xonycutieoxl1

2022-06-29

Triple integration ${\iiint }_{{x}^{2}+{y}^{2}+{z}^{2}\le 2z}{x}^{2}{y}^{2}dxdydz$

Blaine Foster

It's better to use spherical coordinates
The region is ${x}^{2}+{y}^{2}+{z}^{2}\le 2z$ , i.e., ${x}^{2}+{y}^{2}+\left(z-1{\right)}^{2}\le 1$
So we can define the spherical coordinates $r,\theta ,\varphi$ such that
$x=r\mathrm{sin}\theta \mathrm{cos}\varphi$
$y=r\mathrm{sin}\theta \mathrm{sin}\varphi$
$z=1+r\mathrm{cos}\theta$
And the integral becomes
$I={\int }_{r=0}^{1}{\int }_{\theta =0}^{\pi }{\int }_{\varphi =0}^{2\pi }{r}^{6}{\mathrm{sin}}^{5}\theta {\mathrm{cos}}^{2}\varphi {\mathrm{sin}}^{2}\varphi \text{d}\varphi \text{d}\theta \text{d}r$
Integrating with respect r and with respect to $\varphi$ is straight forward, and the integral reduces to
$I=\frac{\pi }{28}{\int }_{\theta =0}^{\pi }si{n}^{5}\theta \text{d}\theta$
The integral with respect to $\theta$ is solved using the substitution $u=\mathrm{cos}\theta$, then
$\int {\mathrm{sin}}^{5}\theta \text{d}\theta =-\int \left(1-{u}^{2}{\right)}^{2}du=-\left(u-\frac{2}{3}{u}^{3}+\frac{{u}^{5}}{5}\right)$
Hence,
$I=\frac{4\pi }{105}$

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