Q 55. - Calculus

Question

Find the volume using integrals:

The region bounded above by the sphere with equation $\rho =R$ and bounded below by the cone with equation $\varphi =\alpha$. Explain why the volume should be zero if

$\alpha =0$ and $\frac{4}{3}\pi R^3$ if $\alpha =\mathrm{\pi }$.

Step-by-Step Solution

Answer

If $\alpha =0$, solid is empty.

If $\alpha =\mathrm{\pi }$, solid is a sphere.

Step 1: Given Information

The given equations are $\rho =R$ bounded above the sphere and $\varphi =\alpha$bounded below the cone.

Step 2: Evaluation of limits and simplification

Limits of spherical coordinates are

$0<\theta <2\pi , 0<\varphi <\alpha , 0<\rho <R$

We will use spherical coordinates to find the required volume as per given conditions.

The relation between rectangular and spherical coordinates is:

$x=\rho \sin \varphi \cos \theta , y=\rho \sin \varphi \sin \theta , z=\rho \cos \varphi$

and $\rho =\sqrt{x^2+y^2+z^2}, \tan \theta =\frac{y}{x}, \cos \varphi =\frac{z}{\rho }, dxdydz={\rho }^2\sin \varphi d\rho d\varphi d\theta$

Step 3: Calculation of Volume

The required volume is

$V={\iiint }_{V}dxdydz$

$V={\int }_{\varphi =0}^a{\int }_{\rho =0}^R{\int }_{\theta =0}^{2\pi }{\rho }^2\sin \varphi d\rho d\varphi d\theta$

$V={\int }_{\varphi =0}^{\alpha }\sin \varphi d\varphi {\int }_{\rho =0}^R{\rho }^{2}d\rho {\int }_{\theta =0}^{2\pi }d\theta$

$V=\left\{{\left.(-\cos \varphi )\right|}_0^{\alpha }\right\}\left\{{\left.\frac{{\rho }^3}{3}\right|}_{\rho =0}^{\rho =R}\right\}\left\{{\left.\theta \right|}_{\theta =0}^{2\pi }\right\}$

Application limits yields

$V=\left\{\right(1-\cos \alpha \left)\right\}\left\{\frac{R^3}{3}\right\}\left\{2\pi \right\}$

$V=\frac{2}{3}\pi R^3(1-\cos \alpha )$

When $\alpha =0, \cos \alpha =\cos 0=1$, volume of solid is zero (empty solid)

When $\alpha =\mathrm{\pi }, \mathrm{cos\alpha }=\mathrm{cos\pi }=-1$, Volume is $\frac{4}{3}{\mathrm{\pi R}}^3$ resulting is solid sphere.