User:Beneficii/hypervolume

${\displaystyle {\begin{aligned}H&=\int _{-r}^{r}{\frac {4}{3}}\pi \left({\sqrt {r^{2}-x^{2}}}\right)^{3}dx&=\int _{-r}^{r}{\frac {4}{3}}\pi r^{3}\left({\sqrt {1-{\frac {x^{2}}{r^{2}}}}}\right)^{3}dx&={\frac {4}{3}}\pi r^{3}\int _{-r}^{r}\left({\sqrt {1-{\frac {x^{2}}{r^{2}}}}}\right)^{3}dx\end{aligned}}}$

Let ${\displaystyle x=r\sin(u).}$ As a result, ${\displaystyle \arcsin \left({\frac {x}{r}}\right)=u}$ and ${\displaystyle dx=r\cos(u)du.}$

${\displaystyle {\begin{aligned}H&={\frac {4}{3}}\pi r^{3}\int _{arcsin\left({\frac {-r}{r}}\right)}^{arcsin\left({\frac {r}{r}}\right)}\left({\sqrt {1-{\frac {(r\sin(u))^{2}}{r^{2}}}}}\right)^{3}r\cos(u)du\\&={\frac {4}{3}}\pi r^{3}(r)\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\left({\sqrt {1-\sin ^{2}(u)}}\right)^{3}\cos(u)du\\&={\frac {4}{3}}\pi r^{4}\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}(cos(u))^{3}\cos(u)du\\&={\frac {4}{3}}\pi r^{4}\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}cos^{4}(u)du\\&={\frac {4}{3}}\pi r^{4}\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\left[{\frac {1}{8}}(3+4\cos(2u)+\cos(4u))\right]du\\&={\frac {4}{3}}\left({\frac {1}{8}}\right)\pi r^{4}\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}(3+4\cos(2u)+\cos(4u))du\\&={\frac {1}{6}}\pi r^{4}\left[3u+2\sin(2u)+{\frac {1}{4}}\sin(4u)\right]_{u=-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\\&={\frac {1}{6}}\pi r^{4}\left[\left(3\left({\frac {\pi }{2}}\right)+2\sin(\pi )+{\frac {1}{4}}\sin(2\pi )\right)-\left(3\left(-{\frac {\pi }{2}}\right)+2\sin(-\pi )+{\frac {1}{4}}\sin(-2\pi )\right)\right]\\&={\frac {1}{6}}\pi r^{4}\left[\left(3\left({\frac {\pi }{2}}\right)\right)-\left(3\left(-{\frac {\pi }{2}}\right)\right)\right]\\&={\frac {1}{6}}\pi r^{4}\left[\left(3\left({\frac {\pi }{2}}\right)\right)+\left(3\left({\frac {\pi }{2}}\right)\right)\right]\\&={\frac {1}{6}}\pi r^{4}\left({\frac {3\pi }{2}}\right)(2)\\&={\frac {1}{6}}\pi r^{4}(3\pi )\\&={\frac {1}{2}}\pi ^{2}r^{4}\end{aligned}}}$