Chapter 7.3, Problem 82E

a.

To determine

To show: The volume of a sphere generated by a region enclosed by a semicircle $y=\sqrt{a^2-x^2}$ and x-axis is revolved about x-axis is $\frac{4\pi a^3}{3}$ .

Explanation of Solution

Given information:

A region enclosed by a semicircle $y=\sqrt{a^2-x^2}$ and x-axis is revolved about x-axis.

Calculation:

The radius of each tiny disk will be $y=\sqrt{a^2-x^2}$ . So, the area of the disk will be $A\left(x\right)=\pi {\left(\sqrt{a^2-x^2}\right)}^2$ .

So, the volume of the sphere will be $V={\displaystyle \underset{-r}{\overset{r}{\int }}\pi {\left(\sqrt{a^2-x^2}\right)}^{2}dx}$ .

Evaluate the integral.

  $\begin{array}{c}V={\displaystyle \underset{-a}{\overset{a}{\int }}\pi {\left(\sqrt{a^2-x^2}\right)}^{2}dx}\\ =\pi {\displaystyle \underset{-a}{\overset{a}{\int }}\left(a^2-x^2\right)dx}\\ =\pi {\left[a^{2}x-\frac{x^3}{3}\right]}_{-a}^a\\ =\pi \left[\left(a^{2}a-\frac{a^3}{3}\right)-\left(-a^{2}a+\frac{a^3}{3}\right)\right]\\ =\pi \left[2\left(a^3-\frac{a^3}{3}\right)\right]\\ =\pi \left[2\left(\frac{2a^3}{3}\right)\right]\\ =\frac{4\pi a^3}{3}\end{array}$

Conclusion:

The volume of the sphere is $\frac{4\pi a^3}{3}$ .

b.

To determine

To find: The volume of a right circular cone of height $h$ and base radius $r$ using calculus.

Answer to Problem 82E

  $\frac{1}{3}\pi r^{2}h$

Explanation of Solution

Given information:

The right circular cone of height $h$ and base radius $r$ .

Calculation:

Draw a right circular cone of height $h$ and base radius $r$ as shown below.

  

A tiny disk of radius $x$ has area $A\left(x\right)=\pi x^2$ .

The relation between $x$ , $y$ , the height $h$ and base radius $r$ is given by $\frac{y}{x}=\frac{h}{r}$ . So, $x=\frac{r}{h}y$ .

Then, the area of the tiny disk is $A\left(y\right)=\pi {\left(\frac{r}{h}y\right)}^2=\pi \frac{r^2}{h^2}{y}^2$ .

So, the volume of the cone is $V={\displaystyle \underset{0}{\overset{h}{\int }}\left(\pi \frac{r^2}{h^2}{y}^2\right)dy}$ .

Evaluate the integral.

  $\begin{array}{c}V={\displaystyle \underset{0}{\overset{h}{\int }}\left(\pi \frac{r^2}{h^2}{y}^2\right)dy}\\ =\pi \frac{r^2}{h^2}{\left[\frac{y^3}{3}\right]}_0^h\\ =\pi \frac{r^2}{h^2}\left(\frac{h^3}{3}-0\right)\\ =\pi \frac{r^2}{h^2}\left(\frac{h^3}{3}\right)\\ =\frac{1}{3}\pi r^{2}h\end{array}$

Conclusion:

The volume of the right circular cone is $\frac{1}{3}\pi r^{2}h$ .