Q 52. - Calculus

Question

Find the volume of solid of region bounded above by the sphere with equation $\rho =2$ and bounded below by the cone with equation $\varphi =\frac{\pi }{3}$.

Step-by-Step Solution

Answer

The required volume is $V=\frac{8}{3}\mathrm{\pi }$ units.

Step 1: Given Information

The given equations are $\rho =2$ and $\varphi =\frac{\pi }{3}$.

Step 2: Evaluation of limits

We know that

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

and

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

Limits of spherical coordinates are

$0<\theta <2\pi ,0<\varphi <\frac{\pi }{3},0<\rho <2$

To find the volume as per given conditions, we will use spherical coordinates.

Step 3: Calculation of Volume

Required Volume is $V={\iiint }_{V}dxdydz$

$V={\int }_{\varphi =0}^{\pi /3}{\int }_{\rho =0}^2{\int }_{\theta =0}^{2\pi }{\rho }^2\sin \varphi d\rho d\varphi d\theta$

$V={\int }_{\varphi =0}^{\pi /3}\sin \varphi d\varphi {\int }_{\rho =0}^2{\rho }^{2}d\rho {\int }_{\theta =0}^{2\pi }d\theta$

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

Application of limits yields

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

Hence, $V=\frac{8}{3}\pi$ units