Chapter 6, Problem 22RE

(a) The base of a solid is a square with vertices located at (1, 0), (0, 1), (−1, 0), and (0, −1). Each cross-section perpendicular 10 the x-axis is a semicircle. Find the volume of the solid.

(b) Show that the volume of the solid of part (a) can be computed more simply by first cutting the solid and rearranging it to form a cone.

To determine

To Find: The volume of solid with vertices located at (1,0), (0,1), (-1,0), (0,-1).

Answer to Problem 22RE

The volume of the solid is $\underset{\_}{\frac{\pi }{3}}$.

Explanation of Solution

Given information:

The base of solid is square with vertices located at (1,0), (0,1), (-1,0), (0,-1).

Calculation:

Consider that the solid to the right of the origin $y=-x+1$.

Radius of the semicircle is $r=-x+1$.

Find the area of the semicircle using the formula.

$A=\frac{1}{2}\pi r^2$.

Here,r is radius of semicircle.

Expression to find the volume of solid as shown below.

Due to symmetry integrate from 0 to 1 as shown below.

$V=2{\displaystyle \underset{a}{\overset{b}{\int }}A\left(x\right)dx}$ (1)

Substitute 0 for a, 1 for b, and $\frac{1}{2}\pi \left(1-x^2\right)$ for $A\left(x\right)$ in Equation (1).

$\begin{array}{c}V=2{\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{2}\pi {\left(1-x\right)}^{2}dx}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{2}\pi {\left(1-x\right)}^{2}dx}\\ ={\left[\frac{\pi {\left(1-x\right)}^3}{3}\right]}_0^1\\ =-\frac{\pi }{3}{\left[\left({\left(1-x\right)}^3\right)\right]}_0^1\end{array}$

$\begin{array}{c}=-\frac{\pi }{3}\left[\left({\left(1-1\right)}^3\right)\right]\\ =\frac{\pi }{3}\end{array}$

Therefore, the volume of the solid is $\underset{\_}{\frac{\pi }{3}}$.

To determine

To find: the volume of the solid.

Answer to Problem 22RE

the volume of the solid is $\underset{\_}{\frac{\pi }{3}}$.

Explanation of Solution

The solid is modified as a cone by cutting method.

Procedure to obtain the volume of solid is explained below.

1. 1. Cut the solid that passes through to the y axis, and perpendicular to the x axis.
2. 2. Fold half of the solid that lies in the region $x\le 0$ , so that the point $(-1,0)$ touched the point (1,0).
3. 3. The final shape of the solid id right circular cone.
4. 4. The radius of the right circular cone is 1, and the vertex $\left(x,y,x\right)=\left(1,0,0\right)$.

Calculation:

Find the volume of the cone as shown below.

$V=\frac{1}{3}\pi r^{2}h$ (2)

Substitute 1 for r and 1 for h in Equation (2).

$\begin{array}{c}V=\frac{1}{3}\pi \left(1\right)\left(1\right)\\ =\frac{\pi }{3}\end{array}$

Therefore, the volume of the cone is $\underset{\_}{\frac{\pi }{3}}$.