Chapter 6, Problem 17RE

a.

To determine

To describe: the solid whose volume is given by the integral.

Answer to Problem 17RE

  $0\le y\le \sqrt{2}\cos x\text{where}0\le x\le \frac{\pi }{2}.$

Explanation of Solution

Given information: Given integral is ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}2\pi }{\cos }^{2}xdx$

Calculation:

Since, the volume will be an area of a curve y=f(x) rotated about the x axis.

Since, the Area would be a washer, would get:

  $\begin{array}{l}A=\pi r^2=\pi y^2=\pi f^2(x).\\ V={\displaystyle \underset{a}{\overset{b}{\int }}Adx=}{\displaystyle \underset{a}{\overset{b}{\int }}\pi f^2(x)dx.}\end{array}$

Given integral is ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}2\pi }{\cos }^{2}xdx$ .

  $\begin{array}{l}V={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\pi (\sqrt{2}}\cos x{)}^{2}dx\\ \text{So}\text{with}:\\ f(x)=\sqrt{2}\cos x,a=0,b=\frac{\pi }{2}.\end{array}$

Therefore, the Solid is the area bounded by the curve $f(x)=\sqrt{2}\cos x$ and the x axis rotated about x axis in the interval $x\in [0,\frac{\pi }{2}]$

b.

To determine

To describe: the solid whose volume is given by the integral.

Answer to Problem 17RE

  $2-\sqrt{x}\le y\le 2-x^2\text{where}0\le x\le 1.$

Explanation of Solution

Given information: Given integral is ${\displaystyle \underset{0}{\overset{1}{\int }}\pi [(2-x^2}{)}^2-{(2-\sqrt{x})}^2]dx$

Calculation:

Since, the volume will be an area of a curve y=f(x) rotated about the x axis.

Since, the Area would be a washer, would get:

  $A=\pi r^2=\pi y^2=\pi f^2(x).$

Now if the Area rotated is the area bound by two functions then the washer would be ,the difference of two areas.

  $A=A_1=A_2$ .

  $\begin{array}{l}{A}_1=\pi r_1{}^2=\pi f^2(x).\\ A_2=\pi r_2{}^2=\pi g^2(x).\\ \text{Now}\text{can}\text{write}:\\ V={\displaystyle \underset{a}{\overset{b}{\int }}\pi (}{f}^2(x)-g^2(x))dx\end{array}$

Given integral is

  $\begin{array}{c}$ {\displaystyle \underset{0}{\overset{1}{\int }}\pi [(2-x^2}{)}^2-{(2-\sqrt{x})}^2]dx.$\text{So}\text{with}:\\ f(x)=2-x^2\\ g(x)=2-\sqrt{x}\\ a=0\\ b=1.\end{array}$

Therefore , the Solid is the Area bound by the curves $f(x)=2-x^2,g(x)=2-\sqrt{x}$ rotated about x axis in the interval $x\in [0,1]$