Chapter 8.3, Problem 74E

a.

To determine

To show : that if you revolve the region enclosed by the semicircle $y=\sqrt{a^2-x^2}$ and the x -axis about the x -axis to generate a solid sphere.

Explanation of Solution

Given information : The calculus formula for volume at the beginning of the section will give $\frac{4}{3}\pi a^3$ for the volume just as should it.

Calculation :

The cross section has radius given by $r=\sqrt{a^2-x^2}$

Therefore the area given by $\pi \left(a^2-x^2\right)$

Hence, the volume is:

  $\begin{array}{l}V={\displaystyle \underset{-a}{\overset{a}{\int }}\pi \left(a^2-x^2\right)}\\ \Rightarrow \frac{4}{3}\pi a^3\end{array}$

b.

To determine

To find : the volume of a right circular cone.

Answer to Problem 74E

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

Explanation of Solution

Given information : Use the calculus formula of a right circular cone of height $h$ and base radius $r$ .

Calculation :

The cross section has radius given by $x=r\left(1-\raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$h$}\right.\right)$ and consequently area given by: $\pi r^2=\pi r^2\left(1-\raisebox{1ex}{$\left(2y\right)$}\!\left/ \!\raisebox{-1ex}{$h$}\right.+\frac{y^2}{h^2}\right)$

Hence, the volume is given by:

  $\begin{array}{l}V={\displaystyle \underset{0}{\overset{h}{\int }}\pi r^2\left(1-\raisebox{1ex}{$\left(2y\right)$}\!\left/ \!\raisebox{-1ex}{$h$}\right.+\frac{y^2}{h^2}\right)}dy\\ V=\frac{1}{3}\pi r^{2}h\end{array}$