Chapter 6, Problem 16RE

a.

To determine

To estimate: the x -coordinates of the points of intersection of the curves.

Answer to Problem 16RE

x −coordinates of the points of intersection are 0 and 0.699.

Explanation of Solution

Given information: Let R be the region bounded by curves $y=1-x^2\text{and}y=x^6-x+1$ .

Calculation:

By using graphing utility, the graph of the curves $y=1-x^2\text{and}y=x^6-x+1$ is shown below.

  

From the graph, the x −coordinates of the points of intersection are 0 and 0.699.

b.

To determine

To estimate: the area of R .

Answer to Problem 16RE

  $Area=0.1216$ .

Explanation of Solution

Given information: Let R be the region bounded by curves $y=1-x^2\text{and}y=x^6-x+1$ .

Calculation:

Find the limits of the integration by solving the system of equations:

  $\begin{array}{c}y=x^6-x+1.\\ y=1-x^2.\\ 1-x^2=x^6-x+1.\\ x^6+x^2-x=0.\\ \text{one}\text{root}\text{is},\\ x_1=0\end{array}$

To find the other root use a graphing system , second root is:

  $x_2=\mathrm{0.7549.}$

Solve the integral for Area:

  $\begin{array}{l}A={\displaystyle \underset{x_1}{\overset{x_2}{\int }}[f(x)-g(x)]dx},\text{Where}f(x)=1-x^2,g(x)=x^6-x+1.\\ A={\displaystyle \underset{0}{\overset{0.7549}{\int }}[1-x^2}-(x^6-x+1)]dx\\ A={\displaystyle \underset{0}{\overset{0.7549}{\int }}[-x^6-}{x}^2+x)]dx\\ A=[-\frac{x^7}{7}-\frac{x^3}{3}+\frac{x^2}{2}]|\begin{array}{c}0.7549\\ 0\end{array}.\\ A=[-\frac{{0.7549}^7}{7}-\frac{{0.7549}^3}{3}+\frac{{0.7549}^2}{2}].\\ A=\mathrm{0.1216.}\end{array}$

c.

To determine

To estimate: the volume generated when R is rotated about the x -axis.

Answer to Problem 16RE

  $Volume=\mathrm{0.1732.}$

Explanation of Solution

Given information: Let R be the region bounded by curves $y=1-x^2\text{and}y=x^6-x+1$ .

Calculation:

Find the limits of the integration by solving the system of equations:

  $\begin{array}{c}y=x^6-x+1.\\ y=1-x^2.\\ 1-x^2=x^6-x+1.\\ x^6+x^2-x=0.\\ \text{one}\text{root}\text{is},\\ x_1=0\end{array}$

To find the other root use a graphing system, second root is:

  $x_2=\mathrm{0.7549.}$

Now, solve the integral for the Area:

  $A={\displaystyle \underset{x_1}{\overset{x_2}{\int }}[A_{f(x)}}-A_{g(x)}]dx\text{Where}f(x)=1-x^2,g(x)=x^6-x+1.$

Now let’s find the Area, the cross section of the Solid in respect to x is the difference :

Between the outer circle $r_{out}=f(x)\text{and}{r}_{in}=g(x).$

  $\begin{array}{c}$ A_{out}=\pi r^2{}_{out}=\pi {(1-x^2)}^2=\pi (1-2x^2+x^4).$$ A_{in}=\pi r^2{}_{in}=\pi {(x^6-x+1)}^2=\pi {(x^6-(x-1))}^2=\pi (x^{12}-2x^7+2x^6+x^2-2x+1)$V=\pi {\displaystyle \underset{0}{\overset{0.7549}{\int }}[(}1-2x^2+x^4)-(x^{12}-2x^7+2x^6+x^2-2x+1)]dx\\ V=\pi {\displaystyle \underset{0}{\overset{0.7549}{\int }}[(}-x^{12}+2x^7-2x^6+x^4-3x^2+2x)]dx\\ V=[-\frac{x^{13}}{13}+\frac{x^8}{4}-\frac{2x^7}{7}+\frac{x^5}{5}-x^3+x^2]|\begin{array}{c}0.7549\\ 0\end{array}\\ V=\mathrm{0.1732.}\end{array}$

d.

To determine

To estimate: the volume generated when R is rotated about the y -axis.

Answer to Problem 16RE

  $Volume=\mathrm{0.3080.}$

Explanation of Solution

Given information: Let R be the region bounded by curves $y=1-x^2\text{and}y=x^6-x+1$ .

Calculation:

Find the limits of the integration by solving the system of equations:

  $\begin{array}{c}y=x^6-x+1.\\ y=1-x^2.\\ 1-x^2=x^6-x+1.\\ x^6+x^2-x=0.\\ \text{one}\text{root}\text{is},\\ x_1=0\end{array}$

To find the other root use a graphing system, second root is:

  $x_2=\mathrm{0.7549.}$

The volume of a solid obtained by rotating about the y −axis the region under the curve.

  $V={\displaystyle \underset{a}{\overset{b}{\int }}[2\pi xf(x)dx}$

  $\begin{array}{l}V=V_{f(x)}-V_{g(x)}.\\ V={\displaystyle \underset{0}{\overset{0.7549}{\int }}[2\pi x(1-x^2})dx-{\displaystyle \underset{0}{\overset{0.7549}{\int }}[2\pi x(x^6}-x+1)]dx.\\ V=2\pi {\displaystyle \underset{0}{\overset{0.7549}{\int }}[(x-x^3}-x^7+x^2-x)]dx\\ V=2\pi {\displaystyle \underset{0}{\overset{0.7549}{\int }}[(-x^7}-x^3+x^2)]dx\\ V=2\pi [-\frac{x^8}{8}-\frac{x^4}{4}+\frac{x^3}{3}]|\begin{array}{c}0.7549\\ 0\end{array}\\ V=\pi [-\frac{{0.7549}^8}{4}-\frac{{0.7549}^4}{2}+\frac{2\times {0.7549}^3}{3}]\\ V=\mathrm{0.3080.}\end{array}$