Q 74. - 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

Volume can be evaluated using the relation between cylindrical and cartesian coordinates.

Step 1: Given Information

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

Step 2: Evaluation of limits

We know that

$r=\sqrt{x^2+y^2}, \tan \theta =\frac{y}{x}, z=z$

and

$x=r\cos \theta , y=r\sin \theta , z=z$

Cylindrical limits are $-R\le r\le R, 0\le \theta \le 2\pi ,-\sqrt{R^2-r^2}\le z\le \sqrt{R^2-r^2}$

Step 3: Calculation of Volume

The volume of sphere is

$V={\iiint }_{V}rdrd\theta dz$

$={\int }_{r=-R}^R{\int }_{\theta =0}^{2\pi }{\int }_{z=-\sqrt{R^2-r^2}}^{\sqrt{R^2-r^2}}rdrd\theta dz$

$={\int }_{r=-R}^R{\int }_{\theta =0}^{2\pi }2\sqrt{R^2-r^2}rdrd\theta$

$=\left\{{\int }_{\theta =0}^{2\pi }d\theta \right\}\times \left\{(-1){\int }_{r=0}^R\sqrt{R^2-r^2}(-2r)dr\right\}$

Apply the limits

$V=\left(2\pi \right)\times {\left.\left\{(-1)\frac{{\left(R^2-r^2\right)}^{3/2}}{3/2}\right\}\right|}_{r=0}^R$

$V=\left(\frac{4\pi }{3}\right)\left(0-\left(-R^3\right)\right)$

$V=\left(\frac{4\pi R^3}{3}\right)$

Hence proved.