### Learning Objectives

- 6.3.1Calculate the volume of a solid of revolution by using the method of cylindrical shells.
- 6.3.2Compare the different methods for calculating a volume of revolution.

In this section, we examine the method of cylindrical shells, the final method for finding the volume of a solid of revolution. We can use this method on the same kinds of solids as the disk method or the washer method; however, with the disk and washer methods, we integrate along the coordinate axis parallel to the axis of revolution. With the method of cylindrical shells, we integrate along the coordinate axis *perpendicular* to the axis of revolution. The ability to choose which variable of integration we want to use can be a significant advantage with more complicated functions. Also, the specific geometry of the solid sometimes makes the method of using cylindrical shells more appealing than using the washer method. In the last part of this section, we review all the methods for finding volume that we have studied and lay out some guidelines to help you determine which method to use in a given situation.

### The Method of Cylindrical Shells

Again, we are working with a solid of revolution. As before, we define a region $R,$ bounded above by the graph of a function $y=f(x),$ below by the $x\text{-axis,}$ and on the left and right by the lines $x=a$ and $x=b,$ respectively, as shown in Figure 6.25(a). We then revolve this region around the *y*-axis, as shown in Figure 6.25(b). Note that this is different from what we have done before. Previously, regions defined in terms of functions of $x$ were revolved around the $x\text{-axis}$ or a line parallel to it.

Figure 6.25 (a) A region bounded by the graph of a function of $x.$ (b) The solid of revolution formed when the region is revolved around the $y\text{-axis}\text{.}$

As we have done many times before, partition the interval $\left[a,b\right]$ using a regular partition, $P=\left\{{x}_{0},{x}_{1}\text{,\u2026},{x}_{n}\right\}$ and, for $i=1,2\text{,\u2026},n,$ choose a point ${x}_{i}^{*}\in \left[{x}_{i-1},{x}_{i}\right].$ Then, construct a rectangle over the interval $\left[{x}_{i-1},{x}_{i}\right]$ of height $f({x}_{i}^{*})$ and width $\text{\Delta}x.$ A representative rectangle is shown in Figure 6.26(a). When that rectangle is revolved around the *y*-axis, instead of a disk or a washer, we get a cylindrical shell, as shown in the following figure.

Figure 6.26 (a) A representative rectangle. (b) When this rectangle is revolved around the $y\text{-axis},$ the result is a cylindrical shell. (c) When we put all the shells together, we get an approximation of the original solid.

To calculate the volume of this shell, consider Figure 6.27.

Figure 6.27 Calculating the volume of the shell.

The shell is a cylinder, so its volume is the cross-sectional area multiplied by the height of the cylinder. The cross-sections are annuli (ring-shaped regions—essentially, circles with a hole in the center), with outer radius ${x}_{i}$ and inner radius ${x}_{i-1}.$ Thus, the cross-sectional area is $\pi {x}_{i}^{2}-\pi {x}_{i-1}^{2}.$ The height of the cylinder is $f({x}_{i}^{*}).$ Then the volume of the shell is

$$\begin{array}{cc}\hfill {V}_{\text{shell}}& =f({x}_{i}^{*})(\pi {x}_{i}^{2}-\pi {x}_{i-1}^{2})\hfill \\ & =\pi f({x}_{i}^{*})\left({x}_{i}^{2}-{x}_{i-1}^{2}\right)\hfill \\ & =\pi f({x}_{i}^{*})\left({x}_{i}+{x}_{i-1}\right)\left({x}_{i}-{x}_{i-1}\right)\hfill \\ & =2\pi f({x}_{i}^{*})\left(\frac{{x}_{i}+{x}_{i-1}}{2}\right)\left({x}_{i}-{x}_{i-1}\right).\hfill \end{array}$$

Note that ${x}_{i}-{x}_{i-1}=\text{\Delta}x,$ so we have

$${V}_{\text{shell}}=2\pi f({x}_{i}^{*})\left(\frac{{x}_{i}+{x}_{i-1}}{2}\right)\text{\Delta}x.$$

Furthermore, $\frac{{x}_{i}+{x}_{i-1}}{2}$ is both the midpoint of the interval $\left[{x}_{i-1},{x}_{i}\right]$ and the average radius of the shell, and we can approximate this by ${x}_{i}^{*}.$ We then have

$${V}_{\text{shell}}\approx 2\pi f({x}_{i}^{*}){x}_{i}^{*}\text{\Delta}x.$$

Another way to think of this is to think of making a vertical cut in the shell and then opening it up to form a flat plate (Figure 6.28).

Figure 6.28 (a) Make a vertical cut in a representative shell. (b) Open the shell up to form a flat plate.

In reality, the outer radius of the shell is greater than the inner radius, and hence the back edge of the plate would be slightly longer than the front edge of the plate. However, we can approximate the flattened shell by a flat plate of height $f({x}_{i}^{*}),$ width $2\pi {x}_{i}^{*},$ and thickness $\text{\Delta}x$ (Figure 6.28). The volume of the shell, then, is approximately the volume of the flat plate. Multiplying the height, width, and depth of the plate, we get

$${V}_{\text{shell}}\approx f({x}_{i}^{*})\left(2\pi {x}_{i}^{*}\right)\text{\Delta}x,$$

which is the same formula we had before.

To calculate the volume of the entire solid, we then add the volumes of all the shells and obtain

$$V\approx {\displaystyle \sum}_{i=1}^{n}\left(2\pi {x}_{i}^{*}f({x}_{i}^{*})\text{\Delta}x\right).$$

Here we have another Riemann sum, this time for the function $2\pi xf(x).$ Taking the limit as $n\to \infty $ gives us

$$V=\underset{n\to \infty}{\text{lim}}{\displaystyle \sum}_{i=1}^{n}\left(2\pi {x}_{i}^{*}f({x}_{i}^{*})\text{\Delta}x\right)={\displaystyle {\int}_{a}^{b}\left(2\pi xf(x)\right)}dx.$$

This leads to the following rule for the method of cylindrical shells.

### Rule: The Method of Cylindrical Shells

Let $f(x)$ be continuous and nonnegative. Define $R$ as the region bounded above by the graph of $f(x),$ below by the $x\text{-axis},$ on the left by the line $x=a,$ and on the right by the line $x=b.$ Then the volume of the solid of revolution formed by revolving $R$ around the *y*-axis is given by

$$V={\displaystyle {\int}_{a}^{b}\left(2\pi xf(x)\right)}dx.$$

(6.6)

Now let’s consider an example.

### Example 6.12

#### The Method of Cylindrical Shells 1

Define $R$ as the region bounded above by the graph of $f(x)=1\text{/}x$ and below by the $x\text{-axis}$ over the interval $\left[1,3\right].$ Find the volume of the solid of revolution formed by revolving $R$ around the $y\text{-axis}.$

#### Solution

First we must graph the region $R$ and the associated solid of revolution, as shown in the following figure.

Figure 6.29 (a) The region $R$ under the graph of $f(x)=1\text{/}x$ over the interval $\left[1,3\right].$ (b) The solid of revolution generated by revolving $R$ about the $y\text{-axis}.$

Then the volume of the solid is given by

$$\begin{array}{cc}\hfill V& ={\displaystyle {\int}_{a}^{b}\left(2\pi xf(x)\right)}dx\hfill \\ & ={\displaystyle {\int}_{1}^{3}\left(2\pi x\left(\frac{1}{x}\right)\right)dx}\hfill \\ & ={\displaystyle {\int}_{1}^{3}2}\pi \phantom{\rule{0ex}{0ex}}dx={2\pi x|}_{1}^{3}=4\pi \phantom{\rule{0ex}{0ex}}{\text{units}}^{3}\text{.}\hfill \end{array}$$

### Checkpoint 6.12

Define *R* as the region bounded above by the graph of $f(x)={x}^{2}$ and below by the *x*-axis over the interval $\left[1,2\right].$ Find the volume of the solid of revolution formed by revolving $R$ around the $y\text{-axis}.$

### Example 6.13

#### The Method of Cylindrical Shells 2

Define *R* as the region bounded above by the graph of $f(x)=2x-{x}^{2}$ and below by the $x\text{-axis}$ over the interval $\left[0,2\right].$ Find the volume of the solid of revolution formed by revolving $R$ around the $y\text{-axis}.$

#### Solution

First graph the region $R$ and the associated solid of revolution, as shown in the following figure.

Figure 6.30 (a) The region $R$ under the graph of $f(x)=2x-{x}^{2}$ over the interval $\left[0,2\right].$ (b) The volume of revolution obtained by revolving $R$ about the $y\text{-axis}.$

Then the volume of the solid is given by

$$\begin{array}{cc}\hfill V& ={\displaystyle {\int}_{a}^{b}\left(2\pi xf(x)\right)}dx\hfill \\ & ={\displaystyle {\int}_{0}^{2}\left(2\pi x\left(2x-{x}^{2}\right)\right)}dx=2\pi {\displaystyle {\int}_{0}^{2}\left(2{x}^{2}-{x}^{3}\right)}dx\hfill \\ & ={2\pi \left[\frac{2{x}^{3}}{3}-\frac{{x}^{4}}{4}\right]\phantom{\rule{0ex}{0ex}}|}_{0}^{2}=\frac{8\pi}{3}\phantom{\rule{0ex}{0ex}}{\text{units}}^{3}\text{.}\hfill \end{array}$$

### Checkpoint 6.13

Define $R$ as the region bounded above by the graph of $f(x)=3x-{x}^{2}$ and below by the $x\text{-axis}$ over the interval $\left[0,2\right].$ Find the volume of the solid of revolution formed by revolving $R$ around the $y\text{-axis}.$

As with the disk method and the washer method, we can use the method of cylindrical shells with solids of revolution, revolved around the $x\text{-axis},$ when we want to integrate with respect to $y.$ The analogous rule for this type of solid is given here.

### Rule: The Method of Cylindrical Shells for Solids of Revolution around the *x*-axis

Let $g(y)$ be continuous and nonnegative. Define $Q$ as the region bounded on the right by the graph of $g(y),$ on the left by the $y\text{-axis},$ below by the line $y=c,$ and above by the line $y=d.$ Then, the volume of the solid of revolution formed by revolving $Q$ around the $x\text{-axis}$ is given by

$$V={\displaystyle {\int}_{c}^{d}\left(2\pi yg(y)\right)}dy.$$

### Example 6.14

#### The Method of Cylindrical Shells for a Solid Revolved around the *x*-axis

Define $Q$ as the region bounded on the right by the graph of $g(y)=2\sqrt{y}$ and on the left by the $y\text{-axis}$ for $y\in \left[0,4\right].$ Find the volume of the solid of revolution formed by revolving $Q$ around the *x*-axis.

#### Solution

First, we need to graph the region $Q$ and the associated solid of revolution, as shown in the following figure.

Figure 6.31 (a) The region $Q$ to the left of the function $g(y)$ over the interval $\left[0,4\right].$ (b) The solid of revolution generated by revolving $Q$ around the $x\text{-axis}.$

Label the shaded region $Q.$ Then the volume of the solid is given by

$$\begin{array}{cc}\hfill V& ={\displaystyle {\int}_{c}^{d}\left(2\pi yg(y)\right)}dy\hfill \\ & ={\displaystyle {\int}_{0}^{4}\left(2\pi y\left(2\sqrt{y}\right)\right)}dy=4\pi {\displaystyle {\int}_{0}^{4}{y}^{3\text{/}2}}dy\hfill \\ & ={4\pi [\frac{2{y}^{5\text{/}2}}{5}\left]\phantom{\rule{0ex}{0ex}}\right|}_{0}^{4}=\frac{256\pi}{5}\phantom{\rule{0ex}{0ex}}{\text{units}}^{3}\text{.}\hfill \end{array}$$

### Checkpoint 6.14

Define $Q$ as the region bounded on the right by the graph of $g(y)=3\text{/}y$ and on the left by the $y\text{-axis}$ for $y\in \left[1,3\right].$ Find the volume of the solid of revolution formed by revolving $Q$ around the $x\text{-axis}.$

For the next example, we look at a solid of revolution for which the graph of a function is revolved around a line other than one of the two coordinate axes. To set this up, we need to revisit the development of the method of cylindrical shells. Recall that we found the volume of one of the shells to be given by

$$\begin{array}{cc}\hfill {V}_{\text{shell}}& =f({x}_{i}^{*})(\pi {x}_{i}^{2}-\pi {x}_{i-1}^{2})\hfill \\ & =\pi f({x}_{i}^{*})\left({x}_{i}^{2}-{x}_{i-1}^{2}\right)\hfill \\ & =\pi f({x}_{i}^{*})\left({x}_{i}+{x}_{i-1}\right)\left({x}_{i}-{x}_{i-1}\right)\hfill \\ & =2\pi f({x}_{i}^{*})\left(\frac{{x}_{i}+{x}_{i-1}}{2}\right)\left({x}_{i}-{x}_{i-1}\right).\hfill \end{array}$$

This was based on a shell with an outer radius of ${x}_{i}$ and an inner radius of ${x}_{i-1}.$ If, however, we rotate the region around a line other than the $y\text{-axis},$ we have a different outer and inner radius. Suppose, for example, that we rotate the region around the line $x=\text{\u2212}k,$ where $k$ is some positive constant. Then, the outer radius of the shell is ${x}_{i}+k$ and the inner radius of the shell is ${x}_{i-1}+k.$ Substituting these terms into the expression for volume, we see that when a plane region is rotated around the line $x=\text{\u2212}k,$ the volume of a shell is given by

$$\begin{array}{cc}\hfill {V}_{\text{shell}}& =2\pi f({x}_{i}^{*})\left(\frac{\left({x}_{i}+k\right)+\left({x}_{i-1}+k\right)}{2}\right)\left(\left({x}_{i}+k\right)-\left({x}_{i-1}+k\right)\right)\hfill \\ & =2\pi f({x}_{i}^{*})\left(\left(\frac{{x}_{i}+{x}_{i-2}}{2}\right)+k\right)\text{\Delta}x.\hfill \end{array}$$

As before, we notice that $\frac{{x}_{i}+{x}_{i-1}}{2}$ is the midpoint of the interval $\left[{x}_{i-1},{x}_{i}\right]$ and can be approximated by ${x}_{i}^{*}.$ Then, the approximate volume of the shell is

$${V}_{\text{shell}}\approx 2\pi \left({x}_{i}^{*}+k\right)f({x}_{i}^{*})\text{\Delta}x.$$

The remainder of the development proceeds as before, and we see that

$$V={\displaystyle {\int}_{a}^{b}\left(2\pi \left(x+k\right)f(x)\right)}dx.$$

We could also rotate the region around other horizontal or vertical lines, such as a vertical line in the right half plane. In each case, the volume formula must be adjusted accordingly. Specifically, the $x\text{-term}$ in the integral must be replaced with an expression representing the radius of a shell. To see how this works, consider the following example.

### Example 6.15

#### A Region of Revolution Revolved around a Line

Define $R$ as the region bounded above by the graph of $f(x)=x$ and below by the $x\text{-axis}$ over the interval $\left[1,2\right].$ Find the volume of the solid of revolution formed by revolving $R$ around the line $x=\mathrm{-1}.$

#### Solution

First, graph the region $R$ and the associated solid of revolution, as shown in the following figure.

Figure 6.32 (a) The region $R$ between the graph of $f(x)$ and the $x\text{-axis}$ over the interval $\left[1,2\right].$ (b) The solid of revolution generated by revolving $R$ around the line $x=\mathrm{-1}.$

Note that the radius of a shell is given by $x+1.$ Then the volume of the solid is given by

$$\begin{array}{cc}\hfill V& ={\displaystyle {\int}_{1}^{2}\left(2\pi \left(x+1\right)f(x)\right)}dx\hfill \\ & ={\displaystyle {\int}_{1}^{2}\left(2\pi \left(x+1\right)x\right)}dx=2\pi {\displaystyle {\int}_{1}^{2}\left({x}^{2}+x\right)}dx\hfill \\ & ={2\pi \left[\frac{{x}^{3}}{3}+\frac{{x}^{2}}{2}\right]\phantom{\rule{0ex}{0ex}}|}_{1}^{2}=\frac{23\pi}{3}\phantom{\rule{0ex}{0ex}}{\text{units}}^{3}\text{.}\hfill \end{array}$$

### Checkpoint 6.15

Define $R$ as the region bounded above by the graph of $f(x)={x}^{2}$ and below by the $x\text{-axis}$ over the interval $\left[0,1\right].$ Find the volume of the solid of revolution formed by revolving $R$ around the line $x=\mathrm{-2}.$

For our final example in this section, let’s look at the volume of a solid of revolution for which the region of revolution is bounded by the graphs of two functions.

### Example 6.16

#### A Region of Revolution Bounded by the Graphs of Two Functions

Define $R$ as the region bounded above by the graph of the function $f(x)=\sqrt{x}$ and below by the graph of the function $g(x)=1\text{/}x$ over the interval $\left[1,4\right].$ Find the volume of the solid of revolution generated by revolving $R$ around the $y\text{-axis}.$

#### Solution

First, graph the region $R$ and the associated solid of revolution, as shown in the following figure.

Figure 6.33 (a) The region $R$ between the graph of $f(x)$ and the graph of $g(x)$ over the interval $\left[1,4\right].$ (b) The solid of revolution generated by revolving $R$ around the $y\text{-axis}.$

Note that the axis of revolution is the $y\text{-axis},$ so the radius of a shell is given simply by $x.$ We don’t need to make any adjustments to the *x*-term of our integrand. The height of a shell, though, is given by $f(x)-g(x),$ so in this case we need to adjust the $f(x)$ term of the integrand. Then the volume of the solid is given by

$$\begin{array}{cc}\hfill V& ={\displaystyle {\int}_{1}^{4}\left(2\pi x\left(f(x)-g(x)\right)\right)}dx\hfill \\ & ={\displaystyle {\int}_{1}^{4}\left(2\pi x\left(\sqrt{x}-\frac{1}{x}\right)\right)dx}=2\pi {\displaystyle {\int}_{1}^{4}\left({x}^{3\text{/}2}-1\right)}dx\hfill \\ & ={2\pi \left[\frac{2{x}^{5\text{/}2}}{5}-x\right]\phantom{\rule{0ex}{0ex}}|}_{1}^{4}=\frac{94\pi}{5}\phantom{\rule{0ex}{0ex}}{\text{units}}^{3}.\hfill \end{array}$$

### Checkpoint 6.16

Define $R$ as the region bounded above by the graph of $f(x)=x$ and below by the graph of $g(x)={x}^{2}$ over the interval $\left[0,1\right].$ Find the volume of the solid of revolution formed by revolving $R$ around the $y\text{-axis}.$

### Which Method Should We Use?

We have studied several methods for finding the volume of a solid of revolution, but how do we know which method to use? It often comes down to a choice of which integral is easiest to evaluate. Figure 6.34 describes the different approaches for solids of revolution around the $x\text{-axis}.$ It’s up to you to develop the analogous table for solids of revolution around the $y\text{-axis}.$

Figure 6.34

Let’s take a look at a couple of additional problems and decide on the best approach to take for solving them.

### Example 6.17

#### Selecting the Best Method

For each of the following problems, select the best method to find the volume of a solid of revolution generated by revolving the given region around the $x\text{-axis},$ and set up the integral to find the volume (do not evaluate the integral).

- The region bounded by the graphs of $y=x,$ $y=2-x,$ and the $x\text{-axis}.$
- The region bounded by the graphs of $y=4x-{x}^{2}$ and the $x\text{-axis}.$

#### Solution

- First, sketch the region and the solid of revolution as shown.
Figure 6.35 (a) The region $R$ bounded by two lines and the $x\text{-axis}.$ (b) The solid of revolution generated by revolving $R$ about the $x\text{-axis}.$

Looking at the region, if we want to integrate with respect to $x,$ we would have to break the integral into two pieces, because we have different functions bounding the region over $\left[0,1\right]$ and $\left[1,2\right].$ In this case, using the disk method, we would have$$V={\displaystyle {\int}_{0}^{1}\left(\pi {x}^{2}\right)}dx+{\displaystyle {\int}_{1}^{2}\left(\pi {(2-x)}^{2}\right)}dx.$$

If we used the shell method instead, we would use functions of $y$ to represent the curves, producing$$\begin{array}{cc}\hfill V& ={\displaystyle {\int}_{0}^{1}\left(2\pi y\left[\left(2-y\right)-y\right]\right)}dy\hfill \\ & ={\displaystyle {\int}_{0}^{1}\left(2\pi y\left[2-2y\right]\right)}dy.\hfill \end{array}$$

(Video) Calculus: Volumes by Cylindrical Shells (Section 6.3)

Neither of these integrals is particularly onerous, but since the shell method requires only one integral, and the integrand requires less simplification, we should probably go with the shell method in this case. - First, sketch the region and the solid of revolution as shown.
Figure 6.36 (a) The region $R$ between the curve and the $x\text{-axis}.$ (b) The solid of revolution generated by revolving $R$ about the $x\text{-axis}.$

Looking at the region, it would be problematic to define a horizontal rectangle; the region is bounded on the left and right by the same function. Therefore, we can dismiss the method of shells. The solid has no cavity in the middle, so we can use the method of disks. Then$$V={\displaystyle {\int}_{0}^{4}\pi}{\left(4x-{x}^{2}\right)}^{2}dx.$$

### Checkpoint 6.17

Select the best method to find the volume of a solid of revolution generated by revolving the given region around the $x\text{-axis},$ and set up the integral to find the volume (do not evaluate the integral): the region bounded by the graphs of $y=2-{x}^{2}$ and $y={x}^{2}.$

### Section 6.3 Exercises

For the following exercises, find the volume generated when the region between the two curves is rotated around the given axis. Use both the shell method and the washer method. Use technology to graph the functions and draw a typical slice by hand.

114.

**[T]** Bounded by the curves $y=3x,x=0,$ and $y=3$ rotated around the $y\text{-axis}.$

115.

**[T]** Bounded by the curves $y=3x,y=0,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=3$ rotated around the $y\text{-axis}.$

116.

**[T]** Bounded by the curves $y=3x,y=0,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}y=3$ rotated around the $x\text{-axis}.$

117.

**[T]** Bounded by the curves $y=3x,y=0,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=3$ rotated around the $x\text{-axis}.$

118.

**[T]** Bounded by the curves $y=2{x}^{3},y=0,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=2$ rotated around the $y\text{-axis}.$

119.

**[T]** Bounded by the curves $y=2{x}^{3},y=0,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=2$ rotated around the $x\text{-axis}.$

For the following exercises, use shells to find the volumes of the given solids. Note that the rotated regions lie between the curve and the $x\text{-axis}$ and are rotated around the $y\text{-axis}.$

120.

$y=1-{x}^{2},x=0,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=1$

121.

$y=5{x}^{3},x=0,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=1$

122.

$y=\frac{1}{x},x=1,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=100$

123.

$y=\sqrt{1-{x}^{2}},x=0,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=1$

124.

$y=\frac{1}{1+{x}^{2}},x=0,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=3$

125.

$y=\text{sin}{x}^{2},x=0,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=\sqrt{\pi}$

126.

$y=\frac{1}{\sqrt{1-{x}^{2}}},x=0,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=\frac{1}{2}$

127.

$y=\sqrt{x},x=0,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=1$

128.

$y={\left(1+{x}^{2}\right)}^{3},x=0,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=1$

129.

$y=5{x}^{3}-2{x}^{4},x=0,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=2$

For the following exercises, use shells to find the volume generated by rotating the regions between the given curve and $y=0$ around the $x\text{-axis}.$

130.

$y=\sqrt{1-{x}^{2}},x=0,\phantom{\rule{0ex}{0ex}}x=1$ and the *x*-axis

131.

$y={x}^{2},x=0,\phantom{\rule{0ex}{0ex}}x=2$ and the *x*-axis

132.

$y=\frac{{x}^{3}}{2},\phantom{\rule{0ex}{0ex}}x=0,\phantom{\rule{0ex}{0ex}}x=2,$ and the *x*-axis

133.

$y=\frac{2}{{x}^{2}},\phantom{\rule{0ex}{0ex}}x=1,\phantom{\rule{0ex}{0ex}}x=2,$ and the *x*-axis

134.

$x=\frac{1}{1+{y}^{2}},y=1,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}y=4$

135.

$x=\frac{1+{y}^{2}}{y},y=1,\phantom{\rule{0ex}{0ex}}y=4,$ and the *y*-axis

136.

$x=\text{cos}\phantom{\rule{0ex}{0ex}}y,x=0,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}y=\pi $

137.

$x={y}^{3}-2{y}^{2},\phantom{\rule{0ex}{0ex}}x=0,\phantom{\rule{0ex}{0ex}}x=9$

138.

$x=\sqrt{y}+1,\phantom{\rule{0ex}{0ex}}x=1,\phantom{\rule{0ex}{0ex}}x=3,$ and the *x*-axis

139.

$x=\sqrt[3]{27y}\text{and}\phantom{\rule{0ex}{0ex}}x=\frac{3y}{4}$

For the following exercises, find the volume generated when the region between the curves is rotated around the given axis.

140.

$y=3-x,y=0,x=0,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=2$ rotated around the $y\text{-axis}.$

141.

$y={x}^{3},x=0,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}y=8$ rotated around the $y\text{-axis}.$

142.

$y={x}^{2},y=x,$ rotated around the $y\text{-axis}.$

143.

$y=\sqrt{x},y=0,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=1$ rotated around the line $x=2.$

144.

$y=\frac{1}{4-x},x=1,\phantom{\rule{0ex}{0ex}}x=2\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}y=0$ rotated around the line $x=4.$

145.

$y=\sqrt{x}\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}y={x}^{2}$ rotated around the $y\text{-axis}.$

146.

$y=\sqrt{x}\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}y={x}^{2}$ rotated around the line $x=2.$

147.

$x={y}^{3},x=\frac{1}{y},x=1,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=2$ rotated around the $x\text{-axis}.$

148.

$x={y}^{2}\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}y=x$ rotated around the line $y=2.$

149.

**[T]** Left of $x=\text{sin}\left(\pi y\right),$ right of $y=x,$ around the $y\text{-axis}.$

For the following exercises, use technology to graph the region. Determine which method you think would be easiest to use to calculate the volume generated when the function is rotated around the specified axis. Then, use your chosen method to find the volume.

150.

**[T]** $y={x}^{2}$ and $y=4x$ rotated around the $y\text{-axis}.$

151.

**[T]** $y=\text{cos}\left(\pi x\right),y=\text{sin}\left(\pi x\right),x=\frac{1}{4},\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=\frac{5}{4}$ rotated around the $y\text{-axis}.$ This exercise requires advanced technique. You may use technology to perform the integration.

152.

**[T]** $y={x}^{2}-2x,x=2,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=4$ rotated around the $y\text{-axis}.$

153.

**[T]** $y={x}^{2}-2x,x=2,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=4$ rotated around the $x\text{-axis}.$

154.

**[T]** $y=3{x}^{3}-2,y=x,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=2$ rotated around the $x\text{-axis}.$

155.

**[T]** $y=3{x}^{3}-2,y=x,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=2$ rotated around the $y\text{-axis}.$

156.

**[T]** $x=\text{sin}\left(\pi {y}^{2}\right)$ and $x=\sqrt{2}y$ rotated around the $x\text{-axis}.$

157.

**[T]** $x={y}^{2},x={y}^{2}-2y+1,\phantom{\rule{0ex}{0ex}}\text{and}\phantom{\rule{0ex}{0ex}}x=2$ rotated around the $y\text{-axis}.$

For the following exercises, use the method of shells to approximate the volumes of some common objects, which are pictured in accompanying figures.

158.

Use the method of shells to find the volume of a sphere of radius $r.$

159.

Use the method of shells to find the volume of a cone with radius $r$ and height $h.$

160.

Use the method of shells to find the volume of an ellipsoid $\left({x}^{2}\text{/}{a}^{2}\right)+\left({y}^{2}\text{/}{b}^{2}\right)=1$ rotated around the $x\text{-axis}.$

161.

Use the method of shells to find the volume of a cylinder with radius $r$ and height $h.$

162.

Use the method of shells to find the volume of the donut created when the circle ${x}^{2}+{y}^{2}=4$ is rotated around the line $x=4.$

163.

Consider he region enclosed by the graphs of $y=f\left(x\right),y=1+f\left(x\right),x=0,y=0,$ and $x=a>0.$ What is the volume of the solid generated when this region is rotated around the $y\text{-axis}?$ Assume that the function is defined over the interval $[0,a].$

164.

Consider the function $y=f\left(x\right),$ which decreases from $f\left(0\right)=b$ to $f\left(1\right)=0.$ Set up the integrals for determining the volume, using both the shell method and the disk method, of the solid generated when this region, with $x=0$ and $y=0,$ is rotated around the $y\text{-axis}.$ Prove that both methods approximate the same volume. Which method is easier to apply? (*Hint:* Since $f\left(x\right)$ is one-to-one, there exists an inverse ${f}^{\mathrm{-1}}(y).)$

## FAQs

### What is the volume of revolution using cylindrical shells? ›

The volume of the cylindrical shell is then **V = 2πrh∆r**. Here the factor 2πr is the average circumference of the cylindrical shell, the factor h is its height, and the factor ∆r is its the thickness.

**How do you find volume in Calc 1? ›**

If the region bounded above by the graph of f, below by the x-axis, and on the sides by x=a and x=b is revolved about the x-axis, the volume V of the generated solid is given by **V=∫abπ[f(x)]2dx**. We can also obtain solids by revolving curves about the y-axis.

**What is the cylindrical shells method in calculus? ›**

The shell method, sometimes referred to as the method of cylindrical shells, is **another technique commonly used to find the volume of a solid of revolution**. So, the idea is that we will revolve cylinders about the axis of revolution rather than rings or disks, as previously done using the disk or washer methods.

**What is cylinder revolution formula? ›**

Rule: The Method of Cylindrical Shells for Solids of Revolution around the x-axis. **V = ∫ c d ( 2 π y g ( y ) ) d y** . V = ∫ c d ( 2 π y g ( y ) ) d y .

**How do I find the volume of the cylinder? ›**

Cylinder's volume is given by the formula, **πr ^{2}h**, where r is the radius of the circular base and h is the height of the cylinder.

**How do you find the volume of a cylindrical hole? ›**

The formula to calculate the volume of a hollow cylinder is given as, **Volume of hollow cylinder = π (R ^{2} - r^{2}) h cubic units**, where, 'R' is the outer radius, 'r' is the inner radius, and, 'h' is the height of the hollow cylinder.

**What are the 3 formulas for volume? ›**

Shapes | Volume Formula | Variables |
---|---|---|

Rectangular Solid or Cuboid | V = l × w × h | l = Length w = Width h = Height |

Cube | V = a^{3} | a = Length of edge or side |

Cylinder | V = πr^{2}h | r = Radius of the circular base h = Height |

Prism | V = B × h | B = Area of base, (B = side^{2} or length.breadth) h = Height |

**What is volume in math grade 1? ›**

What is Volume? Volume is **the amount of space that something takes up**. In our example above, the bigger glass holds more. It has a bigger volume. Tip: Another word for volume is capacity.

**What is the formula for volume volume? ›**

To find the volume of a box, simply **multiply length, width, and height** — and you're good to go! For example, if a box is 5×7×2 cm, then the volume of a box is 70 cubic centimeters. For dimensions that are relatively small whole numbers, calculating volume by hand is easy.

**What is the shell formula in calculus? ›**

ΔV=2πxyΔx. The shell method calculates the volume of the full solid of revolution by summing the volumes of these thin cylindrical shells as the thickness Δ x \Delta x Δx goes to 0 0 0 in the limit: V = ∫ d V = ∫ a b 2 π x y d x = ∫ a b 2 π x f ( x ) d x .

### What is the formula for calculating shells? ›

The formula to find the maximum number of electrons that can be accommodated in a shell = **2n2** where 'n' is the number of the given shell from the nucleus.

**What is cylinder basic formulas? ›**

**Cylinder**

- The volume of a cylinder = Area of the base × Height of the cylinder = πr²h.
- Lateral Surface Area = Perimeter of base × height = 2πrh = πdh.
- Total Surface Area = Lateral Surface Area + Area of bases = 2πrh + 2πr² = 2πr (h+r)

**What is the cylinder equation example? ›**

The formula for the volume of a cylinder is **V=Bh or V=πr2h** . The radius of the cylinder is 8 cm and the height is 15 cm. Substitute 8 for r and 15 for h in the formula V=πr2h .

**What is a revolution in math? ›**

A revolution in math is **a full rotation that is equal to 360 degrees**. A complete circle is also equal to 360 degrees, and a half circle is equal to 180 degrees. Spinning and orbiting are two types of revolution.

**What is the volume of a revolved shape? ›**

Answer: The volume of a solid rotated about the y-axis can be calculated by **V = π∫ ^{d}_{c}[f(y)]^{2}dy**.

**What is volume in math? ›**

In math, volume is **the amount of space in a certain 3D object**. For instance, a fish tank has 3 feet in length, 1 foot in width and two feet in height. To find the volume, you multiply length times width times height, which is 3x1x2, which equals six. So the volume of the fish tank is 6 cubic feet.

**What is an example of a cylinder? ›**

Some real-life examples of cylinder shape are **pipes, fire extinguishers, water tanks, cold-drink cans**, etc.

**What is the new volume of a cylinder? ›**

Therefore, the total formula for the volume of the cylinder is: **V = πr ^{2}h**.

**How do you find the volume of a cylinder with the circumference? ›**

Thus, when the circumference of the base of a cylinder (C) and its height (h) are given, then we first solve the equation C = 2πr for 'r' and then we apply the volume of a cylinder formula, which is, **V = πr ^{2}h**.

**How do you find volume with 3 sides? ›**

The area of a triangle is A=12bh. Essentially, to find to the volume of the triangular prism, you are **multiplying the area of the triangle times the length or depth**. So, the formula for the volume of a triangular prism would be V=12bhl.

### What is volume in math grade 3? ›

Volume is **the amount of space that a solid object takes up**. Finding the volume of a 3-D shape is similar to finding the area of 2-D shapes - but only with 1 additional dimension to include. Take a look at this cube. It has 3 dimensions. To know its volume, we figure out how many cubic units can fit in it.

**What is volume formula easy? ›**

Whereas the basic formula for the area of a rectangular shape is length × width, the basic formula for volume is **length × width × height**.

**What is volume for dummies? ›**

Volume refers to **the amount of space the object takes up**. In other words, volume is a measure of the size of an object, just like height and width are ways to describe size. If the object is hollow (in other words, empty), volume is the amount of water it can hold. Try this at home: Take a large cup and a small cup.

**What is volume in math for kids? ›**

Volume is **the amount of physical space a 3D object takes up**. It is the 3D equivalent of area for a 2D shape. It is measured in cubic measurements, like cm³. This can be found by multiplying its length × height × width.

**What is an example of volume in math? ›**

Volume is the measure of the capacity that an object holds. For example, **if a cup can hold 100 ml of water up to the brim, its volume is said to be 100 ml**.

**What is shell shape in maths? ›**

In geometry, **a spherical shell** is a generalization of an annulus to three dimensions. It is the region of a ball between two concentric spheres of differing radii.

**What is the mathematical pattern of a shell? ›**

Mathematicians have learned to use Fibonacci's sequence to describe certain shapes that appear in nature. These shapes are called **logarithmic spirals**, and Nautilus shells are just one example. You also see logarithmic spiral shapes in spiral galaxies, and in many plants such as sunflowers.

**What are the shell numbers? ›**

**The shell closest to the nucleus, 1n, can hold two electrons, while the next shell, 2n, can hold eight, and the third shell, 3n, can hold up to eighteen**. The number of electrons in the outermost shell of a particular atom determines its reactivity, or tendency to form chemical bonds with other atoms.

**What are the formulas for circles and cylinders? ›**

Answer: The volume of a cylinder can be calculated using the formula **V=πr2(h)**. When determining the volume of a cylinder, you are simply finding the area of the circular base shape and then multiplying this by the height.

**How do you find the volume of a spherical shell? ›**

**Volume of spherical shell =4/3π(R3−r3).**

**R=radius of outer part**

**r=radius of inner part**

- Volume of a spherical shell with outer and inner radii R and r respectively is: ...
- The volume of a spherical shell whose internal and external diameters are 8cm and 10cm respectively (in cubic cm) is:

### How do you calculate shell length? ›

The double-logarithmic regression line shown is represented by the following equations: Loripes : **SL = 10.002*HT 0.950 ( R 2 = 0.84, p < 0.001, n = 106) and SL = 14.807*H 0.811 ( R 2 = 0.79, p < 0.001, n = 106)** and for Dosinia : SL = 7.145*HT 0.891.

**How do you find the volume of a cylindrical cylinder? ›**

A cylinder's volume is **π r² h**, and its surface area is 2π r h + 2π r².

**How do you find the volume of a cylindrical object? ›**

Cylinder's volume is given by the formula, **πr ^{2}h**, where r is the radius of the circular base and h is the height of the cylinder.

**How do you find the volume of a cylindrical block? ›**

The formula for the volume of a cylinder is **V=Bh or V=πr2h** .

**What is the volume of a cylinder example? ›**

Example 1: A cylinder has a radius of 50 cm and a height of 100 cm. How to find the volume of a cylinder? Solution: We know the volume of a cylinder is given by the formula – **π r ^{2} h**, where r is the radius of the cylinder and h is the height. = 3.14 x 50

^{2}x 100 = 785,000 cm

^{3}.

**How to find the volume of a cylinder find the area of its circular base and multiply by its height? ›**

The formula for the volume of a right cylinder is: **V = A * h**, where A is the area of the base, or πr^{2}. Therefore, the total formula for the volume of the cylinder is: V = πr^{2}h.