Chapter 3, Problem 4TP

Solution

(a)

To solve the surface area of a hemisphere formula for r .

  $r=\sqrt{\frac{S}{3\pi }}$

Given:

The surface area of a hemisphere is given by the formula $S=3\pi r^2$ . Two hemispherical snow globes are given in which one globe has twice as much surface area as the other.

Concept Used:

• Power of a power property: ${\left(x^a\right)}^b=x^{a\cdot b}$
• Product of power property: $a^m\cdot a^n=a^{m+n}$
• Quotient of power property: $\frac{a^m}{a^n}=a^{m-n}$
• Product property of radicals: $\sqrt[n]{a}\cdot \sqrt[n]{b}=\sqrt[n]{a\cdot b}$
• Quotient property of radicals: $\frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}}$
• The exponent form of the nth root of a number a is given by $\sqrt[n]{a}=a^{\frac{1}{n}}$

Calculation:

In order to solve the surface area of a hemisphere formula $S=3\pi r^2$ for r , isolate r on one side by performing the basic algebraic operations as shown below

  $\begin{array}{c}S=3\pi r^2\\ \frac{S}{3\pi }=r^2\\ r^2=\frac{S}{3\pi }\\ r=\pm \sqrt{\frac{S}{3\pi }}\\ r=\sqrt{\frac{S}{3\pi }}\left(\text{since radius cannot be negative}\right)\end{array}$

Thus, the formula for r is $r=\sqrt{\frac{S}{3\pi }}$ .

(b)

To substitute the value of r from above part into the volume of a hemisphere formula $V=\frac{2}{3}\pi r^3$ and simplify further.

  $V=\frac{2S^{\frac{3}{2}}}{3^{\frac{5}{2}}{\pi }^{\frac{1}{2}}}$

Given:

Two hemispherical snow globes are given in which one globe has twice as much surface area as the other.

Concept Used:

• Power of a power property: ${\left(x^a\right)}^b=x^{a\cdot b}$
• Product of power property: $a^m\cdot a^n=a^{m+n}$
• Quotient of power property: $\frac{a^m}{a^n}=a^{m-n}$
• Product property of radicals: $\sqrt[n]{a}\cdot \sqrt[n]{b}=\sqrt[n]{a\cdot b}$
• Quotient property of radicals: $\frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}}$
• The exponent form of the nth root of a number a is given by $\sqrt[n]{a}=a^{\frac{1}{n}}$

Calculation:

In order to find the required formula, substitute $r=\sqrt{\frac{S}{3\pi }}$ into the volume formula and simplify further as shown below

  $\begin{array}{c}V=\frac{2}{3}\pi r^3\\ =\frac{2}{3}\pi {\left(\sqrt{\frac{S}{3\pi }}\right)}^3\\ =\frac{2}{3}\pi {\left({\left(\frac{S}{3\pi }\right)}^{\frac{1}{2}}\right)}^3\\ =\frac{2}{3}\pi {\left(\frac{S}{3\pi }\right)}^{\frac{3}{2}}\\ =\frac{2}{3}\pi \left(\frac{S^{\frac{3}{2}}}{3^{\frac{3}{2}}{\pi }^{\frac{3}{2}}}\right)\\ =\frac{2S^{\frac{3}{2}}}{3^{\frac{3}{2}+1}{\pi }^{\frac{3}{2}-1}}\\ =\frac{2S^{\frac{3}{2}}}{3^{\frac{5}{2}}{\pi }^{\frac{1}{2}}}\end{array}$

Thus, the formula for volume in simplified form is $V=\frac{2S^{\frac{3}{2}}}{3^{\frac{5}{2}}{\pi }^{\frac{1}{2}}}$ .