Chapter 7.3, Problem 59E

(a)

To determine

The volume of the solid remains after a round hole was drilled.

Answer to Problem 59E

The volume of the solid remain is $\frac{256}{3}\pi$ .

Explanation of Solution

Given:

A round hole is drilled in spherical solid having radius r and the hole has a height of $\text{4 cm}$ .

Calculation:

The volume can be find out as:

Let us take volume element in cylindrical coordinates:

  $dV=rd\theta drdz$

Limits can be defined as:

  $0\le \theta <2\pi ,r_o\le r\le \sqrt{R^2-z^2},-h\le z\le +h$

In the above formula, R is the radius of original sphere and $r_o$ is the radius of bored out cylinder.

  $r_o=\sqrt{R^2-h^2}$

Now, integrate the upper half of the solid and multiply it by $2$ to get the volume:

  $\begin{array}{l}V=2{\displaystyle \underset{z=0}{\overset{h}{\int }}{\displaystyle \underset{r=r_o}{\overset{\sqrt{R^2-h^2}}{\int }}{\displaystyle \underset{0}{\overset{2\pi }{\int }}dV}}}\\ V=2{\displaystyle \underset{z=0}{\overset{h}{\int }}{\displaystyle \underset{r=r_o}{\overset{\sqrt{R^2-h^2}}{\int }}{\displaystyle \underset{0}{\overset{2\pi }{\int }}rd\theta drdz}}}\\ V=2\pi h\left[R^2-\frac{h^2}{3}-r_o^2\right]\\ V=2\pi h\left[R^2-\frac{h^2}{3}-R^2+h^2\right]\\ V=2\pi h\times \frac{2h^2}{3}\\ V=\frac{4}{3}\pi h^3\\ $ V=\frac{4}{3}\pi {(4)}^3$V=\frac{256}{3}\pi \end{array}$

Conclusion:

The volume of the solid remain is $\frac{256}{3}\pi$ .

(b)

To determine

The unusual about the volume determined.

Answer to Problem 59E

The unusual about volume determined is that it is independent of radius.

Explanation of Solution

Given:

The volume of the solid remain is $\frac{4}{3}\pi h^3$ .

Calculation:

In the volume determined, there is no radius involved. Only height is present in the volume.

This is the unusual fact about the determined volume.

Conclusion:

The unusual about volume determined is that it is independent of radius.