Chapter 8, Problem 53RE

(a)

To determine

To find area of region R

Answer to Problem 53RE

Area of Region R $\approx$ 1.366

Explanation of Solution

Given:

Region R in the first quadrant enclosed by the y-axis and the graphs of $y=2+\sin x$ and $y=\sec x$

Concept Used:

Area between the curves

  $A={\displaystyle \underset{0}{\overset{\frac{153}{125}}{\int }}\left[2+\sin x-\sec x\right]}dx$

  $\text{First, calculating the corresponding indefinete integral}$

  ${\displaystyle \int \left(\sin x-\sec x+2\right)dx=2x-\ln \left(\left|\tan \left(\frac{x}{2}+\frac{\pi }{4}\right)\right|\right)-\cos x}$

  $\text{Acoording to the fundamental theorem of calculas }{\displaystyle {\int }_a^{b}F\left(x\right)dx}=f\left(b\right)-f\left(a\right),$

  $Therefore,$

  $\left(2x-\ln \left(\left|\tan \left(\frac{x}{2}+\frac{\pi }{4}\right)\right|\right)-\cos x\right){|}_{(x=\frac{153}{125})}=-\ln \left(\tan \left(\frac{153}{125}+\frac{\pi }{4}\right)\right)-\cos \left(\frac{153}{125}\right)+\frac{306}{125}$

  $\left(2x-\ln \left(\left|\tan \left(\frac{x}{2}+\frac{\pi }{4}\right)\right|\right)-\cos x\right){|}_{(x=0)}=-\ln \left(\tan \left(0+\frac{\pi }{4}\right)\right)-\cos \left(0\right)+0=-1$

  ${\displaystyle \underset{0}{\overset{\frac{153}{125}}{\int }}\left[2+\sin x-\sec x\right]}dx=\left(2x-\ln \left(\left|\tan \left(\frac{x}{2}+\frac{\pi }{4}\right)\right|\right)-\cos x\right){|}_{(x=\frac{153}{125})}-\left(2x-\ln \left(\left|\tan \left(\frac{x}{2}+\frac{\pi }{4}\right)\right|\right)-\cos x\right){|}_{(x=0)}$

  $\text{ =}-\ln \left(\tan \left(\frac{153}{125}+\frac{\pi }{4}\right)\right)-\cos \left(\frac{153}{125}\right)+\frac{306}{125}-\left(-1\right)$

  $\text{ =}-\ln \left(\tan \left(\frac{153}{125}+\frac{\pi }{4}\right)\right)-\cos \left(\frac{153}{125}\right)+\frac{306}{125}+1$

  $\text{ =}-\ln \left(\tan \left(\frac{153}{125}+\frac{\pi }{4}\right)\right)-\cos \left(\frac{153}{125}\right)+\frac{431}{125}\approx 1.36604180418083$

Given region R is in the first quadrant by the y-axis and the graphs of $y=2+\sin x$ and $y=\sec x$

Getting limits by solving using graphing calculator

  

Therefore,

Limits are $a=0b=1.224=\frac{153}{125}$

Calculation:

Area of region R,

  $A={\displaystyle \underset{0}{\overset{\frac{153}{125}}{\int }}\left[2+\sin x-\sec x\right]}dx$

  $\text{First, calculating the corresponding indefinete integral}$

  ${\displaystyle \int \left(\sin x-\sec x+2\right)dx=2x-\ln \left(\left|\tan \left(\frac{x}{2}+\frac{\pi }{4}\right)\right|\right)-\cos x}$

  $\text{Acoording to the fundamental theorem of calculas }{\displaystyle {\int }_a^{b}F\left(x\right)dx}=f\left(b\right)-f\left(a\right),$

  $Therefore,$

  $\left(2x-\ln \left(\left|\tan \left(\frac{x}{2}+\frac{\pi }{4}\right)\right|\right)-\cos x\right){|}_{(x=\frac{153}{125})}=-\ln \left(\tan \left(\frac{153}{125}+\frac{\pi }{4}\right)\right)-\cos \left(\frac{153}{125}\right)+\frac{306}{125}$

  $\left(2x-\ln \left(\left|\tan \left(\frac{x}{2}+\frac{\pi }{4}\right)\right|\right)-\cos x\right){|}_{(x=0)}=-\ln \left(\tan \left(0+\frac{\pi }{4}\right)\right)-\cos \left(0\right)+0=-1$

  ${\displaystyle \underset{0}{\overset{\frac{153}{125}}{\int }}\left[2+\sin x-\sec x\right]}dx=\left(2x-\ln \left(\left|\tan \left(\frac{x}{2}+\frac{\pi }{4}\right)\right|\right)-\cos x\right){|}_{(x=\frac{153}{125})}-\left(2x-\ln \left(\left|\tan \left(\frac{x}{2}+\frac{\pi }{4}\right)\right|\right)-\cos x\right){|}_{(x=0)}$

  $\text{ =}-\ln \left(\tan \left(\frac{153}{125}+\frac{\pi }{4}\right)\right)-\cos \left(\frac{153}{125}\right)+\frac{306}{125}-\left(-1\right)$

  $\text{ =}-\ln \left(\tan \left(\frac{153}{125}+\frac{\pi }{4}\right)\right)-\cos \left(\frac{153}{125}\right)+\frac{306}{125}+1$

  $\text{ =}-\ln \left(\tan \left(\frac{153}{125}+\frac{\pi }{4}\right)\right)-\cos \left(\frac{153}{125}\right)+\frac{431}{125}\approx 1.36604180418083$

Conclusion:

Area of Region R $\approx$ 1.366

(b)

To determine

To find the volume of the solid generated when region R is revolved about $x-axis$

Answer to Problem 53RE

Volume of the solid generated when region R is revolved about $x-axis$

  $\approx 16.4042$

Explanation of Solution

Given:

Region R in the first quadrant enclosed by the y-axis and the graphs of $y=2+\sin x$ and $y=\sec x$

Concept Used:

The volume V of a solid generated by revolving the region about $x-axis$ is

  $V=\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left({\left[f\left(x\right)\right]}^2-{\left[g\left(x\right)\right]}^2\right)}dx$

Calculation:

  $V=\pi {\displaystyle \underset{0}{\overset{\frac{153}{125}}{\int }}\left({\left[2+\sin \left(x\right)\right]}^2-{\left[\sec \left(x\right)\right]}^2\right)}dx$

  $\text{Firstly solving indefinete integral }{\displaystyle \int \left({\left[2+\sin \left(x\right)\right]}^2-{\left[\sec \left(x\right)\right]}^2\right)}dx$

  ${\displaystyle \int \left({\left[2+\sin \left(x\right)\right]}^2-{\left[\sec \left(x\right)\right]}^2\right)}dx={\displaystyle \int {\left(\sin \left(x\right)+2\right)}^{2}dx-{\displaystyle \int {\sec }^2\left(x\right)}}dx={\displaystyle \int \left({\left(\sin \left(x\right)\right)}^2+4\sin \left(x\right)+4\right)}dx-{\displaystyle \int {\sec }^2\left(x\right)dx}$

  $={\displaystyle \int {\sin }^2\left(x\right)dx}+4{\displaystyle \int \sin \left(x\right)dx+4{\displaystyle \int dx}-}{\displaystyle \int {\sec }^2\left(x\right)dx}$

  $={\displaystyle \int \left(\frac{1-\cos \left(2x\right)}{2}\right)dx-4\cos x+4x-\tan x}+C$

  $={\displaystyle \int \left(\frac{1}{2}-\frac{\cos \left(2x\right)}{2}\right)}dx-4\cos x+4x-\tan x+C$

  $=\frac{x}{2}-\frac{\sin \left(2x\right)}{4}-4\cos x+4x-\tan x+C$

  $=\frac{9x}{2}-\frac{\sin \left(2x\right)}{4}-4\cos x-\tan x+C$

  $Therefore,$

  ${\displaystyle \int \left({\left[2+\sin \left(x\right)\right]}^2-{\left[\sec \left(x\right)\right]}^2\right)}dx=\frac{9x}{2}-\frac{\sin \left(2x\right)}{4}-4\cos x-\tan x+C$

Now finding definite integral,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{\frac{153}{125}}{\int }}\left({\left[2+\sin \left(x\right)\right]}^2-{\left[\sec \left(x\right)\right]}^2\right)\text{dx }}\\ =\left(\frac{9x}{2}-\frac{\sin \left(2x\right)}{4}-4\cos \left(x\right)-\tan \left(x\right)\right){|}_{(x=\frac{153}{125})}-\left(\frac{9x}{2}-\frac{\sin \left(2x\right)}{4}-4\cos x-\tan x\right){|}_{(x=0)}\\ =\left(\frac{1377}{250}-\frac{\sin \left(\frac{306}{125}\right)}{4}-4\cos \left(\frac{153}{125}\right)-\tan \left(\frac{153}{125}\right)\right)-\left(0-0-4-0\right)\\ =-\frac{\sin \left(\frac{306}{125}\right)}{4}-4\cos \left(\frac{153}{125}\right)-\tan \left(\frac{153}{125}\right)+\frac{1377}{250}+4=-\frac{\sin \left(\frac{306}{125}\right)}{4}-4\cos \left(\frac{153}{125}\right)-\tan \left(\frac{153}{125}\right)+\frac{2377}{250}\approx 5.22163\end{array}$

Conclusion:

Required Volume, V= $\pi {\displaystyle \underset{0}{\overset{\frac{153}{125}}{\int }}\left({\left[2+\sin \left(x\right)\right]}^2-{\left[\sec \left(x\right)\right]}^2\right)}dx$

  $V\approx \pi \times 5.22163\approx 16.4042$

c.

To determine

To find volume of the solid whose base is region R and whose cross sections cut by planes perpendicular to $x-axis$ .

Answer to Problem 53RE

Volume of the solid whose base is region R and whose cross sections cut by planes perpendicular to $x-axis$

  $\approx 1.6290$

Explanation of Solution

Given:

The solid whose base is region R and whose cross sections cut by planes perpendicular to $x-axis$ .

Concept Used:

Volume of the solid whose base is region R and whose cross sections cut by planes perpendicular to $x-axis$ = ${\displaystyle \underset{a}{\overset{b}{\int }}{\left(f\left(x\right)-g\left(x\right)\right)}^{2}dx}$

Calculation:

  $Volume\left(V\right)={\displaystyle \underset{0}{\overset{\frac{153}{125}}{\int }}{\left(\left(2+\sin \left(x\right)\right)-\sec \left(x\right)\right)}^{2}dx}$

  $\text{finding indefinete integral }{\displaystyle \int {\left(\left(2+\sin \left(x\right)\right)-\sec \left(x\right)\right)}^{2}dx}$

  ${\displaystyle \int {\left(\left(2+\sin \left(x\right)\right)-\sec x\right)}^{2}dx}$

  $={\displaystyle \int \left({\left(2+\sin \left(x\right)\right)}^2++{\left(\sec \left(x\right)\right)}^2-2\left(2+\sin \left(x\right)\right)\left(\sec \left(x\right)\right)\right)}dx$

  $={\displaystyle \int \left(\left(4+{\sin }^2\left(x\right)+4\sin \left(x\right)\right)+{\sec }^2\left(x\right)-4\sec \left(x\right)-2\sin \left(x\right)\sec \left(x\right)\right)}dx$

  $=4{\displaystyle \int dx+{\displaystyle \int {\sin }^2\left(x\right)dx}+4{\displaystyle \int \sin \left(x\right)dx+{\displaystyle \int {\sec }^2}}}\left(x\right)dx-4{\displaystyle \int \sec \left(x\right)dx}-2{\displaystyle \int \tan \left(x\right)dx}\left(\because \sin x\sec x=\tan x\right)$

  $=4x+\frac{x}{2}+\frac{\sin 2x}{4}-4\cos x+\tan x-4\ln \left(\left|\tan \left(\frac{x}{2}+\frac{\pi }{4}\right)\right|\right)-2\ln \left(\left|\cos \left(x\right)\right|\right)$

  $=\frac{9x}{2}+\frac{\sin 2x}{4}-4\cos x+\tan x-4\ln \left(\left|\tan \left(\frac{x}{2}+\frac{\pi }{4}\right)\right|\right)-2\ln \left(\left|\cos \left(x\right)\right|\right)$

Now finding definite integral by putting values of limits

  ${\displaystyle \underset{0}{\overset{\frac{153}{125}}{\int }}{\left(\left(2+\sin \left(x\right)\right)-\sec \left(x\right)\right)}^{2}dx}$

  $=\left(\frac{9x}{2}+\frac{\sin 2x}{4}-4\cos x+\tan x-4\ln \left(\left|\tan \left(\frac{x}{2}+\frac{\pi }{4}\right)\right|\right)-2\ln \left(\left|\cos \left(x\right)\right|\right)\right){|}_{(x=\frac{153}{125})}-\left(\frac{9x}{2}+\frac{\sin 2x}{4}-4\cos x+\tan x-4\ln \left(\left|\tan \left(\frac{x}{2}+\frac{\pi }{4}\right)\right|\right)-2\ln \left(\left|\cos \left(x\right)\right|\right)\right){|}_{(x=0)}$

  $=\left(\frac{1377}{250}+\frac{\sin \left(\frac{306}{125}\right)}{4}-4\cos \left(\frac{153}{125}\right)+\tan \left(\frac{153}{125}\right)-4\ln \left(\left|\tan \left(\frac{153}{250}+\frac{\pi }{4}\right)\right|\right)-2\ln \left(\left|\cos \left(\frac{153}{125}\right)\right|\right)\right)-\left(0+0-4+0-0\right)$

  $=\frac{1377}{250}+4+\frac{\sin \left(\frac{306}{125}\right)}{4}-4\cos \left(\frac{153}{125}\right)+\tan \left(\frac{153}{125}\right)-4\ln \left(\left|\tan \left(\frac{153}{250}+\frac{\pi }{4}\right)\right|\right)-2\ln \left(\left|\cos \left(\frac{153}{125}\right)\right|\right)$

  $=\frac{2377}{250}+\frac{\sin \left(\frac{306}{125}\right)}{4}-4\cos \left(\frac{153}{125}\right)+\tan \left(\frac{153}{125}\right)-4\ln \left(\left|\tan \left(\frac{153}{250}+\frac{\pi }{4}\right)\right|\right)-2\ln \left(\left|\cos \left(\frac{153}{125}\right)\right|\right)\approx 1.6290$

Conclusion:

Required Volume $\approx 1.6290$