Question

Use a transformation of variables to find the volume of the region bounded by√x+√y+√z= 1 and the coordinate planes.

Answer 1

Denote by D the region volume is to be calculated. Volume D is V_D=$\underset { D }{ \iiint } dxdydz$ . To calculate the integral we use the change of variables :

$\begin{matrix} x=r\cos ^{ 4 }{ \varphi \sin ^{ 4 }{ \theta } } \\ y=r\sin ^{ 4 }{ \varphi \sin ^{ 4 }{ \theta } } \\ z=r\cos ^{ 4 }{ \theta } \end{matrix}$ . (1)

$\sqrt { x } +\sqrt { y } +\sqrt { z } =\sqrt { r } =1$

The image of the region D is the region Δ $=\left\{ \left( r,\varphi ,\theta \right) :\quad 0\le r\le 1,\quad 0\le \varphi \le \frac { \pi }{ 2 } ,0\le \theta \le \frac { \pi }{ 2 } \right\}$

$\underset { D }{ \iiint } dxdydz\quad =\underset { \Delta }{ \iiint } \quad \left| \frac { \partial (x,y,z) }{ \partial (r,\varphi ,\theta ) } \right| drd\varphi d\theta$ , where $\frac { \partial (x,y,z) }{ \partial (r,\varphi ,\theta ) } =det\left( \begin{matrix} \cos ^{ 4 }{ \varphi \sin ^{ 4 }{ \theta } } & -4r\cos ^{ 3 }{ \varphi \sin { \varphi } \sin ^{ 4 }{ \theta } } & 4r\cos ^{ 4 }{ \varphi \sin ^{ 3 }{ \theta } \cos { \theta } \quad } \\ \sin ^{ 4 }{ \varphi \sin ^{ 4 }{ \theta } } & 4r\sin ^{ 3 }{ \varphi \cos { \varphi } \sin ^{ 4 }{ \theta } } & 4r\sin ^{ 4 }{ \varphi \sin ^{ 3 }{ \theta \cos { \theta } } } \\ \cos ^{ 4 }{ \theta } & 0 & -4r\cos ^{ 3 }{ \theta } \sin { \theta } \end{matrix} \right) \quad$ =

$=-16{ r }^{ 2 }\cos ^{ 3 }{ \theta } \cos ^{ 3 }{ \varphi } \sin ^{ 3 }{ \varphi } \sin ^{ 7 }{ \theta }$ .

V_D= $\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \int _{ 0 }^{ \frac { \pi }{ 2 } }{ 16{ r }^{ 2 } } \cos ^{ 3 }{ \theta } \cos ^{ 3 }{ \varphi } \sin ^{ 3 }{ \varphi } \sin ^{ 7 }{ \theta } drd\varphi d\theta \quad =\frac { 16 }{ 3 } \cdot \frac { 1 }{ 12 } \cdot } \frac { 1 }{ 40 } } =\frac { 1 }{ 90 }$ , because

$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos ^{ 3 }{ \varphi } \sin ^{ 3 }{ \varphi } d\varphi } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos ^{ 2 }{ \varphi } \sin ^{ 3 }{ \varphi } d\left( \sin { \varphi } \right) } =\int _{ 0 }^{ 1 }{ \left( 1-{ t }^{ 2 } \right) { t }^{ 3 } } dt=\frac { 1 }{ 4 } -\frac { 1 }{ 6 } =\frac { 1 }{ 12 } \quad \\ \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos ^{ 3 }{ \theta \sin ^{ 7 }{ \theta d\theta } } } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos ^{ 2 }{ \theta \sin ^{ 7 }{ \theta d\left( \sin { \theta } \right) } } } =\int _{ 0 }^{ 1 }{ \left( 1-{ t }^{ 2 } \right) { t }^{ 7 }dt } =\frac { 1 }{ 8 } -\frac { 1 }{ 10 } =\frac { 1 }{ 40 }$

$\int _{ 0 }^{ 1 }{ 16{ r }^{ 2 } } dr=\frac { 16 }{ 3 }$

ANSWER The volume is $\frac{1}{90}$.

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