Chapter 12.6, Problem 11PSB

a.

To determine

To find: The total volume of the submarine formed by joining a hemisphere, a cylinder and a cone.

Answer to Problem 11PSB

The total volume of the submarine is $433.71{\text{units}}^3.$

Explanation of Solution

Given Information:

Radius of a cylinder $\left(r_1\right)=3$

Radius of a cone $\left(r_2\right)=3$

Radius of a hemisphere $\left(r_3\right)=3$

Height of a cylinder $\left(h_1\right)=12$

Height of a cone $\left(h_2\right)=4$

Formula used:

Volume of a cylinder $\left(V_1\right)=\pi {\left(r_1\right)}^2{h}_1$

  $r_1\text{and}{h}_1$ are the radius and height of a cylinder

Volume of a cone $\left(V_2\right)=\frac{1}{3}\pi {\left(r_2\right)}^2{h}_2$

  $r_2\text{and}{h}_2$ are the radius and height of a cone

Volume of a hemisphere $\left(V_3\right)=\frac{2}{3}\pi {\left(r_3\right)}^3$

  $r_3$ is the radius of a hemisphere

Calculation:

Total volume of the submarine $\left(V\right)=$ Volume of a cylinder + Volume of a cone + Volume of a hemisphere

  $\therefore V=V_1+V_2+V_3$

We know that, Volume of a cylinder $\left(V_1\right)=\pi {\left(r_1\right)}^2{h}_1$

Volume of a cone $\left(V_2\right)=\frac{1}{3}\pi {\left(r_2\right)}^2{h}_2$

Volume of a hemisphere $\left(V_3\right)=\frac{2}{3}\pi {\left(r_3\right)}^3$

As $r_1=r_2=r_3=3,h_1=12\text{and}{h}_2=4$

  $\begin{array}{l}V=\frac{22}{7}\times 3^2\times 12+\frac{1}{3}\times \frac{22}{7}\times 3^2\times 4+\frac{2}{3}\times \frac{22}{7}\times 3^3\\ V=\frac{22}{7}\times 3^2\left(12+\frac{4}{3}+2\right)\\ V=\frac{22}{7}\times 9\times \left(\frac{46}{3}\right)\\ V=433.71\end{array}$

Hence, the total volume of the submarine is $433.71{\text{units}}^3.$

b.

To determine

To find: The total surface area of the submarine formed by joining a hemisphere, a cylinder and a cone.

Answer to Problem 11PSB

The total surface area of the submarine is $330{\text{units}}^2.$

Explanation of Solution

Given Information:

Radius of a cylinder $\left(r_1\right)=3$

Radius of a cone $\left(r_2\right)=3$

Radius of a hemisphere $\left(r_3\right)=3$

Height of a cylinder $\left(h_1\right)=12$

Height of a cone $\left(h_2\right)=4$

Formula used:

Curved surface area of a cylinder $\left(S{A}_1\right)=2\pi r_1{h}_1$

  $r_1\text{and}{h}_1$ are the radius and height of a cylinder

Curved surface area of a cone $\left(S{A}_2\right)=\pi r_{2}l$

  $l=\sqrt{h_2{}^2+r_2{}^2}$

  $r_2,h_2\text{and}l$ are the radius and slant height of a cone

Curved surface area of a hemisphere $\left(S{A}_3\right)=2\pi {\left(r_3\right)}^2$

  $r_3$ is the radius of a hemisphere

Calculation:

Total surface area of the submarine $\left(SA\right)=$ Surface area of a cylinder + Surface area of a cone + Surface area of a hemisphere

  $\therefore SA=S{A}_1+S{A}_2+S{A}_3$

We know that, Curved surface area of a cylinder $\left(S{A}_1\right)=2\pi r_1{h}_1$

Curved surface area of a cone $\left(S{A}_2\right)=\pi r_{2}l$

Curved surface area of a hemisphere $\left(S{A}_3\right)=2\pi {\left(r_3\right)}^2$

As $r_1=r_2=r_3=3,h_1=12\text{and}{h}_2=4$

  $l=\sqrt{h_2{}^2+r_2{}^2}=5$

  $\begin{array}{l}\therefore SA=2\pi r_1{h}_1+\pi r_{2}l+2\pi {\left(r_3\right)}^2\\ \therefore SA=2\times \frac{22}{7}\times 3\times 12+\frac{22}{7}\times 3\times 5+2\times \frac{22}{7}\times 3^2\\ \therefore SA=\frac{22}{7}\times 3\left(24+5+6\right)\\ \therefore SA=\frac{22}{7}\times 3\times \left(35\right)\\ \therefore SA=22\times 3\times 5\\ \therefore SA=330{\text{units}}^2\end{array}$

Hence, the total surface area of the submarine is $330{\text{units}}^2.$