Chapter 6, Problem 9RE

a.

To determine

To find: The volume of the solid obtained by rotating the region bounded by the $y=x$ and $y=x^2$ about x-axis.

Answer to Problem 9RE

The required volume is $\frac{2\pi }{15}$ .

Explanation of Solution

Given information :

  $y=x\mathrm{.........}\left(i\right)$

  $y=x^2\mathrm{............}\left(ii\right)$

Formula used :

The volume of the solid found by the curves $f\left(x\right)\&g\left(x\right)$ in between $x=a$ and $x=b$ rotating 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:

  

To find intersecting points we intersect $y=x$ and $y=x^2$

  $x=x^2\Rightarrow x^2-x=0\Rightarrow x=1,x=0$

At $x=0$ , $y=0$ ;

At $x=1,y=1$

Here, $f\left(x\right)=x$ and $g\left(x\right)=x^2$

So,

  $\begin{array}{c}V=\pi {\displaystyle \underset{0}{\overset{1}{\int }}\left({\left[f\left(x\right)\right]}^2-{\left[g\left(x\right)\right]}^2\right)}dx\\ =\pi {\displaystyle \underset{0}{\overset{1}{\int }}\left(x^2-x^4\right)}dx\\ =\pi {\left[\frac{x^3}{3}\right]}_0^1-\pi {\left[\frac{x^5}{5}\right]}_0^1\\ =\frac{\pi }{3}-\frac{\pi }{5}=\frac{2\pi }{15}\end{array}$

b.

To determine

To find: The volume of the solid obtained by rotating the region bounded by the $y=x$ and $y=x^2$ about y-axis.

Answer to Problem 9RE

The required volume is $\frac{\pi }{6}$ .

Explanation of Solution

Given information :

  $y=x\mathrm{.........}\left(i\right)$

  $y=x^2\mathrm{............}\left(ii\right)$

Formula used :

The volume of the solid found by the curves $f\left(x\right)\&g\left(x\right)$ in between $y=a$ and $y=b$ rotating about y-axis is $V=\pi {\displaystyle \underset{a}{\overset{b}{\int }}\left({\left[f\left(y\right)\right]}^2-{\left[g\left(y\right)\right]}^2\right)}dy$

Calculation:

  

To find intersecting points we intersect $y=x$ and $y=x^2$

  $x=x^2\Rightarrow x^2-x=0\Rightarrow x=1,x=0$

At $x=0$ , $y=0$ ;

At $x=1,y=1$

  $y=x^2\Rightarrow x=\sqrt{y}$

Here, $f\left(y\right)=\sqrt{y}$ and $g\left(y\right)=y$

So,

  $\begin{array}{c}V=\pi {\displaystyle \underset{0}{\overset{1}{\int }}\left({\left[\sqrt{y}\right]}^2-y^2\right)}dy\left[asy=0\text{ \&}y=1\text{ are boundaries}\right]\\ =\pi {\displaystyle \underset{0}{\overset{1}{\int }}\left(y-y^2\right)}dy\\ =\pi {\left[\frac{y^2}{2}-\frac{y^3}{3}\right]}_0^1\\ =\pi \left(\frac{1}{2}-\frac{1}{3}\right)=\frac{\pi }{6}\end{array}$

c.

To determine

To find: The volume of the solid obtained by rotating the region bounded by the $y=x$ and $y=x^2$ about $y=2$ .

Answer to Problem 9RE

The required volume is $\frac{\pi }{2}$ .

Explanation of Solution

Given information :

  $y=x\mathrm{.........}\left(i\right)$

  $y=x^2\mathrm{............}\left(ii\right)$

Formula used :

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

Where the axis of rotation is $y=p$ and $a\text{ and }b$ are the upper and lower limit of $x$ .

Calculation:

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

To find intersecting points we intersect $y=x$ and $y=x^2$

  $x=x^2\Rightarrow x^2-x=0\Rightarrow x=1,x=0$

At $x=0$ , $y=0$ ;

At $x=1,y=1$

Here, $x=0\text{ \& }x=1$ are the boundaries.

So,

  $\begin{array}{c}{V}_w=\pi {\displaystyle \underset{0}{\overset{1}{\int }}\left[{\left(2-x\right)}^2-{\left(2-\sqrt{x}\right)}^2\right]dx}\\ =\pi {\displaystyle \underset{0}{\overset{1}{\int }}\left(4-4x+x^2-4+4\sqrt{x}-x\right)dx}\\ =\pi {\displaystyle \underset{0}{\overset{1}{\int }}\left(x^2-5x+4x^{\frac{1}{2}}\right)dx}\\ =\pi {\left[\frac{x^3}{3}-\frac{5x^2}{2}+\frac{4x^{\frac{3}{2}}}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}_0^1\\ =\pi \left(\frac{1}{3}-\frac{5}{2}+\frac{8}{3}\right)\\ =\frac{\pi }{2}\end{array}$