Vi använder samma teknik som i videon med rymdpolära koordinater: \[ V=\iiiint_B1\, dxdydzdw= \] \[ \int_{0}^{2\pi}\!\!\int_{0}^{\pi}\int_{0}^{\pi}\int_0^1 r^3\sin^2\psi \sin\theta drd\psi\theta\phi= \] \[ \int_{0}^{2\pi}d\phi \int_{0}^{\pi} \sin\theta\, d\theta \int_{0}^{\pi}\sin^2\psi\, d\psi \int_0^1r^3\, dr= \] \[ 2\pi \Big[-\cos\theta\Big]_{0}^{\pi}\Big[\frac12(\psi-\sin\psi)\Big]_{0}^{\pi}\Big[\frac14 r^4\Big]_{0}^{1}= \] \[ 2\pi\cdot 2\cdot \frac{\pi}{2}\cdot \frac14=\frac{\pi^2}{2}. \]