Often a given problem can be solved in more than one way. A particular method may be chosen out of convenience, personal preference, or perhaps necessity. Ultimately, it is good to have options.
The previous section introduced the Disk and Washer Methods, which computed the volume of solids of revolution by integrating the cross-sectional area of the solid. This section develops another method of computing volume, the Shell Method. Instead of slicing the solid perpendicular to the axis of rotation creating cross-sections, we now slice it parallel to the axis of rotation, creating “shells.”
Consider Figure 7.3.1, where the region shown in Figure 7.3.1.(a) is rotated around the \(y\)-axis forming the solid shown in Figure 7.3.1.(b). A small slice of the region is drawn in Figure 7.3.1.(a), parallel to the axis of rotation. When the region is rotated, this thin slice forms a cylindrical shell, as pictured in Figure 7.3.1.(c). The previous section approximated a solid with lots of thin disks (or washers); we now approximate a solid with many thin cylindrical shells.
To compute the volume of one shell, first consider the paper label on a soup can with radius \(r\) and height \(h\text{.}\) What is the area of this label? A simple way of determining this is to cut the label and lay it out flat, forming a rectangle with height \(h\) and length \(2\pi r\text{.}\) Thus the area is \(A = 2\pi rh\text{;}\) see Figure 7.3.2.(a).
Do a similar process with a cylindrical shell, with height \(h\text{,}\) thickness \(\Delta x\text{,}\) and approximate radius \(r\text{.}\) Cutting the shell and laying it flat forms a rectangular solid with length \(2\pi r\text{,}\) height \(h\) and depth \(\dx\text{.}\) Thus the volume is \(V \approx 2\pi rh\dx\text{;}\) see Figure 7.3.2.(b). (We say “approximately” since our radius was an approximation.)
By breaking the solid into \(n\) cylindrical shells, we can approximate the volume of the solid as
\begin{equation*}
V \approx \sum_{i=1}^n 2\pi r_ih_i\dx_i\text{,}
\end{equation*}
where \(r_i\text{,}\)\(h_i\) and \(\dx_i\) are the radius, height and thickness of the \(i\)th shell, respectively.
This is a Riemann Sum. Taking a limit as the thickness of the shells approaches 0 leads to a definite integral.
The left side of the image contains a cylinder, and the right side of the image shows a rectangle which comes from unraveling the side of the cylinder. The cylinder has a height \(h \text{,}\) radius \(r \) and is pictured lying on its circular base. Both the circular base, and the top of the cylinder are shaded, while the curved surface is unshaded, but contains a “cut here” instruction. Once the curved surface of the cylinder is cut, it is unraveled into a rectangular region as seen on the right side of the image. This rectangular region has a height \(h \) and a base \(2\pi r \text{.}\) The area of the rectangle is \(A=2\pi r h \text{.}\)
(a)
The left side of the image contains a cylindrical shell, which is just a cylinder with the middle part cut out. The right side of the image shows a rectangular box which comes from unraveling the cylindrical shell. The cylindrical shell has a height \(h \text{,}\) radius \(r \) which spans from the cut out center to the edge of the shell, and the shell has thickness \(\dx \) and is pictured lying on its circular base. The entire cylindrical shell is shaded, and contains a “cut here” instruction along the curved surface as in the previous image. Once the curved surface of the cylinderical shell is cut, the entire shell is unraveled into a rectangular box as seen on the right side of the image. This rectangular box has a height \(h \) and a base length of \(2\pi r \) and a thickness of \(\dx \text{.}\) The volume of the unraveled cylindrical shell is approximated the volume of the box given by \(V=2\pi r h \dx \text{.}\)
(b)
Figure7.3.2.Determining the volume of a thin cylindrical shell
Key Idea7.3.3.The Shell Method.
Let a solid be formed by revolving a region \(R\text{,}\) bounded by \(x=a\) and \(x=b\text{,}\) around a vertical axis. Let \(r(x)\) represent the distance from the axis of rotation to \(x\) (i.e., the radius of a sample shell) and let \(h(x)\) represent the height of the solid at \(x\) (i.e., the height of the shell). The volume of the solid is
\begin{equation*}
V = 2\pi\int_a^b r(x)h(x)\, dx\text{.}
\end{equation*}
Special Cases:
When the region \(R\) is bounded above by \(y=f(x)\) and below by \(y=g(x)\text{,}\) then \(h(x) = f(x)-g(x)\text{.}\)
When the axis of rotation is the \(y\)-axis (i.e., \(x=0\)) then \(r(x) = x\text{.}\)
Let’s practice using the Shell Method.
Example7.3.4.Finding volume using the Shell Method.
Find the volume of the solid formed by rotating the region bounded by \(y=0\text{,}\)\(y=1/(1+x^2)\text{,}\)\(x=0\) and \(x=1\) about the \(y\)-axis.
Solution.
This is the region used to introduce the Shell Method in Figure 7.3.1, but is sketched again in Figure 7.3.5 for closer reference. A line is drawn in the region parallel to the axis of rotation representing a shell that will be carved out as the region is rotated about the \(y\)-axis. (This is the differential element.)
Graph of the region bounded by \(y=0\text{,}\)\(y=1/(1+x^2)\text{,}\)\(x=0\) and \(x=1\text{.}\) The curve \(y=1/(1+x^2)\) begins at the \(y\)-axis at point \((0,1) \) and slopes downwards before ending at the point \((1,\frac12) \text{.}\) The region contains the entire area below the curve, lying above the horizontal line \(y=0 \text{.}\) On the \(x\)-axis, the graph contains an arbitrarily chosen point \(x \text{.}\) The distance between the origin and the point \(x \) is given as the function \(r(x) \text{,}\) which will give us the radius of the cylindrical shell once the region is rotated about the \(y\)-axis. Additionally, coming up from the point \(x \) is a red vertical line, which ends after meeting the curve \(y=1/(1+x^2)\text{.}\) This vertical line is labeled \(h(x)\text{,}\) and will give the us the height of the cylindrical shell.
The distance this line is from the axis of rotation determines \(r(x)\text{;}\) as the distance from \(x\) to the \(y\)-axis is \(x\text{,}\) we have \(r(x)=x\text{.}\) The height of this line determines \(h(x)\text{;}\) the top of the line is at \(y=1/(1+x^2)\text{,}\) whereas the bottom of the line is at \(y=0\text{.}\) Thus \(h(x) = 1/(1+x^2)-0 = 1/(1+x^2)\text{.}\) The region is bounded from \(x=0\) to \(x=1\text{,}\) so the volume is
\begin{align*}
V \amp = 2\pi\int_0^1 \frac{x}{1+x^2}\, dx.\\
\end{align*}
This requires substitution. Let \(u=1+x^2\text{,}\) so \(du = 2x\, dx\text{.}\) We also change the bounds: \(u(0) = 1\) and \(u(1) = 2\text{.}\) Thus we have:
Note: in order to find this volume using the Disk Method, two integrals would be needed to account for the regions above and below \(y=1/2\text{.}\)
With the Shell Method, nothing special needs to be accounted for to compute the volume of a solid that has a hole in the middle, as demonstrated next.
Example7.3.6.Finding volume using the Shell Method.
Find the volume of the solid formed by rotating the triangular region determined by the points \((0,1)\text{,}\)\((1,1)\) and \((1,3)\) about the line \(x=3\text{.}\)
Solution.
The region is sketched in Figure 7.3.7.(a) along with the differential element, a line within the region parallel to the axis of rotation. In Figure 7.3.7.(b), we see the shell traced out by the differential element, and in Figure 7.3.7.(c) the whole solid is shown.
Graph of the triangular region with edges on the points \((0,1)\text{,}\)\((1,1) \) and \((1,3)\text{.}\) The leftmost point of the triangular region begins at the \(y\)-axis at point \((0,1) \) and linearly increases until the point \((1,3) \text{.}\) The equation of the line between the points \((0,1) \) and \((1,3) \) is given to be \(y=2x+1 \text{.}\) The region contains the entire area below the curve, lying above the horizontal line \(y=1 \text{.}\) Above the \(x\)-axis, the graph showcases the distance between an arbitrarily chosen value on the \(x\)-axis, labeled \(x \text{,}\) which is between \(0 \) and \(1\text{,}\) and the axis of rotation at \(x=3 \text{.}\) The distance between these two \(x \) values is given as the function \(r(x) \text{,}\) which will give us the radius of the cylindrical shell once the region is rotated about the vertical line \(x=3\text{.}\) Additionally, coming up from the point \((x,1) \) is a red vertical line, which meets the upper bound of the triangular region, which is given by the line \(y=2x+1 \text{.}\) This vertical line is labeled \(h(x)\text{,}\) and will give the us the height of the cylindrical shell.
The height of the differential element is the distance from \(y=1\) to \(y=2x+1\text{,}\) the line that connects the points \((0,1)\) and \((1,3)\text{.}\) Thus \(h(x) = 2x+1-1 = 2x\text{.}\) The radius of the shell formed by the differential element is the distance from \(x\) to \(x=3\text{;}\) that is, it is \(r(x)=3-x\text{.}\) The \(x\)-bounds of the region are \(x=0\) to \(x=1\text{,}\) giving
When revolving a region around a horizontal axis, we must consider the radius and height functions in terms of \(y\text{,}\) not \(x\text{.}\)
Example7.3.8.Finding volume using the Shell Method.
Find the volume of the solid formed by rotating the region given in Example 7.3.6 about the \(x\)-axis.
Solution.
The region is sketched in Figure 7.3.9.(a) with a sample differential element. In Figure 7.3.9.(b) the shell formed by the differential element is drawn, and the solid is sketched in Figure 7.3.9.(c). (Note that the triangular region looks “short and wide” here, whereas in the previous example the same region looked “tall and narrow.” This is because the bounds on the graphs are different.)
The height of the differential element is an \(x\)-distance, between \(x=\frac12y-\frac12\) and \(x=1\text{.}\) Thus \(h(y) = 1-(\frac12y-\frac12) = -\frac12y+\frac32\text{.}\) The radius is the distance from \(y\) to the \(x\)-axis, so \(r(y) =y\text{.}\) The \(y\) bounds of the region are \(y=1\) and \(y=3\text{,}\) leading to the integral
Graph of the triangular region with edges on the points \((0,1)\text{,}\)\((1,1) \) and \((1,3)\text{.}\) The leftmost point of the triangular region begins at the \(y\)-axis at point \((0,1) \) and linearly increases until the point \((1,3) \text{.}\) The equation of the line between the points \((0,1) \) and \((1,3) \) is given as \(x=\frac12y-\frac12 \text{.}\) The region contains the entire area to the right of the line \(x=\frac12y-\frac12 \) and to the left of the vertical line \(x=1 \text{.}\) Coming up from the \(x\)-axis, the graph contains an arbitrarily chosen value on the \(y\)-axis, labeled \(y \text{,}\) which is between \(1 \) and \(3\text{.}\) The graph showcases the distance between the point labeled \(y\) and the axis of rotation which is \(y=0 \text{.}\) The distance between these two \(y \) values is given as the function \(r(y) \text{,}\) which will give us the radius of the cylindrical shell once the solid is formed by rotating about the \(x\)-axis. Additionally, coming up from the line \(x=\frac12y-\frac12 \) is a red horizontal line, which meets the right side of the triangular region, which is given by the vertical line \(x=1 \text{.}\) This vertical line is labeled \(h(y)\text{,}\) and will give the us the height of the cylindrical shells.
At the beginning of this section it was stated that “it is good to have options.” The next example finds the volume of a solid rather easily with the Shell Method, but using the Washer Method would be quite a chore.
Example7.3.10.Finding volume using the Shell Method.
Find the volume of the solid formed by revolving the region bounded by \(y= \sin(x)\) and the \(x\)-axis from \(x=0\) to \(x=\pi\) about the \(y\)-axis.
Solution.
The region and a differential element, the shell formed by this differential element, and the resulting solid are given in Figure 7.3.11.
Graph of the region bounded by \(y= \sin(x)\) and the \(x\)-axis from \(x=0\) to \(x=\pi\text{.}\) The leftmost point of the triangular region begins at the origin after which we see a singular sine wave ending at the point \((\pi,0) \text{.}\) The region contains the entire area to below the curve \(y= \sin(x)\text{,}\) lying above the \(x\)-axis. The graph contains an arbitrarily chosen value on the \(x\)-axis, labeled \(x \text{,}\) which is between \(0 \) and \(\pi\text{.}\) The graph showcases the distance between the axis of rotation which is \(x=0 \) and the point labeled \(x\) as the function \(r(x) \text{,}\) which will be the radius of the cylindrical shells. Additionally, coming up from the \(x\)-axis is a red vertical line, which meets the curve \(y= \sin(x)\) at the point \((x,\sin(x))\text{.}\) This vertical line is labeled \(h(x)\text{,}\) and will give the us the height of the cylindrical shells.
The radius of a sample shell is \(r(x) = x\text{;}\) the height of a sample shell is \(h(x) = \sin(x)\text{,}\) each from \(x=0\) to \(x=\pi\text{.}\) Thus the volume of the solid is
\begin{align*}
V \amp = 2\pi\int_0^{\pi} x\sin(x) \, dx.\\
\end{align*}
This requires Integration By Parts. Set \(u=x\) and \(dv=\sin(x) \, dx\text{;}\) we leave it to the reader to fill in the rest. We have:
Note that in order to use the Washer Method, we would need to solve \(y=\sin x\) for \(x\text{,}\) requiring the use of the arcsine function. We leave it to the reader to verify that the outside radius function is \(R(y) = \pi-\arcsin y\) and the inside radius function is \(r(y)=\arcsin y\text{.}\) Thus the volume can be computed as
This integral isn’t terrible given that the \(\arcsin^2 y\) terms cancel, but it is more onerous than the integral created by the Shell Method.
We end this section with a table summarizing the usage of the Washer and Shell Methods.
Key Idea7.3.12.Summary of the Washer and Shell Methods.
Let a region \(R\) be given with \(x\)-bounds \(x=a\) and \(x=b\) and \(y\)-bounds \(y=c\) and \(y=d\text{.}\)
Washer Method
Shell Method
Horizontal Axis
\(\ds \pi\int_a^b \big(R(x)^2-r(x)^2\big)\, dx\)
\(\ds 2\pi\int_c^d r(y)h(y)\, dy\)
Vertical Axis
\(\ds\pi \int_c^d\big(R(y)^2-r(y)^2\big)\, dy\)
\(\ds 2\pi\int_a^b r(x)h(x)\, dx\)
As in the previous section, the real goal of this section is not to be able to compute volumes of certain solids. Rather, it is to be able to solve a problem by first approximating, then using limits to refine the approximation to give the exact value. In this section, we approximate the volume of a solid by cutting it into thin cylindrical shells. By summing up the volumes of each shell, we get an approximation of the volume. By taking a limit as the number of equally spaced shells goes to infinity, our summation can be evaluated as a definite integral, giving the exact value.
We use this same principle again in the next section, where we find the length of curves in the plane.
ExercisesExercises
Terms and Concepts
1.
T/F: A solid of revolution is formed by revolving a shape around an axis.
2.
T/F: The Shell Method can only be used when the Washer Method fails.
3.
T/F: The Shell Method works by integrating cross-sectional areas of a solid.
4.
T/F: When finding the volume of a solid of revolution that was revolved around a vertical axis, the Shell Method integrates with respect to \(x\text{.}\)
Problems
Exercise Group.
Use the Shell Method to find the volume of the solid of revolution formed by revolving the given region about the \(y\)-axis.
5.
The region bounded by the curve \(y=3-x^2\text{,}\) the \(x\) axis, and the \(y\) axis:
Graph of the region bounded by the curve \(y=3-x^2\text{,}\) the \(x\)-axis, and the \(y\)-axis. Note that this region can reference both the regions to the left or right of the \(y\)-axis, but we will consider the region to the right of the \(y\)-axis. The curve \(y=3-x^2\) begins at the point \((0,3)\) from which it slopes down until reaching the \(x\)-axis. The region then contains the entire area below this curve, and above the \(x\)-axis.
6.
The region between \(y=5x\) and the \(x\) axis, for \(1\leq x\leq 2\text{:}\)
Graph of the region between \(y=5x\) and the \(x\)-axis, for \(1\leq x\leq 2\text{.}\) The line \(y=5x\) begins at the point \((1,5)\) from which it linearly increases until reaching the rightmost bound which is at the point \((2,10)\text{.}\) The region then contains the entire area below the line \(y=5x\) and above the \(x\)-axis for \(1\leq x\leq 2\text{.}\)
7.
The region between \(y=\cos(x)\) and the \(x\) axis, for \(0\leq x\leq \pi/2\text{:}\)
Graph of the region between \(y=\cos(x)\) and the \(x\)-axis, for \(0\leq x\leq \pi/2\text{.}\) The curve \(y=\cos(x)\) begins at the point \((0,1)\) from which it slopes downwards until ending after reaching the \(x\)-axis at the point \((\pi/2,0)\text{.}\) The region then contains the entire area below the curve \(y=\cos(x)\) and above the \(x\)-axis for \(0\leq x\leq \pi/2\text{.}\) The bound \(0\leq x\leq \pi/2\) can also be interpreted as the leftmost bound of the region being the \(y\)-axis.
8.
The region between the curves \(y=x\) and \(y=\sqrt{x}\text{:}\)
Graph of the region between the curves \(y=x\) and \(y=\sqrt{x}\text{.}\) The curves \(y=x\) and \(y=\sqrt{x}\) both begin at the origin. From this point the curve \(y=\sqrt{x}\) rises above the line \(y=x\) until reaching the point \((1,1)\text{,}\) where the curve once again intersects the line. The region then contains the area below the curve \(y=\sqrt{x}\) and above the line \(y=x\text{.}\)
Exercise Group.
Use the Shell Method to find the volume of the solid of revolution formed by revolving the given region about the \(x\)-axis.
9.
The region between \(y=3-x^2\) and the \(x\) axis:
Graph of the region between \(y=3-x^2\) and the \(x\)-axis. The curve \(y=3-x^2\) begins at \(x\)-axis at the point \((-\sqrt{3},0)\text{.}\) From this point the curve rises until reaching the \(y\)-axis at the point \((0,3)\text{.}\) From this point, the curve slopes downwards until once again reaching the \(x\)-axis at the point \((\sqrt{3},0)\text{.}\) On both the left and right sides of the region, the area is bounded by the curve \(y=3-x^2\text{.}\)
10.
The region between \(y=5x\) and the \(y\) axis, for \(5\leq y\leq 10\text{:}\)
Graph of the region between \(y=5x\) and the \(y\) axis, for \(5\leq y\leq 10\text{.}\) The line \(y=5x\) begins at the point \((1,5)\text{,}\) from which it linearly increases until ending upper bound for \(y\) at the point \((2,10)\text{.}\) The region then contains the entire area to the right of \(x=0\) and to the left of the line \(y=5x\) for \(y\) between \(5\) and \(10\text{.}\)
11.
The region between \(y=\cos(x)\) and the \(x\) axis, for \(0\leq x\leq \pi/2\text{:}\)
Graph of the region between \(y=\cos(x)\) and the \(x\) axis, for \(0\leq x\leq \pi/2\text{.}\) The curve \(y=\cos(x)\) begins at the point \((0,1)\) from which it slopes downwards until ending after reaching the \(x\)-axis at the point \((\pi/2,0)\text{.}\) The region then contains the entire area to the right of \(x=0\) and to the left of of the curve \(y=\cos(x)\) for \(y\) values between \(0\) and \(1\text{.}\)
12.
The region between the curves \(y=x\) and \(y=\sqrt{x}\text{:}\)
Graph of the region between the curves \(y=x\) and \(y=\sqrt{x}\text{.}\) The curves \(y=x\) and \(y=\sqrt{x}\) both begin at the origin. From this point the curve \(y=\sqrt{x}\) rises above the line \(y=x\) until reaching the point \((1,1)\text{,}\) where the curve once again intersects the line. The region consists of the area to the right of the curve \(y=\sqrt{x}\) and the to the left line \(y=x\) for \(y\) values between \(0\) and \(1\text{.}\)
Exercise Group.
Use the Shell Method to find the volume of the solid of revolution formed by revloving the given region about each of the given axes.
13.
Region bounded by: \(y=\sqrt{x}\text{,}\)\(y=0\) and \(x=1\text{.}\)
(a)
Rotate about the \(y\) axis.
(b)
Rotate about \(x=1\text{.}\)
(c)
Rotate about the \(x\) axis.
(d)
Rotate about \(y=1\text{.}\)
14.
Region bounded by: \(y=4-x^2\) and \(y=0\text{.}\)
(a)
Rotate about \(x=2\text{.}\)
(b)
Rotate about \(x=-2\text{.}\)
(c)
Rotate about the \(x\) axis.
(d)
Rotate about \(y=4\text{.}\)
15.
The triangle with vertices \((1,1)\text{,}\)\((1,2)\) and \((2,1)\text{.}\)
(a)
Rotate about the \(y\) axis.
(b)
Rotate about \(x=1\text{.}\)
(c)
Rotate about the \(x\) axis.
(d)
Rotate about \(y=2\text{.}\)
16.
Region bounded by \(y=x^2-2x+2\) and \(y=2x-1\text{.}\)
(a)
Rotate about the \(y\) axis.
(b)
Rotate about \(x=1\text{.}\)
(c)
Rotate about \(x=-1\text{.}\)
17.
Region bounded by \(y=1/\sqrt{x^2+1}\text{,}\)\(x=1\) and the \(x\) and \(y\) axes.
(a)
Rotate about the \(y\) axis.
(b)
Rotate about \(x=1\text{.}\)
18.
Region bounded by \(y=2x\text{,}\)\(y=x\) and \(x=2\text{.}\)
