Chapter 7.3, Problem 36E

a.

To determine

To find:The volume of the solid generated by resolving the region bounded by the curves.

Answer to Problem 36E

The volume of the solid generated is $V=\frac{8\pi }{3}\text{ cu}\text{. units}$ .

Explanation of Solution

Given information:

The indicated axis is $x$ -axis.

Formula used:

There are two formulas that can be applied, depending on the strip utilized or the axis of revolution, to calculate the volume of the solid of revolution using the cylindrical shell method.

Recall that the strip to be utilized in the cylindrical shell method must be parallel to the axis of revolution.

For horizontal strip (rotated about the $x$ -axis),

  $V={\displaystyle {\int }_a^{b}2\pi rhdy}$

For vertical strip (rotated about the $y$ -axis),

  $V={\displaystyle {\int }_a^{b}2\pi rhdx}$

Calculation:

Let's first create a graph showing the region that will be rotated about the $x$ -axis.

  

From the graph, the limits of integration is bounded from $y=0\text{ to }y=2$ and the radius $r=y$ .

The height of the strip is given by,

  $h=\frac{y^2}{2}-\left(\frac{y^4}{4}-\frac{y^2}{2}\right)$

From the given axis let’s use the horizontal strip,

  $\begin{array}{c}V={\displaystyle {\int }_0^{2}2}\pi y\left[\left(\frac{y^2}{2}-\left(\frac{y^4}{4}-\frac{y^2}{2}\right)\right)\right]dy\\ =2\pi {\displaystyle {\int }_0^{2}y}\left[\left(\frac{y^2}{2}-\frac{y^4}{4}+\frac{y^2}{2}\right)\right]dy\\ =2\pi {\displaystyle {\int }_0^{2}y}\left(y^2-\frac{y^4}{4}\right)dy\\ =2\pi {\displaystyle {\int }_0^2\left(y^3-\frac{y^5}{4}\right)}dy\\ ={\left.2\pi \left[\frac{y^4}{4}-\frac{y^6}{24}\right]\right|}_0^2\\ =2\pi \left[\frac{{(2)}^4}{4}-\frac{{(2)}^6}{24}\right]\\ V=\frac{8\pi }{3}\end{array}$

Therefore, the volume of the solid generated is $V=\frac{8\pi }{3}\text{ cu}\text{. units}$ .

b.

To determine

To find: The volume of the solid generated by resolving the region bounded by the curves.

Answer to Problem 36E

The volume of the solid generated is $V=\frac{8\pi }{5}\text{ cu}\text{. units}$ .

Explanation of Solution

Given information:

The line $y=2$ .

Formula used:

There are two formulas that can be applied, depending on the strip utilized or the axis of revolution, to calculate the volume of the solid of revolution using the cylindrical shell method.

Recall that the strip to be utilized in the cylindrical shell method must be parallel to the axis of revolution.

For horizontal strip (rotated about the $x$ -axis),

  $V={\displaystyle {\int }_a^{b}2\pi rhdy}$

For vertical strip (rotated about the $y$ -axis),

  $V={\displaystyle {\int }_a^{b}2\pi rhdx}$

Calculation:

Let's first create a graph showing the region that will be rotated about $y=2$ .

  

From the graph, the limits of integration is bounded from $y=0\text{ to }y=2$ and radius $r=2-y$ .

The height of the strip is given by,

  $h=\frac{y^2}{2}-\left(\frac{y^4}{4}-\frac{y^2}{2}\right)$

From the given axis let’s use the horizontal strip,

  $\begin{array}{c}V={\displaystyle {\int }_0^{2}2}\pi \left(2-y\right)\left[\left(\frac{y^2}{2}-\left(\frac{y^4}{4}-\frac{y^2}{2}\right)\right)\right]dy\\ =2\pi {\displaystyle {\int }_0^2\left(2-y\right)}\left[\left(\frac{y^2}{2}-\frac{y^4}{4}+\frac{y^2}{2}\right)\right]dy\\ =2\pi {\displaystyle {\int }_0^2\left(2-y\right)}\left(y^2-\frac{y^4}{4}\right)dy\\ =2\pi {\displaystyle {\int }_0^2\left(2y^2-\frac{y^4}{2}-y^3+\frac{y^5}{4}\right)}dy\\ ={\left.2\pi \left[\frac{2y^3}{3}-\frac{y^{{}^5}}{10}-\frac{y^4}{4}+\frac{y^6}{24}\right]\right|}_0^2\\ =2\pi \left[{\frac{2\left(2\right)}{3}}^3-\frac{{\left(2\right)}^5}{10}-\frac{{(2)}^4}{4}+\frac{{(2)}^6}{24}\right]\\ V=\frac{8\pi }{5}\end{array}$

Therefore, the volume of the solid generated is $V=\frac{8\pi }{5}\text{ cu}\text{. units}$ .

c.

To determine

To find: The volume of the solid generated by resolving the region bounded by the curves.

Answer to Problem 36E

The volume of the solid generated is $V=8\pi \text{ cu}\text{. units}$ .

Explanation of Solution

Given information:

The line $y=5$ .

Formula used:

There are two formulas that can be applied, depending on the strip utilized or the axis of revolution, to calculate the volume of the solid of revolution using the cylindrical shell method.

Recall that the strip to be utilized in the cylindrical shell method must be parallel to the axis of revolution.

For horizontal strip (rotated about the $x$ -axis),

  $V={\displaystyle {\int }_a^{b}2\pi rhdy}$

For vertical strip (rotated about the $y$ -axis),

  $V={\displaystyle {\int }_a^{b}2\pi rhdx}$

Calculation:

Let's first create a graph showing the region that will be rotated about $y=5$ .

  

From the graph, the limits of integration is bounded from $y=0\text{ to }y=2$ and radius $r=5-y$ .

The height of the strip is given by,

  $h=\frac{y^2}{2}-\left(\frac{y^4}{4}-\frac{y^2}{2}\right)$

From the given axis let’s use the horizontal strip,

  $\begin{array}{c}V={\displaystyle {\int }_0^{2}2}\pi \left(5-y\right)\left[\left(\frac{y^2}{2}-\left(\frac{y^4}{4}-\frac{y^2}{2}\right)\right)\right]dy\\ =2\pi {\displaystyle {\int }_0^2\left(5-y\right)}\left[\left(\frac{y^2}{2}-\frac{y^4}{4}+\frac{y^2}{2}\right)\right]dy\\ =2\pi {\displaystyle {\int }_0^2\left(5-y\right)}\left(y^2-\frac{y^4}{4}\right)dy\\ =2\pi {\displaystyle {\int }_0^2\left(5y^2-\frac{5y^4}{2}-y^3+\frac{y^5}{4}\right)}dy\\ ={\left.2\pi \left[\frac{5y^3}{3}-\frac{y^{{}^5}}{4}-\frac{y^4}{4}+\frac{y^6}{24}\right]\right|}_0^2\\ =2\pi \left[{\frac{5\left(2\right)}{3}}^3-\frac{{\left(2\right)}^5}{4}-\frac{{(2)}^4}{4}+\frac{{(2)}^6}{24}\right]\\ V=8\pi \end{array}$

Therefore, the volume of the solid generated is $V=8\pi \text{ cu}\text{. units}$ .

d.

To determine

To find: The volume of the solid generated by resolving the region bounded by the curves.

Answer to Problem 36E

The volume of the solid generated is $V=4\pi \text{ cu}\text{. units}$ .

Explanation of Solution

Given information:

The line $y=-\frac{5}{8}$ .

Formula used:

There are two formulas that can be applied, depending on the strip utilized or the axis of revolution, to calculate the volume of the solid of revolution using the cylindrical shell method.

Recall that the strip to be utilized in the cylindrical shell method must be parallel to the axis of revolution.

For horizontal strip (rotated about the $x$ -axis),

  $V={\displaystyle {\int }_a^{b}2\pi rhdy}$

For vertical strip (rotated about the $y$ -axis),

  $V={\displaystyle {\int }_a^{b}2\pi rhdx}$

Calculation:

Let's first create a graph showing the region that will be rotated about $y=-\frac{5}{8}$ .

  

From the graph, the limits of integration is bounded from $y=0\text{ to }y=2$ and radius $r=y+\frac{5}{8}$ .

The height of the strip is given by,

  $h=\frac{y^2}{2}-\left(\frac{y^4}{4}-\frac{y^2}{2}\right)$

From the given axis let’s use the horizontal strip,

  $\begin{array}{c}V={\displaystyle {\int }_0^{2}2}\pi \left(y+\frac{5}{8}\right)\left[\left(\frac{y^2}{2}-\left(\frac{y^4}{4}-\frac{y^2}{2}\right)\right)\right]dy\\ =2\pi {\displaystyle {\int }_0^2\left(y+\frac{5}{8}\right)}\left[\left(\frac{y^2}{2}-\frac{y^4}{4}+\frac{y^2}{2}\right)\right]dy\\ =2\pi {\displaystyle {\int }_0^2\left(y+\frac{5}{8}\right)}\left(y^2-\frac{y^4}{4}\right)dy\\ =2\pi {\displaystyle {\int }_0^2\left(y^3-\frac{y^5}{4}+\frac{5y^2}{8}-\frac{5y^4}{32}\right)}dy\\ ={\left.2\pi \left[\frac{y^4}{4}-\frac{y^{{}^6}}{24}+\frac{5y^3}{24}-\frac{y^5}{32}\right]\right|}_0^2\\ =2\pi \left[{\frac{\left(2\right)}{4}}^4-\frac{{\left(2\right)}^6}{24}+\frac{5{(2)}^3}{24}-\frac{{(2)}^5}{32}\right]\\ V=4\pi \end{array}$

Therefore, the volume of the solid generated is $V=4\pi \text{ cu}\text{. units}$ .