Answer to Question #268883 in Calculus for Bhuvana

Question

Perform the integration by transforming the ellipsoidal region of integration into a spherical region of integration and

then evaluating the transformed integral in spherical coordinates.

(a) RRR

G

x

2 dV , where G is the region enclosed by the ellipsoid 9x

2 + 4y

2 + z

2 = 36.

(b) RRR

G

(y

2 + z

2

) dV , where G is the region enclosed by the ellipsoid x

2

a

2

+

y

2

b

2

+

z

2

c

2

= 1

Answer 1

$\int\int\int_G x^2dv$

where G is $9x^2+4y^2+z^2=36$

$\frac{x^2}{4}+\frac{y^2}{9}+\frac{z^2}{36}=1$

let $x=2u, y=3v, z=6w$

$(\frac{x}{2})^2+(\frac{y}{3})^2+(\frac{z}{6})^2=1\implies u^2+v^2+w^2=1$

then

$\int\int\int_G x^2dv=\int_u\int_v\int_w f(u,v,w)|j|dudvdw$

where

$f(u,v,w)=4u^2$

$|j|=\frac{\delta(x,y,z)}{\delta(u,v,w)}=\begin{vmatrix} x_u & x_v & x_w \\ y_u & y_v&y_w\\ z_u & z_v& z_w \end{vmatrix}=\begin{vmatrix} 2 & 0& 0 \\ 0 & 3&0\\ 0 & 0& 6 \end{vmatrix}=36$

$=\int_u\int_v\int_w 144u^2dudvdw$

let $u=rcos\theta sin\phi, v=rsin\theta sin\phi,w=rcos\theta$

$u^2+v^2+w^2=1\implies r^2=1$

$\int_u\int_v\int_w 144u^2dudvdw=\int_r\int_\theta\int_\phi 144r^2cos^2\theta sin^2\phi(r^2sin\theta)drd\theta d\phi$

$$=\int_{r=0}^1\int_{\theta=0}^\pi\int_{\phi=0}^{2\pi} 144r^2(sin\theta cos^2\theta)(sin^2\phi)drd\theta d\phi$$

$$=144[\frac{r^5}{5}]_0^1[\frac{\phi}{2}-\frac{sin2\phi}{4}]_0^{2\pi}(\int_{\theta=0}^\pi(sin\theta cos^2\theta)d\theta)$$

$$=144(\frac{1}{5})(\frac{\pi}{2})(\frac{2}{3})=\frac{144\pi}{15}$$

$$=\frac{144\pi}{15}$$