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 $\pi$?

  • $\begingroup$ Are you actually asking why $\hat r$ is squared? If it is squared then clearly $r\sin\phi$ is squared. $\endgroup$
    – G. Smith
    Commented Apr 8, 2020 at 19:43
  • $\begingroup$ Are you familiar with spherical coordinates? $\endgroup$ Commented Apr 8, 2020 at 19:43
  • $\begingroup$ Why do you have a $V$ in there? $\endgroup$
    – G. Smith
    Commented Apr 8, 2020 at 19:44
  • $\begingroup$ 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. $\endgroup$
    – G. Smith
    Commented Apr 8, 2020 at 19:46
  • $\begingroup$ @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. $\endgroup$ Commented Apr 9, 2020 at 1:44

2 Answers 2


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: $\hat r^2=(r\sin(\phi))^2$.

Regarding the integration limits for this sphere parameterization, think of it like this:

  • 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 $R$.
  • $\begingroup$ Thank you! This answers everything I needed! $\endgroup$ Commented Apr 9, 2020 at 1:46

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.


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