Chapter 7.3, Problem 54E

a.

To determine

To graph: The solid's volume created by rotating the area defined by the $x-axis$ .

Explanation of Solution

Given Information: To determine the solid's formed by rotating volume

Graph: To get the solid's volume created by rotating the area in the $y-axis$ defined by $y=2x-x^2$

  

It then rotates about the $y-axis,$

  

The volume calculation formula.

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

The graph above resembles a cylindrical shell when it rotates at the $y-axis.$

Remember the location of the cylinder-shaped shell.

  $\begin{array}{c}A\left(x\right)=2\pi rh\\ r\left(x\right)=x\\ h\left(x\right)=2x-x^2-x\end{array}$

  

To start, locate the intersections. In place of equation $\left(1\right)$ , write equation $\left(2\right).$

  $\begin{array}{c}y=x\\ 2x-x^2=x\\ 2x-x^2-x=0\\ x-x^2=0\end{array}$

Then use factoring to determine the values of $x.$

  $\begin{array}{c}2x-x^2-x=0\\ x-x^2=0\\ x\left(1-x\right)=0\end{array}$

The points of the $x-axis$ are $0$ and $1$

Substitute the $x-axis$ points in equation after that $\left(1\right)$

  $\begin{array}{c}x=0\\ y=2x-x^2\\ y=2\left(0\right)-0\\ y=0\\ x=1\\ y=2x-x^2\\ y=2\left(1\right)-1^2\\ y=1\end{array}$

The points of intersection are $\left(0,0\right)$ and $\left(1,1\right)$

  

A little area around the cylindrical shell in the graph must be used to solve the volume problem.

  

Substitute equation (4) to equation (3).

\

  $\begin{array}{c}V={\displaystyle \underset{a}{\overset{b}{\int }}A\left(x\right)}dx\\ V=2\pi {\displaystyle \underset{a}{\overset{b}{\int }}rhdx}\\ V=2\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left(x\right)}\left(2x-x^2-x\right)dx\\ V=2\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left(x\right)}(x-x^2)dx\\ V=2\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left(x^2-x^3\right)}dx\end{array}$

The interval $\left[a,b\right]$ is $\left[0,1\right]$

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

Integrate after applying the distributive property.

  $\begin{array}{c}V=2\pi {\displaystyle \underset{0}{\overset{1}{\int }}\left(x^2-x^3\right)}dx\\ V=2\pi {\displaystyle \underset{0}{\overset{1}{\int }}{x}^{2}dx-2\pi {\displaystyle \underset{0}{\overset{1}{\int }}{x}^{3}dx}}\\ V=2\pi \left(\frac{x^3}{3}\right)\left|{}_0^1\right.-2\pi \left(\frac{x^4}{4}\right)\left|{}_0^1\right.\end{array}$

Substitute the results of the above calculation.

  $\begin{array}{c}V=2\pi \left(\frac{x^3}{3}\right)\left|{}_0^1\right.-2\pi \left(\frac{x^4}{4}\right)\left|{}_0^1\right.\\ V=2\pi \left[\left(\frac{1^3}{3}\right)-\left(\frac{0^3}{3}\right)\right]-2\pi \left[\left(\frac{1^4}{4}\right)-\left(\frac{0^4}{4}\right)\right]\\ V=2\pi \left(\frac{1^3}{3}\right)-2\pi \left(\frac{1^4}{4}\right)\\ V=\frac{2\pi }{3}-\frac{2\pi }{4}\\ V=\frac{\pi }{6}\end{array}$

b.

To determine

To graph: To determine the solid's volume created by rotating the area around constrained by $y=2x-x^2$

Explanation of Solution

Given Information: To calculate the volume of the solid produced by rotating the area at the line $x=1$ limited by

  $\begin{array}{l}y=2x-x^2\mathrm{...........}\left(1\right)\\ y=x\mathrm{....................}\left(2\right)\end{array}$

Graph: The previous equation

  

It then rotates at line $x=1.$

  

Remember the volume formula.

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

The graph above resembles a cylindrical shell when it rotates at the $y-axis$ .

Remember the location of the cylinder-shaped shell.

  $\begin{array}{c}A\left(x\right)=2\pi rh\pi \\ r\left(x\right)=1-x\\ h\left(x\right)=2x-x^2-x\end{array}$

  

Find the junction point or points first. In place of equation $(1)$ , write equation $\left(2\right).$

  $\begin{array}{c}y=x\\ 2x-x^2=x\\ 2x-x^2-x=0\\ x-x^2=0\end{array}$

Then use factoring to determine the values of $x.$

  $\begin{array}{c}2x-x^2-x=0\\ x-x^2=0\\ x\left(1-x\right)=0\end{array}$

The points $x-axis$ are then $0$ and $1.$

Then modify equation by substituting the $x-axis$ points (1),

  $\begin{array}{c}x=0\\ y=2x-x^2\\ y=2\left(0\right)-0\\ y=0\end{array}$

  $\begin{array}{c}x=1\\ y=2x-x^2\\ y=2\left(1\right)-1^2\\ y=1\end{array}$

Consequently, $\left(0,0\right)$ and are the places of intersection $\left(1,1\right)$

  

A little section of the graph's cylindrical shell area can be used to solve the volume problem.

Substitute equation $\left(4\right)$ to equation $\left(3\right).$

  $\begin{array}{c}V={\displaystyle \underset{a}{\overset{b}{\int }}A\left(x\right)dx}\\ V=2\pi {\displaystyle \underset{a}{\overset{b}{\int }}rhdx}\\ V=2\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left(1-x\right)}\left(2x-x^2-x\right)dx\\ V=2\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left(1-x\right)\left(x-x^2\right)dx}\\ V=2\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left(x-x^2-x^2+x^3\right)dx}\\ V=2\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left(x-2x^2+x^3\right)dx}\\ V=2\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left(x^3-2x^2+x\right)dx}\end{array}$

See how the interval $\left[a,b\right]$ on the graph is $\left[0,1\right],$ then

  $\begin{array}{c}V=2\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left(x-x^2-x^2+x^3\right)dx}\\ V=2\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left(x-2x^2+x^3\right)dx}\\ V=2\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left(x^3-2x+x\right)dx}\\ V=2\pi {\displaystyle \underset{0}{\overset{1}{\int }}\left(x^3-2x^2+x\right)dx}\end{array}$

Integrate after applying the distributive property.

  $\begin{array}{c}V=2\pi {\displaystyle \underset{0}{\overset{1}{\int }}\left(x^3-2x^2+x\right)dx}\\ V=2\pi {\displaystyle \underset{0}{\overset{1}{\int }}{x}^{3}dx-2\pi {\displaystyle \underset{0}{\overset{1}{\int }}2x^{2}dx+2\pi {\displaystyle \underset{0}{\overset{1}{\int }}xdx}}}\\ V=2\pi \left(\frac{x^4}{4}\right)\left|\frac{1}{0}\right.-2\pi \left(\frac{2x^3}{3}\right)\left|{}_0^1\right.+2\pi \left(\frac{x^2}{2}\right)\left|{}_0^1\right.\end{array}$

After that, swap out the values from the previous result.

  $\begin{array}{c}V=2\pi \left(\frac{x^4}{4}\right)\left|\frac{1}{0}\right.-2\pi \left(\frac{2x^3}{3}\right)\left|{}_0^1\right.+2\pi \left(\frac{x^2}{2}\right)\left|{}_0^1\right.\\ V=2\pi \left[\left(\frac{1^4}{4}\right)-\left(\frac{0^4}{4}\right)\right]-2\pi \left[\left(\frac{2{\left(1\right)}^3}{3}\right)-\left(\frac{2{\left(1\right)}^3}{3}\right)\right]+2\pi \left[\left(\frac{1^2}{2}\right)-\left(\frac{0^2}{2}\right)\right]\\ V=2\pi \left(\frac{1^4}{4}\right)-2\pi \left(\frac{2{\left(1\right)}^3}{3}\right)+2\pi \left(\frac{1^2}{2}\right)\\ V=\frac{\pi }{6}\end{array}$

Consequently, the volume is

  $V=\frac{\pi }{6}$