radius of the base). Here is a picture:

enter image description here

I want to find the moment of inertia around the $z$-axis using calculus. I know the integral I have to evaluate is :$$I_z={\int \int \int}_V \vec{r}^2 dm$$

I decided to switch to cylindrical coordinates and my integrals becomes:

$$I_z={\int_0^H \int_0^{2\pi} \int_0^{\color{red}{?}}} r^3 dr d \theta dz$$

Since my radius is changing with $z$ I need to parameterize $r$ in terms of $z$. By similar triangles we have $$\frac{a}{H}=\frac{r(z)}{z} \iff r(z)=\frac{a z}{H}$$

When I tried to do this exercise without any help I substituted $r^3$ with $r(z)^3=\left( \frac{az}{H}\right)^3$:

$$\implies I_z={\int_0^H \int_0^{2\pi} \int_0^{\color{red}{a}}} \left( \frac{az}{H}\right)^3 dr d \theta dz$$

However, this seems to be wrong and I know the correct way to write the integral is this:

$$I_z={\int_0^H \int_0^{2\pi} \int_0^{\color{red}{\frac{az}{H}}}} r^3 dr d \theta dz$$

I don't understand what's wrong with my method and I don't really understand why the upper bound of the integral w.r.t $r$ is $\frac{az}{H}$? Why is it not just the radius $a$ of the base? What am I getting wrong here?

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For a differential mass $dm$, the moment of inertia is $r^2dm$ or $\rho r^2dV$. In cylindrical coordinates $dV=r \ dr\ d\varphi \ dz$. So:

$$dI=\rho r^3\ dr\ d\varphi \ dz$$

$$I=\rho\int_0^{2\pi} d\varphi\int_0^H dz\int_0^{az \over H} r^3\ dr$$

Why do you have to put $\frac{az}{H}$ as the upper limit? If you put just $a$ you would actually calculate the moment inertia of a cylinder of radius $a$! Look at the range of values that $r$ can take when you move $dm$ inside the cone. At level $z=0$ the maximum distance of $dm$ from the vertical axis is 0. At level $z=H$ the value of $r$ varies between 0 and $a$. But to calculate the integral properly, you need limits for $r$ in an arbitrary position defined by an arbitrary value of $z$ (next variable in the order of integration). For arbitrary height $z$ the distance of $dm$ from the vertical axis varies from 0 to $az/H$ and that is the upper limit of the third integral..

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  • $\begingroup$ Ah of course! That makes perfect sense! Thank you very much for your answer. $\endgroup$ – Nullspace Dec 9 '18 at 18:37

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