Chapter 6.3, Problem 35E

To determine

The volume of the solid (sphere) using cylindrical shell.

Answer to Problem 35E

The volume of the solid (sphere) is $\frac{4\pi r^3}{3}$.

Explanation of Solution

Given:

A sphere of radius $\left(r\right)$.

Calculation:

Consider $f\left(x\right)=\sqrt{r^2-x^2}$ (1)

The region lies between $x=0$ and $x=r$.

Show the equation as below:

$y=\sqrt{r^2-x^2}$ (2)

Plot a graph for the equation $y=\sqrt{r^2-x^2}$ using the calculation as follows:

Calculate x value using Equation (2)

Substitute 0 for x in Equation (2).

$\begin{array}{c}y=\sqrt{r^2-0^2}\\ =\sqrt{r^2}\\ =r\end{array}$

Hence, the co-ordinate of $\left(x,y\right)$ is $\left(0,r\right)$.

Calculate x value using Equation (2)

Substitute r for x in Equation (2).

$\begin{array}{c}y=\sqrt{r^2-r^2}\\ =0\end{array}$

Hence, the co-ordinate of $\left(x,y\right)$ is $\left(r,0\right)$.

Draw the cylindrical shell as shown in Figure 1.

Calculate the volume using the method of cylindricalshell:

$V={\displaystyle \underset{a}{\overset{b}{\int }}2\pi x\left[f\left(x\right)\right]dx}$ (3)

Substitute 0 for a, r for b, and $\sqrt{r^2-x^2}$ for $\left[f\left(x\right)\right]$ in Equation (3).

$\begin{array}{c}V={\displaystyle \underset{a}{\overset{b}{\int }}2\pi x\left[\sqrt{r^2-x^2}\right]dx}\\ =2\pi {\displaystyle \underset{a}{\overset{b}{\int }}x\left[\sqrt{r^2-x^2}\right]dx}\end{array}$ (4)

Consider $u=r^2-x^2$ (5)

Differentiate both sides of the equation.

$\begin{array}{l}du=-2xdx\\ xdx=-\frac{1}{2}du\end{array}$

Calculate the lower limit value of u using Equation (5).

Substitute 0 for x in Equation (5).

$\begin{array}{c}u=r^2-0^2\\ =r^2\end{array}$

Calculate the upper limit value of u using Equation (5).

Substitute r for x in Equation (5).

$\begin{array}{c}u=r^2-r^2\\ =0\end{array}$

Apply lower and upper limits for u in Equation (3).

Substitute u for $\left(r^2-x^2\right)$ and $-\frac{1}{2}du$ for $xdx$ in Equation (3).

$\begin{array}{c}V=2\pi {\displaystyle \underset{r^2}{\overset{0}{\int }}\sqrt{u}\left(-\frac{1}{2}du\right)}\\ =-\frac{2\pi }{2}{\displaystyle \underset{r^2}{\overset{0}{\int }}\sqrt{u}du}\end{array}$

$=-\pi {\displaystyle \underset{r^2}{\overset{0}{\int }}\sqrt{u}du}$ (6)

Integrate Equation (6).

$\begin{array}{c}V=-\pi {\left[\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_{r^2}^0\\ =-\pi {\left[\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]}_{r^2}^0\\ =-\frac{2\pi }{3}\left[{\left(0\right)}^{\frac{3}{2}}-{\left(r^2\right)}^{\frac{3}{2}}\right]\end{array}$

$\begin{array}{c}=-\frac{2\pi }{3}\left(-r^3\right)\\ =\frac{2\pi r^3}{3}\end{array}$ (7)

This is the volume of the hemisphere.

Multiply Equation (7) by 2.

$\begin{array}{c}V=\frac{2\pi r^3}{3}\times 2\\ =\frac{4\pi r^3}{3}\end{array}$

Hence, the volume of the solid is $\frac{4\pi r^3}{3}$.