Chapter 8.4, Problem 26PPSE

Solution

a.

To determine: The expression for the ratio of the volume of the hemispherical tank to its surface area.

The required ratio is $\frac{2}{9}r$ .

Given Information:

The formula for sphere and surface area is defined as,

  $V=\frac{4}{3}\pi r^3$ and $SA=4\pi r^2$

Calculation:

Consider the given formulas,

  $V=\frac{4}{3}\pi r^3$ and $SA=4\pi r^2$

Write the volume of hemisphere.

  $\begin{array}{c}{V}_{\text{hemisphere}}=\frac{V}{2}\\ =\frac{2}{3}\pi r^3\end{array}$

Now, write the surface area of hemisphere.

  $\begin{array}{c}{V}_{\text{hemisphere}}=\frac{1}{2}SA+A\\ =\frac{1}{2}\cdot 4\pi r+\pi r^2\\ =3\pi r^2\end{array}$

Now, divide both the obtained formulas.

  $\begin{array}{c}\frac{V_{\text{hemisphere}}}{V_{\text{hemisphere}}}=\frac{\frac{2}{3}\pi r^3}{3\pi r^2}\\ =\frac{2}{9}r\end{array}$

Therefore, the obtained ratio is $\frac{2}{9}r$ .

b.

To determine: The expression for the ratio of the volume of the cylindrical tank to its surface area.

The obtained ratio is $\frac{r}{4}$ .

Given Information:

The formula for sphere and surface area is defined as,

  $V=\frac{4}{3}\pi r^3$ and $SA=4\pi r^2$

Calculation:

Consider the given information,

  $V=\frac{4}{3}\pi r^3$ and $SA=4\pi r^2$

Write the volume of cylinder.

  $\begin{array}{c}{V}_{\text{cylinder}}=\pi r^2\cdot h\\ =\pi r^2\cdot r\left[\text{given }h=r\right]\\ =\pi r^3\end{array}$

Write the surface area of cylinder.

  $\begin{array}{c}S{A}_{\text{Cylinder}}=2\pi r\cdot h+2\pi r^2\\ =2\pi r\cdot r+2\pi r^2\\ =4\pi r^2\end{array}$

Now, find the ratio of the both formulas.

  $\begin{array}{c}\frac{V_{\text{cylinder}}}{S{A}_{\text{cylinder}}}=\frac{\pi r^3}{4\pi r^2}\\ =\frac{r}{4}\end{array}$

Hence, the obtained result is $\frac{r}{4}$ .

c.

To determine: The comparison between ratios of the volume to surface area for the two tanks.

The ratio for the cylindrical tank is always larger.

Given Information:

The formula for sphere and surface area is defined as,

  $V=\frac{4}{3}\pi r^3$ and $SA=4\pi r^2$

Calculation:

Consider the given information,

Refer the both ratios.

  $\frac{V_{\text{hemisphere}}}{V_{\text{hemisphere}}}=\frac{2}{9}r$

And,

  $\frac{V_{\text{cylinder}}}{S{A}_{\text{cylinder}}}=\frac{r}{4}$

This means that the ratio for the cylindrical tank is always larger.

Therefore, the ratio for the cylindrical tank is always larger.

d.

To determine: The comparison between of the two tanks.

The volume of the cylindrical tank will always be larger.

Given Information:

The formula for sphere and surface area is defined as,

  $V=\frac{4}{3}\pi r^3$ and $SA=4\pi r^2$

Calculation:

Consider the given information,

  $V=\frac{4}{3}\pi r^3$

The volume of cylindrical tank is defined as,

  $\begin{array}{c}{V}_{\text{cylindrical}}=\pi r^{2}h\\ =\pi r^3\left[\text{Given }h=r\right]\end{array}$

So this means that for a given value of r , the volume of the cylindrical tank will always be larger.

Therefore, the volume of the cylindrical tank will always be larger.

e.

To determine: The explanation to use these ratios to compare the volumes of the two tanks and which measurement of the tanks determines the volumes.

The volumes will determines by radius.

Given Information:

The formula for sphere and surface area is defined as,

  $V=\frac{4}{3}\pi r^3$ and $SA=4\pi r^2$

Calculation:

Consider the given information,

As the volume of the tanks is depends on radius as $r$ will decrees then the condition will become opposite. Similarly the radius will increase then cylindrical tank will always be larger.

These volumes can be used for any two solution tanks to compare which one can contained more solution from each other.

As the height and radius is same in both tanks so the volumes will determines by radius.

Therefore, the volumes will determines by radius.