Q 19. - Calculus

Question

Show that the mass of $\in$ is $\frac{1}{6}\mathrm{\pi k}$ by evaluating the integral:

$\int \int {\int }_{\in }kdV={\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^{1}k{\rho }^2\sin \varphi d\rho d\theta d\varphi$

Step-by-Step Solution

Answer

Use spherical coordinates $\left(\rho ,\theta ,\varphi \right)$ while evaluating $dV$ using triple integral.

Step 1: Given Information

Spherical coordinates are $\left(\rho ,\theta ,\varphi \right)$. We need to prove this by solving given integral.

Step 2: Simplification

Taking LHS.

$m=\int \int {\int }_{\in }kdV$

$={\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^{1}k{\rho }^2\sin \varphi d\rho d\theta d\varphi$

$=k{\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^{\frac{\mathrm{\pi }}{2}}\left({\int }_0^1{\rho }^2\sin \varphi d\rho \right)d\theta d\varphi$

$=k{\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^{\frac{\mathrm{\pi }}{2}}{{\left(\frac{{\rho }^3}{3}\right)}^1}_0\sin \varphi d\theta d\varphi$

$=k{\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{3}\sin \varphi d\theta d\varphi$

$=\frac{k}{3}{\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^{\frac{\mathrm{\pi }}{2}}\sin \varphi d\theta d\varphi$

$=\frac{k}{3}{\int }_0^{\frac{\mathrm{\pi }}{2}}{{\left[\theta \right]}^{\frac{\mathrm{\pi }}{2}}}_0\sin \varphi d\varphi$

$=\frac{k}{3}{\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{\mathrm{\pi }}{2}-0\right]\sin \varphi d\varphi$

$=\frac{k\mathrm{\pi }}{6}{{\left[-\cos \varphi \right]}^{\frac{\mathrm{\pi }}{2}}}_0$

$=-\frac{k\mathrm{\pi }}{6}\left[\cos \frac{\mathrm{\pi }}{2}-\cos 0\right]$

$=-\frac{k\mathrm{\pi }}{6}\left[0-1\right]$

$=\frac{\mathrm{\pi k}}{6}$

Hence, the mass is $\frac{1}{6}\mathrm{\pi k}$