Chapter 12, Problem 17RP

To determine

To find: The volume of the resulting solid when a cylindrical part is removed from a cuboid.

Answer to Problem 17RP

The volume of the resulting solid to the nearest cubic inch is $215{\text{in}}^3.$

Explanation of Solution

Given Information:

Diameter of a cylinder $\left(d\right)=2\text{inches}$

Height of a cylinder $\left(h_1\right)=8\text{inches}$

Length of a cuboid $\left(l\right)=8\text{inches}$

Breadth of a cuboid $\left(b\right)=5\text{inches}$

Height of a cuboid $\left(h\right)=6\text{inches}$

Formula used:

Volume of a cuboid $\left(V\right)=lbh$

  $l,b\text{and}h$ are the length, breadth and height of a cuboid respectively

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

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

Calculation:

Now, the given is a cuboid in a cylindrical hole is drilled.

So, the volume of the resulting solid $\left(V\text{'}\right)=$ Volume of cuboid − Volume of cylinder

We know that, Volume of a cuboid $\left(V\right)=lbh$

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

  $\therefore V\text{'}=V-V_1$

As $l=8,b=5,h=6,r=\frac{d}{2}=1\text{and}{h}_1=8$

  $\begin{array}{l}\therefore V\text{'}=lbh-\pi r^2{h}_1\\ \therefore V\text{'}=8\times 5\times 6-\frac{22}{7}\times 1^2\times 8\\ \therefore V\text{'}=240-25.24\\ \therefore V\text{'}=214.86{\text{in}}^3\approx 215{\text{in}}^3\\ \therefore V\text{'}\approx 215{\text{in}}^3\end{array}$

Hence, the volume of the resulting solid to the nearest cubic inch is $215{\text{in}}^3.$