Chapter 6, Problem 8RE

(a)

To determine

To Find: The area of $\Re$ .

Answer to Problem 8RE

The required value of the area is $\frac{5}{12}$ .

Explanation of Solution

Given:

The $y=x^3$ and $y=2x-x^2$ .

Calculation:

The area of the formula by subtracts the two curves but first must find the intersection points,

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

He area of the region that is bounded by the curves is,

  $\begin{array}{c}A={\displaystyle {\int }_0^1\left[2x-x^2-x^3\right]}dx\\ =\left[x^2-\frac{x^3}{3}-\frac{x^4}{4}\right]\\ =\frac{5}{12}\end{array}$

(b)

To determine

To Find: The volume of the obtained by rotating about the x axis.

Answer to Problem 8RE

The volume of the solid is $\frac{41\pi }{105}$ .

Explanation of Solution

Calculation:

Consider the solid obtained by the curves is shown in Figure 1

  

Figure 1

The figure for the graph of the solid that is obtained by the solid rotating about the x axis and the region bounded the given curves is shown in Figure 2

  

Figure 2

The two diagrams shows that the curves intersect at the points $\left(0,1\right),\left(0,0\right)$ .

Then, the area of the cross section is,

  $\begin{array}{c}A\left(x\right)=\pi {\left(2x-x^2\right)}^2-\pi {\left(x^3\right)}^2\\ =\pi \left({\left(2x-x^2\right)}^2\right)-x^6\end{array}$

Then, the volume is,

  $\begin{array}{c}V={\displaystyle {\int }_0^1\pi A\left(x\right)dx}\\ ={\displaystyle {\int }_0^1\left[{\left(2x-x^2\right)}^2-x^6\right]dx}\\ =\pi {\left[\frac{4x^3}{3}-x^4+\frac{x^5}{5}-\frac{x^7}{7}\right]}_0^1\\ =\frac{41\pi }{105}\end{array}$

(c)

To determine

To Find: The volume of the obtained by rotating about the y axis.

Answer to Problem 8RE

The volume of the solid is $\frac{13\pi }{30}$ .

Explanation of Solution

Calculation:

Consider the solid obtained by the curves is shown in Figure 1

  

Figure 1

The figure for the graph of the solid that is obtained by the solid rotating about the y axis and the region bounded the given curves is shown in Figure 2

  

Figure 2

The two diagrams shows that the curves intersect at the points $\left(0,1\right),\left(0,0\right)$ .

Then, the volume is,

  $\begin{array}{c}V={\displaystyle {\int }_0^{1}2\pi f\left(x\right)dx}\\ =2\pi {\displaystyle {\int }_0^{1}x\left[\left(2x-x^2\right)-x^3\right]dx}\\ =\pi {\left[\frac{2x^3}{3}-\frac{x^4}{4}+\frac{x^4}{4}-\frac{x^5}{5}\right]}_0^1\\ =\frac{13\pi }{30}\end{array}$