Answer to Question #259142 in Calculus for Debraj Acharjee

Question

Q.1: Apply Gauss's Divergence theorem to evaluate double integration S (lx^ 2 +my^ 2 +nz^ 2 )dS taken over the sphere (x-a)^ 2 +(y-b)^ 2 +(z-c)^ 2 = rho^ 2 ;l,m,n being the direction cosines of the external normal to the sphere.

Answer 1

By Gauss divergence theorem\\$\iint_s\bar{F.\bar{n}}\space ds=\iiint_v\space div\space dv$

where $\bar{F}.\bar{n}=lx^2+my^2+nz^2$

let $\bar{n}=li+mj+nk\space then \space \bar{F}=xi^2+y@j+n62\bar{k}$

so that div $\bar{F}=\frac{\delta}{\delta x}(x^2)+\frac{\delta}{\delta y}(y^2)+\frac{\delta}{\delta z}(z^2)$

$=2x+2y+2z\\=2(x+y+z)$

$\iiint_v div \bar{F} div=\iiint_v 2(x+y+z)dxdydz....(1)$

now consider $\iiint_v 2x\space dx\space dy\space dz$

use spherical coordinate

$x-a=sin \theta\space cos\theta\\ y-b=sin \theta\space sin\theta\\z-c=cos\theta$

$dx\space dy\space dz=sin \theta\space cos \theta$

and $0≤\theta≤\pi,0=\theta≤2\pi,0≤e$

$\iiint_v 2x\space dx\space dy\space dz=\int^{\pi }_0 \int ^{2\pi}_0 \int ^e_0 2(a+r sin \theta cos \theta)r^2 sin\theta dvd \theta d \theta$

$\int^{\pi }_0 \int ^{2\pi}_0 \int ^e_0 2ar^2+ sin \theta dvd \theta d \theta+ \int^{\pi }_0 \int ^{2\pi}_0 \int ^e_0 r^3+ sin ^2\theta dvd \theta d \theta$

$=-cos\int ^{\pi}_0 (2ae^3)\int ^{2\pi}_0 d\theta+{\int\pi}_0(\frac{1-cos \theta}{2}) d\theta(sin\theta)\int^{2\pi}_0(\frac{e^4}{4})$

$=2\frac{2ae^3}{3}(2\pi-0)\\=\frac{8}{3}a\pi e^3$

similarly$\iiint 2y\space dx\space dy\space dz=\frac{8}{3}b \pi e^3$

and $\iiint 2z\space dx\space dy\space dz=\frac{8}{3}c \pi e^3$

by adding these three integrals we get

$\iint_s\bar{F.\bar{n}}\space ds=\iiint_v\space div\space \bar{F} dv=8\pi e^3(a+b+c)$