Moment of Inertia

Starting from the definition $$I = \int r^2\,\mathrm{d}m$$ we replace $\mathrm{d}m$ with $\varrho\,\mathrm{d}V$. Additionally, we can insert $r^2\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\varphi$ for $\mathrm{d}V$ for the case of a sphere which leads to $$I = \varrho \int_0^R\,\mathrm{d}r \int_0^\pi\mathrm{d}\theta\int_0^{2\pi}\mathrm{d}\varphi\,r^4 \sin^3\theta$$ Calculating the first and last integral leads to $$I = \frac{2}{5}\pi\varrho R^5\int_0^\pi \sin^3\theta \,\mathrm{d}\theta$$ Now we can substitute $\cos\theta$ with $u$ which results in $\mathrm{d}u = -\sin\theta\,\mathrm{d}\theta$. This then finally leads to $$I = \frac{2}{5}\frac{3}{4}\pi\varrho R^5$$ After using the formula of the volume of a sphere given by $V = \frac{4}{3}\pi R^3$ we obtained $$I = \frac{2}{5}MR^2$$ for the momentum of inertia for a solid sphere.