Chapter 3.3, Problem 2MRPS

Solution

(a)

Simplify the right side of the formula $V=3^{-1}{(4\pi )}^{-1/2}{(S^3)}^{1/2}$

  $V=\frac{S^3{}^{/2}}{6{(\pi )}^{1/2}}$

Given:

The formula for the volume V of a sphere in terms of its surface area S is $V=3^{-1}{(4\pi )}^{-1/2}{(S^3)}^{1/2}$

Concept Used:

• The composition function $f\circ g$ can be found by replacing x by $g\left(x\right)$ in the function $f\left(x\right)$
• The domain of a function $y=f\left(x\right)$ is the set of all x values for which the function is defined.
• A rational function $y=f\left(x\right)$ is undefined for the values of x , for which the denominator is zero.
• A radical function is defined for positive values greater than or equal to zero.
• Zero product property: if $a\cdot b=0\text{then}a=0\text{or}b=0$

Calculation:

As given $V=3^{-1}{(4\pi )}^{-1/2}{(S^3)}^{1/2}$

Simplifying the right side of the formula

  $\begin{array}{c}$ V=3^{-1}{(4\pi )}^{-1/2}{(S^3)}^{1/2}$=\frac{{(S^3)}^{1/2}}{3{(4\pi )}^{1/2}}\\ =\frac{S^3{}^{/2}}{3{(4\pi )}^{1/2}}\\ =\frac{S^3{}^{/2}}{3\sqrt{4\pi }}\\ =\frac{S^3{}^{/2}}{3\cdot 2\sqrt{\pi }}\\ =\frac{S^3{}^{/2}}{6\sqrt{\pi }}\\ =\frac{S^3{}^{/2}}{6{(\pi )}^{1/2}}\end{array}$

(b)

A candlepin bowling ball has a surface area of about 79 square inches. What is its volume?

  $V=\frac{{79}^3{}^{/2}}{6{(\pi )}^{1/2}}$

Given:

The formula for the volume V of a sphere in terms of its surface area S is $V=3^{-1}{(4\pi )}^{-1/2}{(S^3)}^{1/2}$

Concept Used:

• The composition function $g\circ f$ can be found by replacing x by $g\left(x\right)$ in the function $f\left(x\right)$
• The domain of a function $y=f\left(x\right)$ is the set of all x values for which the function is defined.
• A rational function $y=f\left(x\right)$ is undefined for the values of x , for which the denominator is zero.
• A radical function is defined for positive values greater than or equal to zero.
• Zero product property: if $a\cdot b=0\text{then}a=0\text{or}b=0$

Calculation:

The surface area of candlepin bowling ball (S) = 79 square inches

Volume of candlepin bowling ball $V=3^{-1}{(4\pi )}^{-1/2}{(S^3)}^{1/2}$

  $\begin{array}{c}$ V=3^{-1}{(4\pi )}^{-1/2}{(S^3)}^{1/2}$=\frac{{({79}^3)}^{1/2}}{3{(4\pi )}^{1/2}}\\ =\frac{{79}^3{}^{/2}}{3{(4\pi )}^{1/2}}\\ =\frac{{79}^3{}^{/2}}{3\sqrt{4\pi }}\\ =\frac{{79}^3{}^{/2}}{3\cdot 2\sqrt{\pi }}\\ =\frac{{79}^3{}^{/2}}{6\sqrt{\pi }}\\ =\frac{{79}^3{}^{/2}}{6{(\pi )}^{1/2}}\end{array}$

(c)

A 10-pin bowling ball has a surface area of about 232 square inches. What is its volume?

  

  $V=\frac{116{(232)}^1{}^{/2}}{3{(\pi )}^{1/2}}$

Given:

The formula for the volume V of a sphere in terms of its surface area S is $V=3^{-1}{(4\pi )}^{-1/2}{(S^3)}^{1/2}$

Concept Used:

• The composition function $g\circ f$ can be found by replacing x by $g\left(x\right)$ in the function $f\left(x\right)$
• The domain of a function $y=f\left(x\right)$ is the set of all x values for which the function is defined.
• A rational function $y=f\left(x\right)$ is undefined for the values of x , for which the denominator is zero.
• A radical function is defined for positive values greater than or equal to zero.
• Zero product property: if $a\cdot b=0\text{then}a=0\text{or}b=0$

Calculation:

The surface area of 10-pin bowling ball (S) = 232 square inches

Volume of 10-pin bowling ball $V=3^{-1}{(4\pi )}^{-1/2}{(S^3)}^{1/2}$

  $\begin{array}{c}$ V=3^{-1}{(4\pi )}^{-1/2}{(S^3)}^{1/2}$=\frac{{({232}^3)}^{1/2}}{3{(4\pi )}^{1/2}}\\ =\frac{{232}^3{}^{/2}}{3{(4\pi )}^{1/2}}\\ =\frac{{232}^3{}^{/2}}{3\sqrt{4\pi }}\\ =\frac{{232}^3{}^{/2}}{3\cdot 2\sqrt{\pi }}\\ =\frac{{232}^3{}^{/2}}{6\sqrt{\pi }}\\ =\frac{{232}^3{}^{/2}}{6{(\pi )}^{1/2}}\\ =\frac{232({232}^1{}^{/2})}{6{(\pi )}^{1/2}}\\ =\frac{116{(232)}^1{}^{/2}}{3{(\pi )}^{1/2}}\end{array}$

(d)

Compare the surface areas and volumes of the two bowling balls.

Surface area of 10-pin bowling ball > Surface area of candlepin bowling ball.

Volume of 10-pin bowling ball > Volume of candlepin bowling ball.

Given:

The formula for the volume V of a sphere in terms of its surface area S is $V=3^{-1}{(4\pi )}^{-1/2}{(S^3)}^{1/2}$

Concept Used:

• The composition function $g\circ f$ can be found by replacing x by $g\left(x\right)$ in the function $f\left(x\right)$
• The domain of a function $y=f\left(x\right)$ is the set of all x values for which the function is defined.
• A rational function $y=f\left(x\right)$ is undefined for the values of x , for which the denominator is zero.
• A radical function is defined for positive values greater than or equal to zero.
• Zero product property: if $a\cdot b=0\text{then}a=0\text{or}b=0$

Calculation:

Surface area of candlepin bowling ball = 79 square inches And surface area of 10-pin bowling ball = 232 square inches

So the surface area of 10-pin bowling ball > the surface area of candlepin bowling ball

232 square inches > 79 square inches

Volume of candlepin bowling ball $=\frac{{79}^3{}^{/2}}{6{(\pi )}^{1/2}}$

Volume of 10-pin bowling ball $=\frac{116{(232)}^1{}^{/2}}{3{(\pi )}^{1/2}}$

Volume of 10-pin bowling ball > Volume of candlepin bowling ball

  $\frac{116{(232)}^1{}^{/2}}{3{(\pi )}^{1/2}}$ >$\frac{{79}^3{}^{/2}}{6{(\pi )}^{1/2}}$