Chapter 6.3, Problem 24E

a.

To determine

Find the volume of the solid.

Answer to Problem 24E

The volume of $y$ the given solid integral $y=0\text{ and }y=2$

Explanation of Solution

Consider the integral, $2\pi {\displaystyle \underset{0}{\overset{2}{\int }}\frac{y}{1+y^2}dy}$

Calculation:

When a curve is rotate above $y-$ axis, the volume of the solid obtained is $V={\displaystyle \underset{a}{\overset{b}{\int }}2\pi xydx}$

When a curve is rotate above $x-$ axis, the volume of the solid obtained is $V={\displaystyle \underset{a}{\overset{b}{\int }}2\pi xydy}$

The given integral $2\pi {\displaystyle \underset{0}{\overset{2}{\int }}\frac{y}{1+y^2}dy}$

Re −arrange the terms as $2\pi {\displaystyle \underset{0}{\overset{2}{\int }}\frac{1}{\left(1+y^2\right)}ydy}={\displaystyle \underset{0}{\overset{3}{\int }}2\pi xydy}$ Where $x=\frac{1}{1+y^2}$

Therefore this represent the volume of solid obtained after rotation about $x-$ axis of the curve $x=\frac{1}{1+y^2}$ Between $y=0\text{ and }y=2$

  

Thus, the volume of $y$ the given solid integral $y=0\text{ and }y=2$

b.

To determine

Find the volume of the solid.

Answer to Problem 24E

The volume enclosed by the given cylinder is $x=\pi$.

Explanation of Solution

  $V=2\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(\pi -x\right)\left(\cos x-\sin x\right)dy}$

One possible interpretation of the integral would be the region bounded by $y=\cos x\text{, }y=\sin x,x=0,$ about the line $x=\pi$

Alternatively replace $y=\cos x,y=\sin x$ with $y=\cos x-\sin x,y=0.$ the shaded part would be flipped horizontally left to right.

Example cylinder shown for $x=0.5$

  

The region bounded by $y=\cos x\text{, }y=\sin x,x=0,$ about the line $x=\pi$

Thus, the volume enclosed by the given cylinder $x=\pi$.