Chapter 7.3, Problem 4QQ

a.

To determine

To find: The area of the region $R$ .

Answer to Problem 4QQ

  $\text{0}\text{.16152}$

Explanation of Solution

Given information:

The region $R$ is bounded by the graphs of $y=\sqrt{x}$ , $y=e^{-x}$ and $y$ -axis.

Calculation:

The graph of $y=\sqrt{x}$ and $y=e^{-x}$ intersects at $x=0.426$ .

So, the area of the region $R$ is $A={\displaystyle \underset{0}{\overset{0.426}{\int }}\left(e^{-x}-\sqrt{x}\right)dx}$ .

Now, evaluate the integral.

  $\begin{array}{c}A={\displaystyle \underset{0}{\overset{0.426}{\int }}\left(e^{-x}-\sqrt{x}\right)dx}\\ ={\left[-e^{-x}-\frac{x^{3/2}}{3/2}\right]}_0^{0.426}\\ =\text{0}\text{.16152}\end{array}$

Conclusion:

The area of the region is $\text{0}\text{.16152}$ .

b.

To determine

To find: The volume of the solid generated when $R$ is revolved around the horizontal line $y=-1$ .

Answer to Problem 4QQ

  $1.63055$

Explanation of Solution

Given information:

The region $R$ is bounded by the graphs of $y=\sqrt{x}$ , $y=e^{-x}$ and $y$ -axis.

Calculation:

The cross-sections of the solid perpendicular to x-axis are washers with inner radius $1+\sqrt{x}$ and outer radius $1+e^{-x}$ . The area of each cross-section is $A\left(x\right)=\pi {\left(1+e^{-x}\right)}^2-\pi {\left(1+\sqrt{x}\right)}^2$ .

Therefore, the volume of the solid is $\text{V=}{\displaystyle \underset{0}{\overset{0.426}{\int }}\left(\pi {\left(1+e^{-x}\right)}^2-\pi {\left(1+\sqrt{x}\right)}^2\right)dx}$ .

Evaluate the integral to find the volume.

  $\begin{array}{c}V={\displaystyle \underset{0}{\overset{0.426}{\int }}\left(\pi {\left(1+e^{-x}\right)}^2-\pi {\left(1+\sqrt{x}\right)}^2\right)dx}\\ =\pi {\displaystyle \underset{0}{\overset{0.426}{\int }}\left(1+e^{-2x}+2e^{-x}-1-x-2\sqrt{x}\right)dx}\\ =\pi {\displaystyle \underset{0}{\overset{0.426}{\int }}\left(e^{-2x}+2e^{-x}-x-2\sqrt{x}\right)dx}\\ =\pi \left(-{\displaystyle {\int }_0^{0.426}x}dx-{\displaystyle {\int }_0^{0.426}2}\sqrt{x}dx+{\displaystyle {\int }_0^{0.426}{e}^{-2x}}dx+{\displaystyle {\int }_0^{0.426}2}{e}^{-x}dx\right)\\ =\pi \left(-0.090738-0.37072+0.28671+0.69376\right)\\ =1.63055\end{array}$

Conclusion:

The volume of the solid is $1.63055$ .

c.

To determine

To find: The volume of the solid whose base is the region $R$ and each cross-section perpendicular to x-axis is a semicircle whose diameter runs from the graph of $y=\sqrt{x}$ to the graph of $y=e^{-x}$ .

Answer to Problem 4QQ

  $0.03476$

Explanation of Solution

Given information:

The region $R$ is bounded by the graphs of $y=\sqrt{x}$ , $y=e^{-x}$ and $y$ -axis.

Calculation:

The diameter of each semicircle is $e^{-x}-\sqrt{x}$ . Then, the area of each semicircle is $A\left(x\right)=\frac{1}{2}\pi {\left(\frac{e^{-x}-\sqrt{x}}{2}\right)}^2=\frac{\pi }{8}{\left(e^{-x}-\sqrt{x}\right)}^2$ .

So, the volume of the solid is $V={\displaystyle \underset{0}{\overset{0.426}{\int }}\left(\frac{\pi }{8}{\left(e^{-x}-\sqrt{x}\right)}^2\right)dx}$ .

Evaluate the integral.

  $\begin{array}{c}V={\displaystyle \underset{0}{\overset{0.426}{\int }}\left(\frac{\pi }{8}{\left(e^{-x}-\sqrt{x}\right)}^2\right)dx}\\ =\frac{\pi }{8}{\displaystyle \underset{0}{\overset{0.426}{\int }}\left(e^{-2x}+x-2\sqrt{x}{e}^{-x}\right)dx}\\ =\frac{\pi }{8}\left({\displaystyle {\int }_0^{0.426}{e}^{-2x}}dx-{\displaystyle {\int }_0^{0.426}2}{e}^{-x}\sqrt{x}dx+{\displaystyle {\int }_0^{0.426}x}dx\right)\\ =\frac{\pi }{8}\left(0.28671-0.28892+0.090738\right)\\ =0.03476\end{array}$

Conclusion:

The volume of the solid is $0.03476$ .