Chapter 6.3, Problem 23E

a.

To determine

Find the volume of the solid.

Answer to Problem 23E

The volume enclosed by the given cylinder $y=3$

Explanation of Solution

Consider the equation ${\displaystyle \underset{0}{\overset{3}{\int }}2\pi x^{5}dx}$

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 ${\displaystyle \underset{0}{\overset{3}{\int }}2\pi x^{5}dx}$

Re −arrange the terms as ${\displaystyle \underset{0}{\overset{3}{\int }}2\pi \left(x^4\right)dx=}{\displaystyle \underset{0}{\overset{3}{\int }}2\pi xydx}$

Where $y=x^4$

Therefore this represent the volume of solid obtained after rotation about $y-$ axis of the curve $y=x^4$

Between $x=0\text{ and }x=3$

Thus, the volume of the solid by the given integral $x=0\text{ and }x=3$

b.

To determine

Find the volume of the solid.

Answer to Problem 23E

The volume enclosed by the given cylinder is $V=\pi r^{2}h$.

Explanation of Solution

  $V={\displaystyle \underset{0}{\overset{1}{\int }}2\pi \left(3-y\right)\left(1-y^2\right)dy}$

One possible interpretation of the integral would be the region bounded by

  $x=1-y^2\text{, }x=0,0\le y\le 1,$ About the line $y=3$

Alternatively replace $x=1-y^2$ with $x=y^2\text{,}$

  $x=1$ the shaded part would be flipped horizontally left to right.

Or another would be instead of $x=1-y^2$ you have $y=\sqrt{1-x}\text{,}$ which is only the upper half of the parabola. So would not need to specify $0\le y\le 1,$ since that is understood from the range of $y=\sqrt{1-x}\text{,}$ because it can’t be less than $\text{0}$ .

Example cylinder shown for $y=0.5$

  

The region bounded by $x=1-y^2\text{, }x=0,0\le y\le 1,$ about the line $y=3$

Thus, the volume enclosed by the given cylinder $y=3$