We use the procedure of “Slice, Approximate, Integrate" to develop the washer
method to compute volumes of solids of revolution.
1 The washer method
We can slice a solid of revolution perpendicular to the axis of rotation. We saw that
we could generate the solid of revolution by considering the corresponding slices in
the region of revolution in the \(xy\)-plane. To illustrate the details, we start with a
motivating example.
Consider the region in the
\(xy\)-plane bounded by
\(y=x^2-4\),
\(x=1\), and
\(y=5\), which is shown below.
A solid of revolution is formed by revolving this region about the \(y\)-axis.
Let’s try to apply the “Slice, Approximate, Integrate" procedure to find the volume
of the solid.
Step 1: Slice
The geometry of the base region suggests that it is advantageous to use horizontal
slices, so we should integrate with respect to \(x\)\(y\).
We draw a slice of thickness \(\Delta y\) at a fixed, but unspecified \(y\)-value on the region of
revolution.
Step 2: Approximate
We approximate the slice in the region by a rectangle.
The result of rotating the slice appears on the solid just as before.
Notice that the outer radius and inner radius are finite, but the thickness
\(\Delta y\) is thought of as quite small. To find the volume of the hollow cylinder,
recall
\begin{align*} \left < \textrm {Volume of the hollow cylinder } \right > &= \\ \left < \textrm { Volume of the outer } \right . & \left . \textrm {cylinder } \right > - \left < \textrm { Volume of the inner cylinder } \right >. \end{align*}
The outer cylinder has radius \(R(y)\) and its volume is \(\Delta V_{outer} = \pi [R(y)]^2 \Delta y\), while the volume of the inner
cylinder has radius \(r(y)\), so its volume is \(\Delta V_{inner} = \pi [r(y)]^2 \Delta \). Here, we have explicitly noted that these radii
will certainly depend at which \(y\)-value they are drawn, but to make the notation
cleaner the rest of the way, we will only write \(R\) and \(r\) instead of \(R(y)\) and \(r(y)\). The volume of
our hollow cylinder is thus
\[ \Delta V = \pi R^2 \Delta y -\pi r^2 \Delta y=\pi \left (R^2-r^2\right ) \Delta y \]
Our goal is now to express both \(R\) and \(r\) in terms of the unspecified \(y\)-value
of the slice. Using the picture is helpful to think about how to find \(R\) and
\(r\).
- The outer radius \(R\) is the distance from the axis of rotation to the outer curveinner curve.
- The inner radius \(r\) is the distance from the axis of rotation to the outer curveinner curve.
Also, both of these distances are horizontalvertical distances.
We now label these on the image of the region of rotation.
We can find both \(R\) and \(r\) now the way we always find horizontal distances.
For the outer radius \(R\):
- The righthand curve is given by \(x_{right} = \sqrt {y+4}\)\(x_{right} = 1\)\(x_{right} = 0\)
- The lefthand curve is given by \(x_{left} = \sqrt {y+4}\)\(x_{left} = 1\)\(x_{left} = 0\)
Thus \(R = x_{right}-x_{left} = \answer [given]{\sqrt {y+4}-0}\).
Following similar logic for the inner radius \(r\) gives \(r = x_{right}-x_{left} = \answer [given]{1-0}\).
The volume of our single approximate slice is thus
\[ \Delta V = \pi \left [\left (\sqrt {y+4}\right )^2-(1)^2\right ]\Delta y = \pi (y+3)\Delta y. \]
and the approximate total volume using \(n\) slices is found by adding up the volumes of
each slice, so
\[ V = \sum _{k=1}^n \pi (y_k+3)\Delta y_k, \]
where \(y_k\) is the \(y\)-value chosen for the \(k\)-th slice and \(\Delta y_k\) is the thickness of the slice.
Step 3: Integrate
In order to find the exact volume, we simultaneously must shrink the width of our
slices while adding all of the volumes together. As usual, the definite integral allows
us to do this, and we may write
\[ V= \int _{y=0}^{y=5} \pi (y+3) \d y \]
Evaluating this integral gives that the total volume is
\(\answer [given]{\frac {55}{2}\pi }\).
Let’s generalize the ideas in the above example. First, note that we slice the region of
revolution perpendicular to the axis of revolution, and we approximate each slice by a
rectangle. We call the slice obtained this way a washer. If the washer is not hollow
(i.e. \(r=0\)), it is sometimes referred to as a disk. Washers are characterized by finite inner
and outer radii but infinitesimal height. We now summarize the results of the above
argument.
At the risk of belaboring an important point, the variable of integration is
determined by the requirement that the slices be perpendicular to the axis of
rotation. Indeed, this requirement will allow us to determine if the slices are vertical
or horizontal. Just as in the previous applications, if we have vertical slices, we will
integrate with respect to \(x\), and if we have horizontal slices, we will integrate with
respect to \(y\).
Now, let’s see how this formula works in action by considering an example where we
take the same region and revolve it about different lines.
Let \(R\) be the region in the \(xy\)-plane bounded by \(y=0\), \(y=\sqrt {x}\), and \(x=2\).
Use the Washer Method to set up an integral that gives the volume of the solid of
revolution when
\(R\) is revolved about the following line
\(x=4\).
When we use the Washer Method, the slices are
perpendicularparallel to the axis
of rotation. This means that the slices are horizontal and we must integrate with
respect to
\(y\). We draw and label a picture, making sure to describe all curves using
functions of
\(y\).
Since we must integrate with respect to \(y\), we will use the result
\[V = \int _{y=c}^{y=d} \pi \left (R^2-r^2\right ) \d y \]
to set up the volume. We must now find the limits of integration and express
the outer radius \(R\) and inner radius \(r\) in terms of the variable of integration
\(y\).
The limits of integration are: \(c = \answer [given]{0}\) and \(d = \answer [given]{\sqrt {2}}\).
We see from the picture that both \(R\) and \(r\) are verticalhorizontal distances an we
can compute both by taking \(x_{right}-x_{left}\). This gives \(R= \answer [given]{4-y^2}\) and \(r= \answer [given]{4-2}\).
Thus, the integral that expresses the volume of the solid of revolution is
\[ V= \int _{y=\answer [given]{0}}^{y=\answer [given]{\sqrt {2}}} \answer [given]{\pi \left [(4-y^2)^2-(2)^2\right ]}\d y. \]
Set up an integral that gives the volume when the region is revolved about the line
\(y=-1\).
When we use the Washer Method, the slices are
perpendicularparallel to the axis
of rotation. This means that the slices are
verticalhorizontal and we should
integrate with respect to
\(x\)\(y\).
We draw and label a picture, making sure to describe all curves by functions of
\(x\).
Since we must integrate with respect to \(x\), we will use the result
\[V = \int _{x=a}^{x=b} \pi \left (R^2-r^2\right ) \d x. \]
We must now find the limits of integration as express the outer radius \(R\) and the inner
radius \(r\) in terms of the variable of integration \(x\).
The limits of integration are: \(a = \answer [given]{0}\) and \(b = \answer [given]{2}\).
We see from the picture that both \(R\) and \(r\) are verticalhorizontal distances, so we
can compute both by taking \(y_{top}-y_{bot}\).
Since the bottom \(y\)-values lie on the axis of rotation \(y=-1\), \(R= \answer [given]{\sqrt {x}-(-1)}\) and \(r= \answer [given]{0-(-1)}\).
Thus, the volume integral to be evaluated is
\[ V= \int _{x=\answer [given]{0}}^{x=\answer [given]{2}} \answer [given]{\pi \left [\left (1+\sqrt {x}\right )^2-(1)^2\right ]}\d x. \]
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