Chapter 6.4, Problem 46PPS

(a)

To determine

To Find:

The radius for volumes of

  $\begin{array}{l}{V}_1=1000c{m}^3\\ V_2=8000c{m}^3\\ V_3=64000c{m}^3\end{array}$

Answer to Problem 46PPS

  $\begin{array}{l}(1000,6.2)\\ (8000,12.4)\\ (64000,24.8)\end{array}$

Explanation of Solution

Given Information:

The radius r of a sphere with volume V can be found using the formula

  $r=\sqrt[3]{\frac{3V}{4\pi }}$

Calculation:

• Given

  $r=\sqrt[3]{\frac{3V}{4\pi }}$

And

  $\begin{array}{l}{V}_1=1000c{m}^3\\ V_2=8000c{m}^3\\ V_3=64000c{m}^3\end{array}$

• Substituting the above values in the equation one by one to find r :
• $\begin{array}{l}{r}_1=\sqrt[3]{\frac{3V_1}{4\pi }}\\ V_1=1000\\ \Rightarrow r_1=\sqrt[3]{\frac{3\times 1000}{4\pi }}\\ Now,\text{ we use the calculator to find the approximate solution }\text{.}\end{array}$

  

So, we have

• $(V_1,r_1)=(1000,6.20)$
• $\begin{array}{l}{r}_2=\sqrt[3]{\frac{3V_2}{4\pi }}\\ V_2=8000\\ \Rightarrow r_2=\sqrt[3]{\frac{3\times 8000}{4\pi }}\\ Now,\text{ we use the calculator to find the approximate solution }\text{.}\end{array}$

  

So, we have

• $(V_2,r_2)=(8000,12.41)$
• $\begin{array}{l}{r}_3=\sqrt[3]{\frac{3V_3}{4\pi }}\\ V_3=64000\\ \Rightarrow r_3=\sqrt[3]{\frac{3\times 64000}{4\pi }}\\ Now,\text{ we use the calculator to find the approximate solution }\text{.}\end{array}$

  

So, we have

• $(V_3,r_3)=(64000,24.81)$

(b)

To determine

To Explain:

How does the volume of the sphere changes if the radius is doubled.

Answer to Problem 46PPS

The volume of the sphere whose radius is doubled becomes 8 times the volume of the original sphere.

Explanation of Solution

Given Information:

The radius of the sphere is doubled.

Formula Used:

  $\begin{array}{l}Volume\text{ of the sphere , whose radius is 'r' :}\\ \text{V=}\frac{4}{3}\pi r^3\end{array}$

Calculation:

  $\begin{array}{l}Volume\text{ of the sphere , whose radius is 'r' :}\\ \text{V=}\frac{4}{3}\pi r^3\\ Let\text{ the initial radius of the sphere is 'r'}\text{.}\\ \Rightarrow V_{initial}=\frac{4}{3}\pi r^3\\ Since\text{ the radius is now double = '2r'}\\ \Rightarrow V_{final}=\frac{4}{3}\pi {(}^2=\frac{4}{3}\pi 8r^3=8(\frac{4}{3}\pi r^3)=8\times V_{initial}\\ \Rightarrow V_{final}=8V_{initial}\end{array}$

So, the volume of the sphere whose radius is doubled becomes 8 times the volume of the original sphere.