Q 76. - Calculus

Question

Let $R$ be the radius of a sphere. Use cylindrical coordinates to set up and evaluate a triple integral proving that the volume of the sphere is $\frac{4}{3}{\mathrm{\pi R}}^3$.

Step-by-Step Solution

Answer

To evaluating the given integral, use the relation between cartesian and cylindrical coordinates.

Step 1: Given Information

It is given that radius of sphere is $R$.

Step 2: Evaluation of limits

To determine the above volume, we will use spherical coordinates.

The limits are $0<\theta <2\pi , 0<\varphi <\pi , 0<\rho <R$

Step 3: Calculation of Volume

We know the relation

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

and

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

Volume is evaluated as $V={\iiint }_{V}dxdydz$

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

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

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

After applying limits

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

$V=\frac{4}{3}\pi R^3$

Hence proved.