Question

Use a double integral in polar coordinates to prove that the volume of a sphere with radius $R$is$\frac{4}{3}\pi R^3$.

Short answer

The volume of a sphere is $V=\frac{4}{3}\pi R^3$

Step-by-step solution

Given information

The objective of this problem is to use double integral to prove that the volume of the sphere with radius $R$is$\frac{4}{3}\pi R^3$.

Calculation

In Cartesian system the equation of a sphere with radius $R$is

$$\begin{gathered} x^2+y^2+z^2=R^2 \\ {z}^2=R^2-x^2-y^2 \\ z=\sqrt{R^2-x^2-y^2} \end{gathered}$$

Put $x=r\cos \theta$and $y=r\sin \theta$

$z=\sqrt{R^2-r^2}$

Volume of solid in double integration

$V=\iint zdxdy$

In polar form

$V={\int }_0^{2\pi }{\int }_{-R}^2\sqrt{R^2-r^2}rdrd\theta$

Volume of a sphere is symmetrical about any axis.

Therefore,

$V=2{\int }_0^{2\pi }{\int }_0^z\sqrt{R^2-r^2}rdrd\theta$

Here, ${\theta }_1=0,{\theta }_2=2\pi$and $r_1=-R,r_2=R$

Put $R^2-r^2=t^2$

$$\begin{gathered} -2rdr=2tdt \\ rdr=-tdt \end{gathered}$$

For $r=0,t=R$and for $r=R,t=0$

Then$$\begin{gathered} V=2{\int }_0^{2z}{\int }_{R^0}^0{t}^2(-dt)d\theta \\ V=2{\int }_0^{2\pi }\left[{\int }_0^{\pi }{t}^{2}dtd\theta \right. \\ V=2{\int }_0^{2\pi }{\left[\frac{t^3}{3}\right]}_0^{R}d\theta \\ V=2{\int }_0^{2\pi }\left[\frac{R^3}{3}\right]\theta \\ V=2\frac{R^3}{3}{\int }_0^{2r}d\theta \\ V=\frac{2R^3}{3}[\theta {]}_0^{2\pi } \\ V=\frac{4}{3}\pi R^3 \end{gathered}$$

Thus, the volume of a sphere is

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