Chapter 12.6, Problem 31PPS

a.

To determine

Tocalculate : The volume and surface area of the top.

Answer to Problem 31PPS

$S=748.5c{m}^2$ and $V=1282.7c{m}^3$ .

Explanation of Solution

Given information : A top having a sphere and a cone. The diameter of each is $d=14cm$ and the height of the cone is $h=11cm$ . The radius of each will be $r=7cm$

Formula used : The surface area of the top would be $2\pi r^2+\pi r\left(r+\sqrt{h^2+r^2}\right)$ . And, the volume of the top would be $V=\frac{2\pi r^3}{3}+\frac{\pi r^{2}h}{3}$ .

Find surface area and volume of the top.

Calculation : Calculate the surface area of the top.

$\begin{array}{l}S=2\pi r^2+\pi r\left(r+\sqrt{h^2+r^2}\right)\\ =2\pi \times 7^2+7\pi \left(7+\sqrt{{11}^2+7^2}\right)\\ =307.8+440.7\\ S=748.5c{m}^2\end{array}$

Now, calculate the volume of the top.

  $V=\frac{2\pi r^3}{3}+\frac{\pi r^{2}h}{3}$

  $\begin{array}{l}V=\frac{2\pi \times 7^3}{3}+\frac{\pi \times 7^2\times 11}{3}\\ V=718.3+564.4\\ V=1282.7c{m}^3\end{array}$

b.

To determine

Tocalculate : The surface area and volume of another model, whose dimensions are half of the previous dimensions.

Answer to Problem 31PPS

$S=187.2c{m}^2$ and $V=160.3c{m}^3$ .

Explanation of Solution

Given:

A model whose dimensions are half of the previous dimensions. That is $h=5.5cm$ and the radius of each will be $r=3.5cm$ .

Calculation : Calculate the surface area of the top.

$\begin{array}{l}S=2\pi r^2+\pi r\left(r+\sqrt{h^2+r^2}\right)\\ =2\pi \times {3.5}^2+3.5\pi \left(3.5+\sqrt{{5.5}^2+{3.5}^2}\right)\\ =77+110.2\\ S=187.2c{m}^2\end{array}$

Now, calculate the volume of the top.

  $\begin{array}{l}V=\frac{2\pi r^3}{3}+\frac{\pi r^{2}h}{3}\\ =\frac{2\pi \times {3.5}^3}{3}+\frac{\pi \times {3.5}^2\times 5.5}{3}\\ =89.8+70.5\\ V=160.3c{m}^3\end{array}$