The shell method

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We use the procedure of “Slice, Approximate, Integrate” to develop the shell method to compute volumes of solids of revolution.

The shell method

Some volumes of revolution require more than one integral using the washer method. We study such an example now.

Consider the solid formed when the region $R$ bounded by the curves $y=2-x^2$ , $x=0$ , $x=1$ , and $y=0$ is revolved about the $y$ -axis.

If we insist on using the Washer Method, the slices must be perpendicular to the axis of rotation. This means that the slices will be horizontal, but the righthand curve will change so we will need $\answer [given]{2}$ integrals with respect to $y$ to compute the volume. Rather than being locked into the choice of method, recall that we can generate solids of revolution by rotating slices in the region of integration about the axis of revolution. The region in this example is clearly easier to treat if we use vertical slices. Let’s apply the “Slice, Approximate, Integrate” procedure and see what happens.

We indicate a slice of thickness $\Delta x$ at an arbitrary but fixed $x$ -value in the region of revolution.

Step 2: Approximate We approximate the slice on the base by a rectangle.

The solid of revolution and the result of rotating the slice appear below.

The result of revolving the slice produced another hollow cylinder. This solid is now built by nesting larger shells inside of smaller ones (rather than by stacking washers on top of each other). Recall from earlier that the volume of a hollow cylinder is

$V= \pi (R^2-r^2)h,$

and our goal is to find $R$ , $r$ , and $h$ in terms of the variable of integration. Let’s examine the hollow cylinder again.

The rectangle that generates this cylinder is shown as well because this is ultimately what we will need to use for our analysis.

As before, as the width of the slice becomes smaller, the rectangle approximates the actual slice better. As $\Delta x$ becomes arbitrarily small, the height $h$ outer radius $R$ inner radius $r$ the difference between the outer and inner radius, $R-r$ becomes arbitrarily small.

In order to understand how to write the volume $\Delta V$ of a slice, we note that $R-r = \Delta x$ , so we write

$\Delta V = \pi (R^2-r^2)h = \pi (R+r)(R-r)h = \pi (R+r)h \Delta x.$

As the slice width shrinks, we see that $R$ and $r$ become less distinguishable. We split the difference and replace them with their average value $\rho = \frac {R+r}{2}$ .

$\Delta V = \pi (R+r)h \Delta x = \pi \left (2 \cdot \frac {R+r}{2} \right )h \Delta x = 2\pi \rho h \Delta x.$

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

$V= \int _{x=0}^{x=1} 2 \pi \rho (x) h(x) \d x .$

Here, we have written $\rho (x)$ and $h(x)$ to draw explicit attention to the fact that the radius $\rho$ and height $h$ of a slice can depend on the $x$ -value where the slice is chosen. To finish off the problem, we must express these geometric quantitiesfor our arbitrary slice in terms of the variable of integration.

To find $\rho (x)$ , look at the previous images and note that $\rho (x)$ is the height of the rectangledistance from the axis of rotation to the slice .

This distance is a horizontal distance and can be found using by $\rho = x_{right}-x_{left}$ . Noting that the arbitrary slice occurs at an unspecified $x$ -value, we have $x_{right} =$ and $x_{left}=$ $1$ $x$ $2-x^2$ $0$ .

If you chose $2-x^2$ for an answer above, consider the image below and think about what this represents.

The quantity $2-x^2$ is not how far the slice is from the axis of rotation; it is the height of the rectangle.

Thus, $\rho (x) = x_{right}-x_{left} = \answer [given]{x-0}$ .

To find $h(x)$ , look at the previous images and note that $h$ is the height of the rectangledistance from the axis of rotation to the slice .

This height is a vertical distance and can be found using by $\rho = y_{top}-y_{bot} = \answer [given]{2-x^2-0}$ .

Thus, the volume of the solid of revolution is

$V = \int _{x=0}^{x=1} 2 \pi \rho (x) h(x) \d x = \int _{x=0}^{x=1} 2 \pi x(2-x^2) \d x.$

The integral is perhaps easier to evaluate if we factor out the $\pi$ and expand the integrand.

$\begin{align*} V &= \pi \int_{x=0}^{x=1} 2x(2-x^2) \d x \\ &= \pi \int_{x=0}^{x=1} (4x-2x^3) \d x \\ &= \pi \eval{\answer[given]{2x^2-\frac{1}{2}x^4}}_0^1 \end{align*}$

By evaluating this integral we find that the total volume is $\answer [given]{\frac {3}{2}\pi }$ .

The shell method formula

Let’s generalize the ideas in the above example. First, note that we slice the region of revolution parallel to the axis of revolution, and we approximate each slice by a rectangle. We call the slice obtained this way a shell. Shells are characterized as hollow cylinders with an infinitesimal difference between the outer and inner radii and a finite height. We now summarize the results of the above argument.

Suppose that a region in the $xy$ -plane has a continuous boundary and that a solid of revolution is formed by revolving the region about a vertical or horizontal line in the $xy$ -plane that does not intersect the region.

If slices taken parallel to the axis of revolution are vertical, then the volume of the solid of revolution is given by $V=\int _{x=a}^{x=b} 2\pi \rho h \d x .$
If slices taken parallel to the axis of revolution are horizontal, then the volume of the solid of revolution is given by $V=\int _{y=c}^{y=d} 2 \pi \rho h \d y .$

In both cases, the radius $\rho$ is the distance from the axis of rotation to the slice and $h$ is the length of the slice. These must be expressed with respect to the variable of integration.

To find $\rho$ and $h$ , draw an arbitrary slice in the region according to the variable of integration (vertical if integrating with respect to $x$ , horizontal if integrating with respect to $y$ ). Then, $\rho$ is the distance from the axis of rotation to the slice and $h$ is the “height” of the slice.

Once again, the variable of integration is chosen by requiring that the slices be parallel to the axis of rotation. We find $\rho$ and $h$ by drawing a picture and interpreting them as horizontal or vertical distances. Note that it does not matter in which quadrant the axis of rotation is or in which quadrants the region lies since we always find vertical and horizontal distances the same way.

Now that we have established the above result, we do not have to go through the “Slice, Approximate, Integrate” procedure for every example. Let’s see now how the formula works in action.

Let $R$ be the region in the $xy$ -plane bounded by $y=4-x^2$ , $y=2x+4$ , and $y=0$ shown below.

Suppose that this region is now revolved about the line $y=-2$ .

We will need a minimum of $\answer [given]{2}$ integrals with respect to $x$ to express the volume of the region, but we only need $\answer [given]{1}$ integral with respect to $y$ . As such, we choose to integrate with respect to $y$ .

Since we integrate with respect to $y$ , the slices should be verticalhorizontal . These slice are thus parallelperpendicular to the axis of rotation, so we should use the shellwasher method.

Since we must integrate with respect to $y$ , we will use the result

$V = \int _{y=c}^{y=d}2\pi \rho h \d y.$

Let’s start by expressing the curves as functions of $y$ .

For the curve described by $y=2x+4$ , we find $x= \answer [given]{\frac {1}{2}y-2}$ .
For the curve described by $y=4-x^2$ , we find $x= \answer [given]{\sqrt {4-y}}$ .

We must now find the limits of integration as express the radius $\rho$ and the height $h$ in terms of the variable of integration $y$ .

The limits of integration are: $c= \answer [given]{0}$ and $d = \answer [given]{4}$ .

To find $\rho$ and $h$ , we draw a helpful picture of the region $R$ below.

We see from the picture that $\rho$ is a verticalhorizontal distance.

Since $\rho$ is the distance from the axis of rotation to the slice, and this is a vertical distance, we find $\rho = y_{top}-y_{bot}$ . Noting that $y_{top}=$ and $y_{bot}=$ , we find $\rho = \answer [given]{y-(-2)}$ .

To find $h$ , note that from the picture $h$ is a distance, and we can write $h = x_{right}-x_{left}$ . By noting that $x_{right}=$ and $x_{left} =$ , we find $h= \answer [given]{ \sqrt {4-y} - \left (\frac {1}{2}y-2\right )}$ .

Using the shell method result $V = \int _{y=c}^{y=d} 2\pi \rho h \d y$ , we find that an integral that gives the volume of the solid of revolution is $V= \int _{y=\answer [given]{0}}^{y=\answer [given]{4}} 2 \pi \answer [given]{(y+2)\left ( \sqrt {4-y} - \frac {1}{2}y +2 \right )}\d y.$