# Moment of Inertia of a Solid Sphere

I just seen the following derivation of I for a solid sphere about an axis through the center of the sphere:

$$I= \int V\hat{r}^2dm =\int_0^\pi\int^{2\pi}_0\int^{R}_0\rho(r\sin\phi)^2r^2\sin\phi\,drd\theta d\phi =\frac{2\pi\rho}{5}R^5\int^\pi_0\sin^3(\phi) d\phi$$

Where $$\hat{r} = r\sin\phi$$ is defined as the distance of any point from the axis. My question is, why is $$r\sin\phi$$ squared and why is the bounds on $$\phi$$ from $$0$$ to $$\pi$$?

• Are you actually asking why $\hat r$ is squared? If it is squared then clearly $r\sin\phi$ is squared. Commented Apr 8, 2020 at 19:43
• Are you familiar with spherical coordinates? Commented Apr 8, 2020 at 19:43
• Why do you have a $V$ in there? Commented Apr 8, 2020 at 19:44
• Does your book really use $\phi$ for the polar angle and $\theta$ for the azimuthal one? This is backwards from what I am used to seeing. Commented Apr 8, 2020 at 19:46
• @G.Smith My book uses them the other way around as well, this is a proof I actually found online, sorry! I am asking why $\hat{r}$ is squared, and the V is meant to be for volume I think, but I couldn't decode this proof well enough. Commented Apr 9, 2020 at 1:44

A point-particle has the moment-of-inertia $$I=m \hat r^2$$, where $$m$$ is the particle's mass and $$\hat r$$ the distance from the rotational axis.
Your integral sums up all the values of $$I$$ for each of the infinitely many point-particles that the sphere consists of. Since $$\hat r=r\sin(\phi)$$, then when plugged into the formula for $$I$$ you get it squared: $$\hat r^2=(r\sin(\phi))^2$$.
• We integrate over the parameter $$\theta$$ from $$0$$ to $$2\pi$$, to draw a full circle.
• Then we "flip" that circle over in order to form / sweep out a sphere (a spherical shell). If you have a circle, you only have to rotate it (about an axis through the circle centre and parallel to a tangent) half a round in order to have swept through a spherical space. So, we only integrate from $$0$$ to $$\pi$$, which is half a round.
• Finally, the parameter $$r$$ takes care of the radius, and by integrating from $$0$$ to $$R$$, we "fill out" the sphere up to the radius $$R$$.
They are using spherical coordinates to do the integral, not cylindrical ones. The notation is a bit confusing as they use the same letter r, to denote distance from origin in $$r$$ and to denote distance from the rotation axis in $$\hat{r}$$. The distance from the rotation axis is $$rsin\phi$$, which is squared. The other term ($$r^2sin\phi$$) is the Jacobian.