Chapter 5.4, Problem 51PS

Solution

a.

To prove: The height $h$ of the cylindrical tank is $\frac{4}{3}r$ .

Given information:

Volume of the spherical tank is equal to Volume of the cylindrical tank.

Formula used:

Volume of the sphere: $\frac{4}{3}\pi r^3$

Volume of the cylinder: $\pi r^{2}h$

Proof:

Given the:

Volume of the sphere is equal to Volume of the cylinder

Know that the;

Volume of the sphere: $\frac{4}{3}\pi r^3$

Volume of the cylinder: $\pi r^{2}h$

So,

  $\frac{4}{3}\pi r^3=\pi r^{2}h$

Factor out $r^2$ ;

  $\frac{4}{3}\pi r=\pi h$

Factor out $\pi$ ;

  $\frac{4}{3}r=h$

Here, prove that the height $h$ of the cylindrical tank is $\frac{4}{3}r$ .

b.

To calculate: The surface area of each tank in terms of $r$ .

The surface area of spherical tank in terms of $r$ is $4\pi r^2$ .

The surface area of cylindrical tank in terms of $r$ is $\frac{14}{3}\pi r^2$ .

Given information:

Volume of the spherical tank is equal to Volume of the cylindrical tank.

Formula used:

The surface area of a sphere: $4\pi r^2$

The surface area of a cylinder: $2\pi r^2+2\pi rh$

Calculation:

The expression to find the surface area of a sphere, the tower on the left;

The surface area of a sphere: $4\pi r^2$

The surface area of a cylinder, the tower on the right;

The surface area of a cylinder: $2\pi r^2+2\pi rh$

The volumes of each water tower are equal:

  $\frac{4}{3}\pi r^3=\pi r^{2}h$

Divide both sides by $\pi r^2$ ;

  $h=\frac{4}{3}r$

So,

  $\begin{array}{l}2\pi r^2+2\pi r\left(\frac{4}{3}r\right)\\ 2\pi r^2+\frac{8}{3}\pi r^2\\ \frac{14}{3}\pi r^2\end{array}$

Conclusion:

The surface area of each tank in terms of $r$ : $4\pi r^2;\frac{14}{3}\pi r^2$

c.

To calculate: The ratio of the surface area of the spherical tank to the surface area of the cylindrical tank. Explain what the ratio tells you about which water tower would take less material to build.

The ratio of the surface area of the spherical tank to the surface area of the cylindrical tank is $6:7$ . This tells that when building two tanks with the same volume and radius, but with one as a sphere and one as a cylinder, the spherical one will take up less material to build.

Given information:

Volume of the spherical tank is equal to Volume of the cylindrical tank.

Concept used:

The surface area of a sphere: $4\pi r^2$

The surface area of a cylinder: $2\pi r^2+2\pi rh$

Calculation:

The surface area of spherical tank = $4\pi r^2$ .

The surface area of cylindrical tank = $\frac{14}{3}\pi r^2$

Now,

The ratio of the surface area of the spherical tank to the surface area of the cylindrical tank:

  $\frac{4\pi r^2}{\frac{14}{3}\pi r^2}$

Cancel $\pi r^2$ from the numerator and denominator and simplify;

  $\frac{4}{\left(\frac{14}{3}\right)}=\frac{3\times 4}{14}=\frac{12}{14}=\frac{6}{7}$

Conclusion:

The obtained ratio by calculation is $6:7$ .