Chapter 7, Problem 26RE

To determine

To find: The volume of the piece cut.

Answer to Problem 26RE

The volume of the piece cut is $V=\frac{28\pi }{3}f{t}^3$ .

Explanation of Solution

Given information:

The radius of the round hole = $\sqrt{3}$

The radius of the sphere = $2\text{ feet}$

Calculation:

Let's graph thegiven area and draw some conclusions based on it.

  

The height of the cylinder is $2\text{h}$ .

So,

  $\begin{array}{c}{h}^2+{\left(\sqrt{3}\right)}^2=2^{{}^2}\\ h=1\end{array}$

The components that were taken from the sphere are a cylinder and two caps. So, using the next method, we can determine how much material has been removed.

  $V=V_c+2V_{ca}$

Here,

  $V_c$ = Volume of the cylinder

  $2V_{ca}$ = Volume of the cap

The volume of the cylinder is given by,

  $V_c=\left(2h\right)\pi {\left(\sqrt{3}\right)}^2$

Using the disk method to find the volume of the cap

From $x^2+y^2=2^2$ it is known that $R\left(x\right)=\pm \sqrt{4-y^2}$

The volume of the cap is,

  $\begin{array}{c}{V}_{ca}={{\displaystyle {\int }_1^2\pi \left[R\left(x\right)\right]}}^{2}dy\\ ={\displaystyle {\int }_1^2\pi \left(4-y^2\right)}dy\\ =\pi {\left[4y-\frac{y^3}{3}\right]}_1^2\\ =\pi \left[\left(8-\frac{8}{3}\right)-\left(4-\frac{1}{3}\right)\right]\\ =\frac{5\pi }{3}{\text{ ft}}^3\end{array}$

Now, the volume of the removed material is

  $\begin{array}{c}V=V_c+2V_{ca}\\ =6\pi +\frac{10\pi }{3}\\ V=\frac{28\pi }{3}{\text{ ft}}^3\end{array}$

Therefore, the volume of the removed material is $V=\frac{28\pi }{3}{\text{ ft}}^3$ .