The moment of inertia of a shape about an axis of rotation is equals to the integral of the density \( \rho \) times the squared distance of each point \( \mathbf{p} \) from the axis over the entire volume of the shape. Let \( \mathbf{r} \) be the vector connecting \( \mathbf{p} \) to its closest point on the axis of rotation, then
\[\iiint \rho (x,y,z) \, \| \mathbf{r} (x,y,z) \|^2 \, \mathrm{d}V\]Consider a cylinder of radius \( R \) and length \( L \) centered at the origin with its main axis (i.e. axis of rotational symmetry) aligned with the \( x \) axis on a right-hand coordinate system where \( y \) points up.
The density \( \rho \) is assumed to be constant.
A point on the volume of this cylinder in the three-dimensional Euclidean space can be described in cylindrical coordinates:
\[\mathbf{p}(r,\theta,\ell) = (p_x, p_y, p_z) = (\ell,r \sin \theta, r \cos \theta)\]The squared length of the vector connecting \( \mathbf{p} \) to its closest point on the axis of rotation \( x \) is
\[\| \mathbf{r} \|^2 = p_y^2 + p_z^2\]Thus, the moment of inertia along the \( x \) axis is:
\[I_{xx} = \rho \int _{-{\frac {L}{2}}}^{\frac {L}{2}} \int _{0}^{2\pi} \int _{0}^{R} \left( p_y^2 + p_z^2 \right) \, r \, \mathrm{d}r \, \mathrm{d}\theta \, \mathrm{d}\ell\]Note the use of \( r \, \mathrm{d}r \, \mathrm{d}\theta \) for the integral in polar coordinates. Missing \( r \) here is a common mistake. Developing the integral further, we get
\[\rho \int _{-{\frac {L}{2}}}^{\frac {L}{2}} \int _{0}^{2\pi} \int _{0}^{R} \left( r^2 \cos^2 \! \theta + r^2 \sin^2 \! \theta \right) \, r \, \mathrm{d}r \, \mathrm{d}\theta \, \mathrm{d}\ell\] \[\rho \int _{-{\frac {L}{2}}}^{\frac {L}{2}} \int _{0}^{2\pi} \int _{0}^{R} r^2 \left(\cos^2 \! \theta + \sin^2 \! \theta \right) \, r \, \mathrm{d}r \, \mathrm{d}\theta \, \mathrm{d}\ell\]It’s well known that
\[\cos^2 \! \theta + \sin^2 \! \theta = 1\]thus
\[\rho \int _{-{\frac {L}{2}}}^{\frac {L}{2}} \int _{0}^{2\pi} \int _{0}^{R} r^3 \, \mathrm{d}r \, \mathrm{d}\theta \, \mathrm{d}\ell\] \[\rho \int _{-{\frac {L}{2}}}^{\frac {L}{2}} \int _{0}^{2\pi} \left. \frac {r^4}{4} \right|_{0}^{R} \, \mathrm{d}\theta \, \mathrm{d}\ell\] \[\rho \int _{-{\frac {L}{2}}}^{\frac {L}{2}} \int _{0}^{2\pi} \frac {R^4}{4} \, \mathrm{d}\theta \, \mathrm{d}\ell\] \[\rho \int _{-{\frac {L}{2}}}^{\frac {L}{2}} \left. \frac {\theta R^4}{4} \right|_{0}^{2\pi} \, \mathrm{d}\ell\] \[\rho \int _{-{\frac {L}{2}}}^{\frac {L}{2}} \frac {2 \pi R^4}{4} \, \mathrm{d}\ell\] \[\rho \left. \frac {\pi R^4 \ell}{2} \right|_{-{\frac {L}{2}}}^{\frac {L}{2}}\] \[\rho \, \frac {\pi R^4}{2} \left( \frac {L}{2} - \left(-{\frac {L}{2}}\right) \right)\] \[\rho \, \frac {\pi R^4}{2} \left( \frac {L}{2} + {\frac {L}{2}} \right)\] \[\rho \, \frac {\pi R^4 L}{2}\]The volume of the cylinder is the area of its base \( \pi R^2 \) times its length \( L \)
\[V = \pi R^2 L\]then
\[\rho \, \pi R^2 L \frac {R^2}{2}\] \[\rho \, V \frac {R^2}{2}\]The mass of the cylinder is
\[M = \rho V\]Finally
\[I_{xx} = \frac {M R^2}{2}\]The moment of inertia along the other two principal axes \( I_{yy} \) and \( I_{zz} \) are equal due to symmetry. Lets calculate \( I_{zz} \).
\[I_{zz} = \rho \int _{-\frac {L}{2}}^{\frac{L}{2}} \int _{0}^{2\pi} \int _{0}^{R} \left( p_x^2 + p_y^2 \right) \, r \, \mathrm{d}r \, \mathrm{d}\theta \, \mathrm{d}\ell\] \[\rho \int _{-\frac {L}{2}}^{\frac{L}{2}} \int _{0}^{2\pi} \int _{0}^{R} \left( \ell^2 + r^2 \sin^2 \! \theta \right) \, r \, \mathrm{d}r \, \mathrm{d}\theta \, \mathrm{d}\ell\] \[\rho \int _{-\frac {L}{2}}^{\frac{L}{2}} \int _{0}^{2\pi} \int _{0}^{R} \ell^2 r + r^3 \sin^2 \! \theta \, \mathrm{d}r \, \mathrm{d}\theta \, \mathrm{d}\ell\] \[\rho \int _{-\frac {L}{2}}^{\frac{L}{2}} \int _{0}^{2\pi} \left. \frac {\ell^2 r^2}{2} + \frac{r^4}{4} \sin^2 \! \theta \, \right|_{0}^{R} \, \mathrm{d}\theta \, \mathrm{d}\ell\] \[\rho \int _{-\frac {L}{2}}^{\frac{L}{2}} \int _{0}^{2\pi} \frac{\ell^2 R^2}{2} + \frac{R^4}{4} \sin^2 \! \theta \, \mathrm{d}\theta \, \mathrm{d}\ell\]The integral of \( \sin^2 \! \theta \) is (from Wikipedia)
\[\int \sin^2 \! \theta \, \mathrm{d}\theta = \frac{\theta}{2} - \frac{1}{4} \sin 2 \theta\]thus
\[\rho \int _{-\frac {L}{2}}^{\frac{L}{2}} \left. \frac{\ell^2 R^2 \theta}{2} + \frac{R^4}{4} \left( \frac{\theta}{2} - \frac{1}{4} \sin 2 \theta \right) \right|_{0}^{2\pi} \, \mathrm{d}\ell\] \[\rho \int _{-\frac {L}{2}}^{\frac{L}{2}} \frac{2 \pi R^2 \ell^2}{2} + \frac{R^4}{4} \left( \frac{2 \pi}{2} - \frac{1}{4} \sin 4 \pi \right) \, \mathrm{d}\ell\]Given that \( \sin 4 \pi = 0 \), we continue like so
\[\rho \int _{-\frac {L}{2}}^{\frac{L}{2}} \pi R^2 \ell^2 + \frac{\pi R^4}{4} \, \mathrm{d}\ell\] \[\rho \left( \left. \frac{\pi R^2 \ell^3}{3} + \frac{\pi R^4 \ell}{4} \right|_{-\frac {L}{2}}^{\frac{L}{2}} \right)\] \[\rho \left( \frac{\pi R^2}{3} \left( \left( \frac{L}{2} \right)^3 - \left( -\frac{L}{2} \right)^3 \right) + \frac{\pi R^4}{4} \left(\frac{L}{2} - \left( -\frac{L}{2} \right) \right) \right)\] \[\rho \left( \frac{\pi R^2}{3} \left( \frac{L^3}{8} - \left( -\frac{L^3}{8} \right) \right) + \frac{\pi R^4}{4} \left(\frac{L}{2} + \frac{L}{2} \right) \right)\] \[\rho \left( \frac{\pi R^2}{3} \left( \frac{L^3}{8} + \frac{L^3}{8} \right) + \frac{\pi R^4 L}{4} \right)\] \[\rho \left( \frac{\pi R^2}{3} \frac{L^3}{4} + \frac{\pi R^4 L}{4} \right)\] \[\rho \left( \frac{\pi R^2 L^3}{12} + \frac{\pi R^4 L}{4} \right)\] \[\rho \frac{\pi R^2 L^3 + 3 \pi R^4 L}{12}\] \[\rho \pi R^2 L \frac{L^2 + 3 R^2}{12}\]Finally, using our knowledge about the cylinder volume and mass from above
\[I_{zz} = I_{yy} = \frac{M}{12} \left( L^2 + 3 R^2 \right)\]And we’re done. I hope it was clear.