2.3: Volumes of Revolution - Cylindrical Shells

Note to the Instructor (click to expand) This is the typical "cylindrical shells" topic. To succeed in this section, you'll need to use some skills from previous courses. While you should already know them, this is the first time they've been required. You can review these skills in CRC's Corequisite Codex. If you have a support class, it might cover some, but not all, of these topics. Prerequisite Skills and Support Topics (click to expand) Geometry Common Volumes: The volume of a cylindrical shell is derived by considering it as a cylinder whose height is the height of the function and whose base is an annulus. The "unrolling" analogy uses the volume of a rectangular prism (a flat plate). The following is a list of learning objectives for this section. Learning Objectives (click to expand) Calculate the volume of a solid of revolution using the Method of Cylindrical Shells. Compare the different methods for calculating the volume of revolution.

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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.

The Method of Cylindrical Shells

Again, we are working with a solid of revolution. As before, we define a region \(\mathbf{R}\), bounded above by the graph of a function \(y=f(x)\), below by the \(x\)-axis, and on the left and right by the lines \(x=a\) and \(x=b\), respectively, as shown in Figure \(\PageIndex{1(a)}\). We then revolve this region around the \(y\)-axis, as shown in Figure \(\PageIndex{1(b)}\). Note that this is different from what we have done before. Previously, regions defined in terms of functions of \(x\) were revolved around the \(x\)-axis or a line parallel to it.

Figure \(\PageIndex{1(a)}\): A region bounded by the graph of a function of \(x\).
Figure \( \PageIndex{ 1(b) } \): The solid of revolution formed when the region revolves around the \(y\)-axis.

As we have done many times before, partition the interval \([a,b]\) using a regular partition, \(P=\{x_0,x_1, \ldots ,x_n\}\) and, for \(i=1,2, \ldots ,n\), choose a point \(x^∗_i \in [x_{i−1},x_i]\). Then, construct a rectangle over the interval \([x_{i−1},x_i]\) of height \(f(x^∗_i)\) and width \( \Delta x\). A representative rectangle is shown in Figure \(\PageIndex{2(a)}\). When that rectangle is revolved around the \(y\)-axis, instead of a disk or a washer, we get a cylindrical shell, as shown in Figure \(\PageIndex{2(b)}\).

Figure \(\PageIndex{2(a)}\): A representative rectangle.
Figure \( \PageIndex{ 2(b) } \): When this rectangle revolves around the \(y\)-axis, the result is a cylindrical shell.
Figure \( \PageIndex{ 2(c) } \): When we put all the shells together, we get an approximation of the original solid.

To calculate the volume of this shell, consider Figure \(\PageIndex{3}\).

Figure \(\PageIndex{3}\): 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 \( \pi x^2_i− \pi x^2_{i−1}\). The height of the cylinder is \(f(x^∗_i).\) Then the volume of the shell is\[ \begin{array}{rcl}
V_{\text{shell}} & = & f(x^∗_i)( \pi \,x^2_{i}− \pi \,x^2_{i−1}) \\
\\
& = & \pi \,f(x^∗_i)(x^2_i−x^2_{i−1}) \\
\\
& = & \pi \,f(x^∗_i)(x_i+x_{i−1})(x_i−x_{i−1}) \\
\\
& = & 2 \pi \,f(x^∗_i)\left(\dfrac {x_i+x_{i−1}}{2}\right)(x_i−x_{i−1}). \\
\end{array} \nonumber\]Note that \(x_i−x_{i−1}= \Delta x\), so we have\[V_{\text{shell}}=2 \pi \,f(x^∗_i)\left(\dfrac {x_i+x_{i−1}}{2}\right)\, \Delta x. \nonumber \]Furthermore, \(\frac {x_i+x_{i−1}}{2}\) is both the midpoint of the interval \([x_{i−1},x_i]\) and the average radius of the shell. We can approximate this by \(x^∗_i\). We then have\[V_{\text{shell}} \approx 2 \pi \,f(x^∗_i)x^∗_i\, \Delta x. \nonumber \]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 \(\PageIndex{4}\)).

Figure \(\PageIndex{4(a)}\): Make a vertical cut in a representative shell.
Figure \( \PageIndex{ 4(b) } \): Open the shell up to form a flat plate.

In reality, the outer radius of the shell is greater than the inner radius. 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 \( \Delta x\) (Figure \( \PageIndex{4(b)} \)). Then, the shell's volume is approximately the volume of the flat plate. Multiplying the height, width, and depth of the plate, we get\[V_{\text{shell}} \approx f(x^∗_i)(2 \pi \,x^∗_i)\, \Delta x, \nonumber \]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\[V \approx \sum_{i=1}^n(2 \pi \,x^∗_i f(x^∗_i)\, \Delta x). \nonumber \]Here we have another Riemann sum, this time for the function \(2 \pi \,x\,f(x).\) Taking the limit as \(n \to \infty \) gives us\[V=\lim_{n \to \infty }\sum_{i=1}^n(2 \pi \,x^∗_if(x^∗_i)\, \Delta x)=\int ^b_a(2 \pi \,x\,f(x))\,dx. \nonumber \]This leads to the following rule for the Method of Cylindrical Shells.

Theorem: The Method of Cylindrical Shells About the \( y \)-axis

Let \(f(x)\) be continuous and nonnegative. Define \(\mathbf{R}\) as the region bounded above by the graph of \(f(x)\), below by the \(x\)-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 \(\mathbf{R}\) around the \(y\)-axis is given by\[V=\int ^b_a(2 \pi \,x\,f(x))\,dx. \nonumber \]

As stated in the previous section, you want to understand how this works rather than memorizing the formula. We will work hard to showcase the thought process you should follow when using the Method of Cylindrical Shells.

Example \(\PageIndex{1}\)

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

Solution

First, we must graph the region \(\mathbf{R}\) and the associated solid of revolution, as shown in Figure \(\PageIndex{5}\). You should also get used to graphing a representative shell (use the CalcPlot3D applet below to visualize this for this example).

Figure \(\PageIndex{5(a)}\): The region \(\mathbf{R}\) under the graph of \(f(x)=1/x\) over the interval \([1,3]\).
Figure \( \PageIndex{ 5(b) } \): The solid of revolution generated by revolving \(\mathbf{R}\) about the \(y\)-axis.

Interactive Object

Figure \(\PageIndex{5c}\) Visualizing the solid of revolution with CalcPlot3D.

If we had not been told to use the Method of Cylindrical Shells, we would have a choice to make. If you can visualize horizontal slices being rotated about the \( y \)-axis, you would see that the outer radius changes functions at \( y = 1/3 \). This means that you would need two groups of integrals (one for the bottom set of washers, and one for the washers starting at a height of \( y = 1/3 \)). This is highly inefficient.

On the other hand, choosing to make vertical slices and rotating those about the \( y \)-axis shows that the top function of each slice is always \( f(x) = 1/x \) and the bottom function is always \( y = 0 \). Therefore, vertical slicing is a more attractive option.

The volume of the \( i^{\text{th}} \) slice is given in words by\[ V_i = \left( \text{Circumference of the rotated slice} \right) \left( \text{Height of the rotated slice} \right) \left( \text{Thickness of the rotated slice} \right). \nonumber \]As per our usual approach, we let \( r(x_i^*) \) be the radius of rotation. We will also let \( h(x_i^*) \) be the height of the slice and naturally select \( \Delta x \) as the thickness. Then our language formula transforms to\[ \begin{array}{ccccc}
& & \text{Circumference} & \text{Height} & \text{Thickness} \\
V_i & = & 2 \pi r(x_i^*) & h(x_i^*) & \Delta x. \\
\end{array} \nonumber \]The radius of rotation is the distance between the slice and the axis of rotation (the \( y\)-axis). Since this is a horizontal distance, we measure it as \( x_R - x_L \). In this case, \( x_R = x_i^* \) and \( x_L \) is the \( y \)-axis. That is, or \( x_L = 0 \). The height of the slice is a vertical distance, so this should be \( y_T - y_B \). The top of the slice is \( f(x_i^*) = \frac{1}{x_i^*} \) and the bottom is \( y = 0 \). Putting this altogether, we get\[ \begin{array}{ccccc}
& & \text{Circumference} & \text{Height} & \text{Thickness} \\
V_i & = & 2 \pi r(x_i^*) & h(x_i^*) & \Delta x \\
& = & 2 \pi x_i^* & \frac{1}{x_i^*} & \Delta x. \\
\end{array} \nonumber \]Hence, the true volume is\[ \begin{array}{rcl}
V & = & \displaystyle \int^{x = 3}_{x = 1} \left(2 \pi \,x\left(\dfrac {1}{x}\right)\right)\,dx \\
\\
& = & \displaystyle \int^{x =3}_{x = 1} 2 \pi \,dx \\
\\
& = & 2 \pi \,x\bigg|^{x = 3}_{x = 1} \\
\\
& = & 4 \pi \,\text{units}^3. \\
\end{array} \nonumber\]

Checkpoint \(\PageIndex{1}\)

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

Answer

\(\frac{15 \pi }{2} \, \text{units}^3 \)

Example \(\PageIndex{2}\)

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

Solution

First graph the region \(\mathbf{R}\) and the associated solid of revolution, as shown in Figure \(\PageIndex{6}\).

Figure \(\PageIndex{6(a)}\): The region \(\mathbf{R}\) under the graph of \(f(x)=2x−x^2\) over the interval \([0,2].\)
Figure \( \PageIndex{ 6(b) } \): The volume of revolution obtained by revolving \(\mathbf{R}\) about the \(y\)-axis.

If we chose horizontal slices, we would get washers; however, each washer's outer and inner edges would be described by the same function. While we could find a way to get the volume of this washer, there are more efficient methods than this one. Instead, let's try slicing vertically.

Figure \( \PageIndex{6} \) shows that vertical slices would result in cylindrical shells. From Example \( \PageIndex{1} \), we know the volume of the \( i^{\text{th}} \) such shell would be\[ \begin{array}{ccccc}
& & \text{Circumference} & \text{Height} & \text{Thickness} \\
V_i & = & 2 \pi r(x_i^*) & h(x_i^*) & \Delta x, \\
\end{array} \nonumber \]where \( r(x_i^*) = x_i^*\) and \( h(x_i^*) = 2x_i^* - (x_i^*)^2 \). Hence,\[ \begin{array}{ccccc}
& & \text{Circumference} & \text{Height} & \text{Thickness} \\
V_i & = & 2 \pi r(x_i^*) & h(x_i^*) & \Delta x \\
& = & 2 \pi x_i^* & \left( 2x_i^* - (x_i^*)^2 \right) & \Delta x. \\
\end{array} \nonumber \]Thus,\[\begin{array}{rcl}
V & = & \displaystyle \int ^2_0(2 \pi \,x(2x−x^2))\,dx \\
\\
& = & 2 \pi \displaystyle \int ^2_0(2x^2−x^3)\,dx \\
\\
& = & 2 \pi \left[\dfrac {2x^3}{3}−\dfrac {x^4}{4}\right]\bigg|^2_0 \\
\\
& = & \dfrac {8 \pi }{3}\,\text{units}^3 \\
\end{array} \nonumber \]

Checkpoint \(\PageIndex{2}\)

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

Answer

\(8 \pi \, \text{units}^3 \)

As with the Disk Method and the Washer Method, we can use the Method of Cylindrical Shells with solids of revolution, revolved around the \(x\)-axis, when we want to integrate with respect to \(y\). The analogous rule for this type of solid is given here. To be clear, you should not memorize this formula without truly understanding how to derive it yourself. Committing this formula to memory is worthless if you genuinely understand the derivation of the process.

Theorem: The Method of Cylindrical Shells About the \( x \)-axis

Let \(g(y)\) be continuous and nonnegative. Define \(\mathbf{Q}\) as the region bounded on the right by the graph of \(g(y)\), on the left by the \(y\)-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 \(\mathbf{Q}\) around the \(x\)-axis is given by\[V=\int ^d_c(2 \pi \,y\,g(y))\,dy. \nonumber \]

Example \(\PageIndex{3}\)

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

Solution

First, we need to graph the region \(\mathbf{Q}\) and the associated solid of revolution, as shown in Figure \(\PageIndex{7}\).

Figure \(\PageIndex{7(a)}\): The region \(\mathbf{Q}\) to the left of the function \(g(y)\) over the interval \([0,4]\).
Figure \( \PageIndex{ 7(b) } \): The solid of revolution generated by revolving \(\mathbf{Q}\) around the \(x\)-axis.

This is an excellent example of how we could easily use either the Washer Method or the Method of Cylindrical Shells. It's advisable to set up both integrals for practice and to see which one looks easier to evaluate.

VERTICAL SLICES: If we choose vertical slices, we will get washers, which implies the Washer Method. Moreover, each washer will have width \( \Delta x \). This informs us that all of our work should eventually be in terms of \( x \). Let's state the required information first.\[ r_O(x_i^*) = y_T - y_B = 4 - 0 = 4. \nonumber \]Finding \( r_I(x_i^*) \) requires us to solve \( x = 2 \sqrt{y} \) for \( y \). This gives \( \frac{x^2}{4} = y \). Therefore,\[ r_I(x_i^*) = y_T - y_B = \frac{x^2}{4} - 0 = \frac{x^2}{4}. \nonumber \]The volume of the \( i^{\text{th}} \) slice is\[ \begin{array}{rcl}
V_i & = & \pi \left[ r_O(x_i^*) \right]^2 \Delta x - \pi \left[ r_I(x_i^*) \right]^2 \Delta x \\
\\
& = & \pi \left( 16 - \dfrac{x^4}{16} \right) \Delta x \\
\end{array} \nonumber \]Thus,\[ V = \pi \int_{x = 0}^{x = 4} 16 - \dfrac{x^4}{16} \, dx. \nonumber \]

HORIZONTAL SLICES: If, on the other hand, we decide on horizontal slices, the rotation will result in shells. Hence, we will use the Method of Cylindrical Shells. The thickness of each shell will be \( \Delta y \). This informs us that all of our work should eventually be only in terms of \( y \).

The radius of rotation is \( y_i^* \) and the "height" of each shell is \( x_R - x_L = 2\sqrt{y_i^*} - 0 = 2\sqrt{y_i^*} \). Therefore, the volume of the \( i^{\text{th}} \) slice is\[ \begin{array}{ccccc}
& & \text{Circumference} & \text{Height} & \text{Thickness} \\
V_i & = & 2 \pi r(y_i^*) & h(y_i^*) & \Delta y \\
& = & 2 \pi y_i^* & \left( 2 \sqrt{y_i^*} \right) & \Delta y. \\
\end{array} \nonumber \]Then the volume of the solid is given by\[ V = 4 \pi \int_{y = 0}^{y = 4} y^{3/2} dy. \nonumber \]You be the judge of which integral looks nicer. While they will both yield the same result, I am choosing the second because... well, it's nicer.\[ \begin{array}{rcl}
V & = & \displaystyle 4 \pi \int^{y = 4}_{y =0}y^{3/2}\,dy \\
\\
& = & 4 \pi \left[\dfrac {2y^{5/2}}{5}\right]\bigg|^4_0 \\
\\
& = & \dfrac {256 \pi }{5}\, \text{units}^3 \\
\end{array} \nonumber \]

Example \( \PageIndex{3} \) showcases two fundamental concepts. First, you should refrain from getting married to a method. That is, always be willing to try both horizontal and vertical slices. This takes little time once you get used to things and can make an impossible problem simple. The second important concept is based on notation. If you look back through Example \( \PageIndex{3} \), you will notice that I wrote the limits of integration as \( x = 0 \) or \( y =0 \), and \( x = 4 \) or \( y = 4 \), accordingly. While the limits didn't change in this example, they often are not the same, and writing \( x = \) or \( y = \) will keep your work "honest" and remind you that, yes, you already changed those limits into the proper variable.

Checkpoint \(\PageIndex{3}\)

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

Answer

\(12 \pi \) units³

Example \(\PageIndex{4}\)

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

Solution

First, graph the region \(\mathbf{R}\) and the associated solid of revolution, as shown in Figure \(\PageIndex{8}\).

Figure \(\PageIndex{8(a)}\): The region \(\mathbf{R}\) between the graph of \(f(x)\) and the \(x\)-axis over the interval \([1,2]\).
Figure \( \PageIndex{ 8(b) } \): The solid of revolution generated by revolving \(\mathbf{R}\) around the line \(x=−1.\)

HORIZONTAL SLICES: If we chose to slice \(\mathbf{R}\) into horizontal slices, we can easily see that we would need two sets of computations - one for the region where the left edge is \( x=1 \) and the right edge is \( x = 2 \), and one for the region where the left edge is \( f(x)=x \) and the right edge is \( x=2 \). This should motivate us to try vertical slices.

VERTICAL SLICES: Slicing the region \(\mathbf{R}\) into vertical slices means that each has a width of \( \Delta x \), and so all of our work needs to be in terms of \( x \). Moreover, a rotation about the vertical line \( x = -1 \) means we are creating shells. The radius of rotation to the \( i^{\text{th}} \) slice is \( x_R - x_L = x_i^* - (-1) = x_i^* + 1 \). The height of the slice is \( h(x_i^*) = y_T - y_B = x_i^* - 0 = x_i^* \). Therefore, the volume of the \( i^{\text{th}} \) slice is\[ V_i = 2 \pi r(x_i^*) h(x_i^*) \Delta x = 2 \pi (x_i^* + 1) x_i^* \Delta x. \nonumber \]Thus, the volume of the solid is given by\[\begin{array}{rcl}
V & = & 2 \pi \displaystyle \int^{x = 2}_{x = 1} x^2+x \, dx \\
\\
& = & 2 \pi \left[\dfrac{x^3}{3}+\dfrac{x^2}{2}\right]\bigg|^2_1 \\
\\
& = & \dfrac{23 \pi }{3} \, \text{units}^3 \\
\end{array} \nonumber\]

Checkpoint \(\PageIndex{4}\)

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

Answer

\(\frac {11 \pi }{6}\) units³

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

Example \(\PageIndex{5}\)

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

Solution

First, graph the region \(\mathbf{R}\) and the associated solid of revolution, as shown in Figure \(\PageIndex{9}\). During this process, you will need to find the point of intersection of these two curves, which is \( \left( 1,1 \right) \).

Figure \(\PageIndex{9(a)}\): The region \(\mathbf{R}\) between the graph of \(f(x)\) and the graph of \(g(x)\) over the interval \([1,4]\).
Figure \( \PageIndex{ 9(b) } \): The solid of revolution generated by revolving \(\mathbf{R}\) around the \(y\)-axis.

HORIZONTAL OR VERTICAL SLICES? Since we are rotating about the \( y \)-axis, a quick inspection reveals that horizontal slices yield two separate regions (one below \( y=1 \) and one above \( y = 1 \)), so this is likely not the best choice.

Now that we know to slice vertically, we also know that each slice has width \( \Delta x \). Again, this means all of our eventual work must be in terms of \( x \). Moreover, vertical slices rotated about the \( y \)-axis leads to shells. Hence, we are using the Method of Cylindrical Shells. The radius of rotation for the \( i^{\text{th}} \) shell is \( x_R - x_L = x_i^* - 0 = x_i^* \). The height of this slice is \( h(x_i^*) = y_T - y_B = \sqrt{x_i^*} - \frac{1}{x_i^*} \). Combining this information, we setup the volume of the \( i^{\text{th}} \) slice to be\[ V_i = 2 \pi r(x_i^*) h(x_i^*) \Delta x = 2 \pi x_i^* \left( \sqrt{x_i^*} - \dfrac{1}{x_i^*} \right) \Delta x. \nonumber \]Then the volume of the solid is given by\[\begin{array}{rcl}
V & = & \displaystyle \int^{x = 4}_{x = 1}\left(2 \pi \,x\left(\sqrt{x}−\dfrac {1}{x}\right)\right)\,dx \\
\\
& = & 2 \pi \displaystyle \int^{x = 4}_{x = 1}(x^{3/2}−1) \, dx \\
\\
& = & 2 \pi \left[\dfrac {2x^{5/2}}{5}−x\right]\bigg|^4_1 \\
\\
& = & \dfrac {94 \pi }{5} \, \text{units}^3. \\
\end{array} \nonumber \]

Checkpoint \(\PageIndex{5}\)

Define \(\mathbf{R}\) 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 \([0,1]\). Find the volume of the solid of revolution formed by revolving \(\mathbf{R}\) around the \(y\)-axis.

Answer

\(\frac { \pi }{6}\) units³