# Why $\int_m \left( y^2 + z^2 \right)dm \neq \frac{1}{3}mL^2$?

for a cylinder, let's say with constant density with radius $$3$$ and height $$10$$ so

$$\rho(r, \theta, y)=1$$

so $$dm = \rho\left(r, \theta, y\right)r \, dr \, d\theta \, dy = r \, dr \, d\theta \, dy$$

in that case the inertia at the end of the rod is (from here)

$$I_{xx} = \int_m \left( y^2 + z^2 \right) \, dm=10060.91$$

where $$z = rsin(\theta)$$

but (from here)

$$\frac{1}{3}mL^2 = 9424.77$$

but

$$I_{xx} = \int_m \left( y^2 + z \right) \, dm=9424.77$$

if $$z=sin(\theta)$$

why is this?

• In cylindrical coordinates, $z$ should be a direct variable, and the only angle should be azimuthal, so that $x=\rho \cos(\alpha)$ and $y=\rho \sin(\alpha)$. – FGSUZ Jul 31 at 19:33
• A typesetting details here is that you want the differentials to be identifiable as distinct entities. So you set them off with a thin space (dr\, d\theta to get $dr\, d\theta$). I'm one of the people who also like to typeset the "d" upright (so \mathrm{d}r to get $\mathrm{d}r$), but that is much less universal. – dmckee Jul 31 at 19:33
• @FGSUZ in my case $y$ is the direct variable and $x, z$ are not. As in this post math.stackexchange.com/questions/3259247/…. – fullnitrous Jul 31 at 19:43
• Oh, could you then draw the axes and the limits of integration? Thanks. By the way, I think there's a missing squaring in your last formula – FGSUZ Jul 31 at 19:49
• @FGSUZ on purpose because it works – fullnitrous Jul 31 at 20:13

So it looks like you are doing the integral $$\int_0^{10}\int_0^{2\pi}\int_0^3\left(y^2+r^2\sin^2\theta\right)r\,\text dr\,\text d\theta\,\text d y$$

So you are assuming your cylinder is oriented along the y-axis, and $$\theta=0$$ lines up with the positive x-axis. The value $$y^2+z^2$$ is the distance a point is from the x-axis. Therefore, based on your limits of integration your cylinder's base is in the x-z plane, and you are calculating the moment of inertia about the x-axis.

Looking at your link then, you want to use the equation in the lower left corner along with the parallel axis theorem, not $$I=1/3ML^2$$ for the thin rod. More explicitly: $$I_{xx}=\frac14MR^2+\frac1{12}ML^2+M\left(\frac12L\right)^2=\frac14MR^2+\frac13ML^2$$

This is the equation you want for your moment of inertia. Plug in your numbers and it all works out.

Why does your integral with the $$z$$ instead of $$z^2$$ work out to be $$1/3ML^2$$ like the rod on one end? Well if you don't square $$z$$ in the integrand, then this term integrates to $$0$$ due to the $$\sin\theta$$ term from $$0$$ to $$2\pi$$. Therefore the integral is equivalent to just integrating over $$y^2$$. This would be the same as if you were looking at a rod for $$R=0$$. This is why you get $$1/3ML^2$$ for your incorrect integral

• no you don't understand, the problem is that the integration gives another value that the equation for a special case. i need the integration to give me the same answer. i do not care about the special case equation, i am just using it for verification. also it is not the equation in the lower left for the origin is not in the middle. – fullnitrous Jul 31 at 22:17
• here is how the cylinder looks like imgur.com/V70wTa9 – fullnitrous Jul 31 at 22:22
• @fullnitrous Yes I deduced that figure from the integral. My answer is what you want. Please read it more carefully. – Aaron Stevens Jul 31 at 22:24
• but the equation in the bottom left has the rotation axis in the center which i do not. – fullnitrous Jul 31 at 22:25
• @fullnitrous Which is why you use that with the parallel axis theorem. As stated in my answer. – Aaron Stevens Jul 31 at 22:29