Chapter 10, Problem 32CP

To determine

To calculate the volume of the mailbox

Answer to Problem 32CP

1841 $i{n}^3$

Explanation of Solution

Given information:

The give figure consist of two parts The upper part which is in the shape of a half cylinder with radius $r=6$$inches$

The lower part which is in the shape of 3 dimensional prism with length $l=18$$inches$ breadth $b=12$$inches$ and height $h=8$$inches$

Formula used:

Volume of the cylinder, $\frac{1}{2}\times v=\frac{1}{2}\pi r^{2}h$

The cylinder is half, therefore dividing the volume by 2

Volume of the prism, $v=l\times b\times h$

Calculation:

Volume of upper part which is in the shape of half cylinder

  $v=\frac{1}{2}\times \pi r^{2}h$

Substituting the value of $\pi$ , $r$ and $h$$v=\frac{1}{2}\times \frac{22}{7}\times {(6)}^2\times 18$

Volume of half cylinder

  $v=113.04$$i{n}^3$

Also volume of 3 dimensional prism

  $v=l\times b\times h$

Substituting values of $l$ , $b$ and $h$$v=18\times 12\times 8$

Volume of prism

  $v=1728$$i{n}^3$

Volume of the mailbox = volume of half cylinder + volume of prism

  $v=1728+113$

Volume of mailbox

  $v=1841$$i{n}^3$