Chapter 8.3, Problem 33E

a.

To determine

To find : the volume of the solid about the indicated axis.

Answer to Problem 33E

The volume of the solid is $\frac{6\pi }{5}$ .

Explanation of Solution

Given information : The x -axis the graph provided in the question is shown below:

  

Calculation :

By cylindrical shell method.

The shell height is just given by $12\left(y^2-y^3\right)$

So, integral is,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{1}{\int }}\left(2\pi \right)\left(y\right)}12\left(y^2-y^3\right)dy\\ \Rightarrow \frac{6\pi }{5}\end{array}$

b.

To determine

To find : the volume of the solid about the indicated axis.

Answer to Problem 33E

The volume of the solid is $\frac{4\pi }{5}$ .

Explanation of Solution

Given information : About the line $y=1$

the graph provided in the question is shown below:

  

Calculation :

By cylindrical shell method.

The shell height is just given by $1-y$

So, the integral is,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{1}{\int }}\left(2\pi \right)\left(1-y\right)}12\left(y^2-y^3\right)dy\\ \Rightarrow \frac{4\pi }{5}\end{array}$

c.

To determine

To find : the volume of the solid about the indicated axis.

Answer to Problem 33E

The volume of the solid is $2\pi$ .

Explanation of Solution

Given information : About the line $y=\raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$

the graph provided in the question is shown below:

  

Calculation :

By cylindrical shell method.

The shell height is just given by $\frac{8}{5}-y$

So, the integral is,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{1}{\int }}\left(2\pi \right)\left(\frac{8}{5}-y\right)}12\left(y^2-y^3\right)dy\\ \Rightarrow 2\pi \end{array}$

d.

To determine

To find : the volume of the solid about the indicated axis.

Answer to Problem 33E

The volume of the solid is $2\pi$ .

Explanation of Solution

Given information : About the line $y=-\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$ the graph provided in the question is shown below

  

Calculation :

By cylindrical shell method.

The shell height is just given by $y+\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$

So, the integral is,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{1}{\int }}\left(2\pi \right)\left(y+\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)}12\left(y^2-y^3\right)dy\\ \Rightarrow 2\pi \end{array}$