Shell Potential Energy

Consider the given scenario which is analysed in spherical coordinates. Any point on the surface of the sphere is:

$x = \sin{\theta}\cos{\phi}$ $y = \sin{\theta}\sin{\phi}$ $z = \cos{\theta}$

A surface area element is:

$dS = \sin{\theta} \ d\theta \ d\phi$

Mass per unit area is:

$\rho = (xyz)^2 = \sin^4{\theta} \cos^2{\theta} \sin^2{\phi} \cos^2{\phi}$

Therefore the mass of an elementary surface area is:

$dm = \rho \ dS = \sin^5{\theta} \cos^2{\theta} \sin^2{\phi} \cos^2{\phi} \ d\theta \ d\phi$

The potential energy due to this mass is:

$dV = dm \ gz$

$dV = 10 \sin^5{\theta} \cos^3{\theta} \sin^2{\phi} \cos^2{\phi} \ d\theta \ d\phi$

$V = \int_{0}^{\pi/2} \int_{0}^{2\pi} 10 \sin^5{\theta} \cos^3{\theta} \sin^2{\phi} \cos^2{\phi} \ d\theta \ d\phi$

Since the limits of integration are constant:

$V = 10 \left(\int_{0}^{\pi/2} \sin^5{\theta} \cos^3{\theta} \ d\theta\right)\left(\int_{0}^{2\pi} \sin^2{\phi} \cos^2{\phi} \ d\phi\right)$ $\implies V = 10\left(\int_{0}^{\pi/2} \sin^5{\theta} \cos^3{\theta} \ d\theta\right)\left(4\int_{0}^{\pi/2} \sin^2{\phi} \cos^2{\phi} \ d\phi\right)$

Using the following formula:

$\int_{0}^{\pi/2} \sin^m{\theta} \cos^n{\theta} \ d\theta = \frac{\Gamma\left(\frac{m+1}{2}\right)\Gamma\left(\frac{n+1}{2}\right)}{2\Gamma\left(\frac{m+n+2}{2}\right)}$

Where $\Gamma (n)$ is the Gamma function. Applying the above gives:

$V = 40\left(\frac{\Gamma\left(\frac{5+1}{2}\right)\Gamma\left(\frac{3+1}{2}\right)}{2\Gamma\left(\frac{5+3+2}{2}\right)}\right)\left(\frac{\Gamma\left(\frac{2+1}{2}\right)\Gamma\left(\frac{2+1}{2}\right)}{2\Gamma\left(\frac{2+2+2}{2}\right)}\right)$

$\implies V = 40 \left(\frac{\Gamma\left(3\right)\Gamma\left(2\right)}{2\Gamma\left(5\right)}\right)\left(\frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{3}{2}\right)}{2\Gamma\left(3\right)}\right)$

Using the facts that for integers $\Gamma(n+1) = n!$ , and in general $\Gamma(n+1) = n\Gamma(n)$ and also the fact that $\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$ . Applying these results and simplifying gives the answer:

$\boxed{V = \frac{5\pi}{48} \approx 0.32725}$

$\textcolor{magenta}{Fun}$ $\textcolor{magenta}{Problem}$

$x = \cos \alpha \sin \beta \\ y = \sin \alpha \sin \beta \\ z = \cos \beta \\ 0 \leq \alpha \leq 2 \pi \\ 0 \leq \beta \leq \pi/2 \\ dS = \sin \beta \, d \alpha d \beta \\ dm = \rho \, dS$ Consider $G$ as Gravitational potential energy.