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2022-06-29

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

Blaine Foster

Beginner2022-06-30Added 33 answers

It's better to use spherical coordinates

The region is ${x}^{2}+{y}^{2}+{z}^{2}\le 2z$ , i.e., ${x}^{2}+{y}^{2}+(z-1{)}^{2}\le 1$

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

$I={\displaystyle {\int}_{r=0}^{1}{\int}_{\theta =0}^{\pi}{\int}_{\varphi =0}^{2\pi}{r}^{6}{\mathrm{sin}}^{5}\theta {\mathrm{cos}}^{2}\varphi {\mathrm{sin}}^{2}\varphi \text{d}\varphi \text{d}\theta \text{d}r}$

Integrating with respect r and with respect to $\varphi $ is straight forward, and the integral reduces to

$I={\displaystyle \frac{\pi}{28}}{\displaystyle {\int}_{\theta =0}^{\pi}si{n}^{5}\theta \text{d}\theta}$

The integral with respect to $\theta $ is solved using the substitution $u=\mathrm{cos}\theta $, then

$\int {\mathrm{sin}}^{5}\theta \text{d}\theta =-\int (1-{u}^{2}{)}^{2}du=-(u-{\displaystyle \frac{2}{3}}{u}^{3}+{\displaystyle \frac{{u}^{5}}{5}})$

Hence,

$I={\displaystyle \frac{4\pi}{105}}$

The region is ${x}^{2}+{y}^{2}+{z}^{2}\le 2z$ , i.e., ${x}^{2}+{y}^{2}+(z-1{)}^{2}\le 1$

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

$I={\displaystyle {\int}_{r=0}^{1}{\int}_{\theta =0}^{\pi}{\int}_{\varphi =0}^{2\pi}{r}^{6}{\mathrm{sin}}^{5}\theta {\mathrm{cos}}^{2}\varphi {\mathrm{sin}}^{2}\varphi \text{d}\varphi \text{d}\theta \text{d}r}$

Integrating with respect r and with respect to $\varphi $ is straight forward, and the integral reduces to

$I={\displaystyle \frac{\pi}{28}}{\displaystyle {\int}_{\theta =0}^{\pi}si{n}^{5}\theta \text{d}\theta}$

The integral with respect to $\theta $ is solved using the substitution $u=\mathrm{cos}\theta $, then

$\int {\mathrm{sin}}^{5}\theta \text{d}\theta =-\int (1-{u}^{2}{)}^{2}du=-(u-{\displaystyle \frac{2}{3}}{u}^{3}+{\displaystyle \frac{{u}^{5}}{5}})$

Hence,

$I={\displaystyle \frac{4\pi}{105}}$

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