Question

Find r.ns where s is the surface of the sphere x^2+y^2+z^2=a

Answer 1

$$Solution$$

To find r.ns where s is the surface of the sphere $x^2+y^2+z^2=a$

We know that $x^2+y^2+z^2=a^2$

Let $a=1$ because $a^2=a=1$

We will compute the surface integral $\intop \intop_s x^2 \ ds$ where s is the unit sphere.

Apply the parametric representation

x=sin\ \theta\ cos\ \phi;\ y=sin\ \theta\ sin\ \phi;\ z=cos\ \phi\\ 0 \le \phi \le\pi,\ 0 \le \theta \le2\pi,\

That is

$$r(\phi, \theta)=sin\ \phi\ cos\ \theta\ i+sin\ \phi\ sin\ \theta\ j+cos\ \phi\ k$$

We then compute

$|r_{\phi} \times r_{\theta}|=sin\ \theta\\ \therefore \intop \intop_Sx^2 dS\\ \implies \intop \intop_D(sin \phi cos \theta)^2|r_{\phi} \times r_{\theta}|dA\\ \implies \intop_0^{2 \pi} \intop_0^{\pi}(sin^2 \phi cos^2 \theta sin \phi)d \phi d\theta\\ \implies \intop_0^{2 \pi}cos^2 \theta d\theta \intop_0^{\pi}(sin^3 \phi )d \phi\\ \implies \intop_0^{2 \pi}\frac12(1+cos2 \theta) d\theta \intop_0^{\pi}(sin \phi-sin \phi\ cos^2 \phi )d \phi\\ \implies \frac12[\theta+\frac12 sin \theta]_0^{2\pi}[-cos \phi + \frac13cos \phi]_0^{\pi}\\ \therefore \frac{4 \pi}{3}$