Chapter 6, Problem 6P

To determine

To calculate: The value of radius r.

Answer to Problem 6P

The value of r is $\underset{\_}{\frac{2}{\sqrt{5}}}$.

Explanation of Solution

Given information:

A sphere of radius 1 overlaps a smaller sphere of radius r in such a way that their intersection is a circle of radius r.

Calculation:

Sketch the sphere of radius 1 overlaps a smaller sphere of radius r as shown in Figure 1.

Refer to Figure 1.

The volume of the shaded region V is the volume of the hemisphere of radius r minus the volume of the spherical cap with height $h=1-\sqrt{1-r^2}$ and radius 1.

Find the volume of the shaded region as shown below.

$\begin{array}{c}V=\frac{1}{2}\times \frac{4}{3}\pi r^3-\frac{1}{3}\times \pi {\left(1-\sqrt{1-r^2}\right)}^2\left(3\times 1-\left(1-\sqrt{1-r^2}\right)\right)\\ =\frac{1}{3}\pi \left(2r^3-\left(1+\left(1-r^2\right)-2\sqrt{1-r^2}\right)\left(2+\sqrt{1-r^2}\right)\right)\\ =\frac{1}{3}\pi \left(2r^3-\left(2-r^2-2\sqrt{1-r^2}\right)\left(2+\sqrt{1-r^2}\right)\right)\\ =\frac{1}{3}\pi \left(2r^3-4-2\sqrt{1-r^2}+2r^2+r^2\sqrt{1-r^2}+4\sqrt{1-r^2}+2\left(1-r^2\right)\right)\end{array}$

$\begin{array}{c}=\frac{1}{3}\pi \left(2r^3-2+2\sqrt{1-r^2}+r^2\sqrt{1-r^2}\right)\\ =\frac{1}{3}\pi \left(2r^3-2+\left(r^2+2\right)\sqrt{1-r^2}\right)\end{array}$

Differentiate both sides of the Equation.

$\begin{array}{c}{V}^{\prime }=\frac{1}{3}\pi \left(6r^2+\left(r^2+2\right)\frac{-r}{\sqrt{1-r^2}}+\sqrt{1+r^2}\left(2r\right)\right)\\ =\frac{1}{3}\pi \left(\frac{6r^2\sqrt{1-r^2}-r^3-2r+\left(1-r^2\right)\left(2r\right)}{\sqrt{1-r^2}}\right)\\ =\frac{1}{3}\pi \left(\frac{6r^2\sqrt{1-r^2}-r^3-2r+2r-2r^3}{\sqrt{1-r^2}}\right)\\ =\pi \left(\frac{2r^2\sqrt{1-r^2}-r^3}{\sqrt{1-r^2}}\right)\end{array}$

Equate the above Equation to be zero

$\begin{array}{l}{V}^{\prime }=0\\ \pi \left(\frac{2r^2\sqrt{1-r^2}-r^3}{\sqrt{1-r^2}}\right)=0\\ 2r^2\sqrt{1-r^2}-r^3=0\\ r^2\left(2\sqrt{1-r^2}-r\right)=0\end{array}$

$\begin{array}{l}2\sqrt{1-r^2}-r=0\\ 2\sqrt{1-r^2}=r\\ 4\left(1-r^2\right)=r^2\\ 4-4r^2=r^2\end{array}$

$\begin{array}{l}4=5r^2\\ r^2=\frac{4}{5}\\ r=\frac{2}{\sqrt{5}}\end{array}$

Therefore, the value of r is $\underset{\_}{\frac{2}{\sqrt{5}}}$.