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The Washer Method

Some solids of revolution have cavities in the middle; they are not solid all the way to the axis of revolution. Sometimes, this is just a result of the way the region of revolution is shaped with respect to the axis of revolution. In other cases, cavities arise when the region of revolution is defined as the region between the graphs of two functions. A third way this can happen is when an axis of revolution other than the \(x -\text{axis}\) or \(y -\text{axis}\) is selected.

When the solid of revolution has a cavity in the middle, the slices used to approximate the volume are not disks, but washers (disks with holes in the center). For example, consider the region bounded above by the graph of the function \(f \left(\right. x \left.\right) = \sqrt{x}\) and below by the graph of the function \(g \left(\right. x \left.\right) = 1\) over the interval \(\left[\right. 1 , 4 \left]\right. .\) When this region is revolved around the \(x -\text{axis},\) the result is a solid with a cavity in the middle, and the slices are washers. The graph of the function and a representative washer are shown in Figure 6.22(a) and (b). The region of revolution and the resulting solid are shown in Figure 6.22(c) and (d).

This figure has four graphs. The first graph is labeled “a” and has the two functions f(x)=squareroot(x) and g(x)=1 graphed in the first quadrant. f(x) is an increasing curve starting at the origin and g(x) is a horizontal line at y=1. The curves intersect at the ordered pair (1,1). In between the curves is a shaded rectangle with the bottom on g(x) and the top at f(x). The second graph labeled “b” is the same two curves as the first graph. The shaded rectangle between the curves from the first graph has been rotated around the x-axis to form an open disk or washer. The third graph labeled “a” has the same two curves as the first graph. There is a shaded region between the two curves between where they intersect and a line at x=4. The fourth graph is the same two curves as the first with the region from the third graph rotated around the x-axis forming a solid region with a hollow center. The hollow center is represented on the graph with broken horizontal lines at y=1 and y=-1.
Figure 6.22 (a) A thin rectangle in the region between two curves. (b) A representative washer formed by revolving the rectangle about the \(x -\text{axis} .\) (c) The region between the curves over the given interval. (d) The resulting solid of revolution.

The cross-sectional area, then, is the area of the outer circle less the area of the inner circle. In this case,

\[A \left(\right. x \left.\right) = \pi \left(\right. \sqrt{x} \left.\right)^{2} - \pi \left(\right. 1 \left.\right)^{2} = \pi \left(\right. x - 1 \left.\right) .\]

Then the volume of the solid is

\[\begin{aligned} V & = \int_{a}^{b} A \left(\right. x \left.\right) d x \\ & = \int_{1}^{4} \pi \left(\right. x - 1 \left.\right) d x = \left(\pi \left[\right. \frac{x^{2}}{2} - x \left]\right. \left|\right.\right)_{1}^{4} = \frac{9}{2} \pi \text{units}^{3} . \end{aligned}\]

Generalizing this process gives the washer method.

Rule: The Washer Method

Suppose \(f \left(\right. x \left.\right)\) and \(g \left(\right. x \left.\right)\) are continuous, nonnegative functions such that \(f \left(\right. x \left.\right) \geq g \left(\right. x \left.\right)\) over \(\left[\right. a , b \left]\right. .\) Let \(R\) denote the region bounded above by the graph of \(f \left(\right. x \left.\right) ,\) below by the graph of \(g \left(\right. x \left.\right) ,\) 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 \(x -\text{axis}\) is given by

\[V = \int_{a}^{b} \pi \left[\right. \left(\right. f \left(\right. x \left.\right) \left.\right)^{2} - \left(\right. g \left(\right. x \left.\right) \left.\right)^{2} \left]\right. d x .\]

Example 6.10

Using the Washer Method

Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of \(f \left(\right. x \left.\right) = x\) and below by the graph of \(g \left(\right. x \left.\right) = 1 / x\) over the interval \(\left[\right. 1 , 4 \left]\right.\) around the \(x -\text{axis} .\)

Solution

The graphs of the functions and the solid of revolution are shown in the following figure.

This figure has two graphs. The first graph is labeled “a” and has the two curves f(x)=x and g(x)=1/x. They are graphed only in the first quadrant. f(x) is a diagonal line starting at the origin and g(x) is a decreasing curve with the y-axis as a vertical asymptote and the x-axis as a horizontal asymptote. The graphs intersect at (1,1). There is a shaded region between the graphs, bounded to the right by a line at x=4. The second graph is the same two curves. There is a solid formed by rotating the shaded region from the first graph around the x-axis.
Figure 6.23 (a) The region between the graphs of the functions \(f \left(\right. x \left.\right) = x\) and \(g \left(\right. x \left.\right) = 1 / x\) over the interval \(\left[\right. 1 , 4 \left]\right. .\) (b) Revolving the region about the \(x -\text{axis}\) generates a solid of revolution with a cavity in the middle.

We have

\[\begin{aligned} V & = \int_{a}^{b} \pi \left[\right. \left(\right. f \left(\right. x \left.\right) \left.\right)^{2} - \left(\right. g \left(\right. x \left.\right) \left.\right)^{2} \left]\right. d x \\ & = \pi \int_{1}^{4} \left[\right. x^{2} - \left(\right. \frac{1}{x} \left.\right)^{2} \left]\right. d x = \left(\pi \left[\right. \frac{x^{3}}{3} + \frac{1}{x} \left]\right. \left|\right.\right)_{1}^{4} = \frac{81 \pi}{4} \text{units}^{3} . \end{aligned}\]

Checkpoint 6.10

Find the volume of a solid of revolution formed by revolving the region bounded by the graphs of \(f \left(\right. x \left.\right) = \sqrt{x}\) and \(g \left(\right. x \left.\right) = 1 / x\) over the interval \(\left[\right. 1 , 3 \left]\right.\) around the \(x -\text{axis} .\)

As with the disk method, we can also apply the washer method to solids of revolution that result from revolving a region around the y-axis. In this case, the following rule applies.

Rule: The Washer Method for Solids of Revolution around the y-axis

Suppose \(u \left(\right. y \left.\right)\) and \(v \left(\right. y \left.\right)\) are continuous, nonnegative functions such that \(v \left(\right. y \left.\right) \leq u \left(\right. y \left.\right)\) for \(y \in \left[\right. c , d \left]\right. .\) Let \(Q\) denote the region bounded on the right by the graph of \(u \left(\right. y \left.\right) ,\) on the left by the graph of \(v \left(\right. y \left.\right) ,\) 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 \(y -\text{axis}\) is given by

\[V = \int_{c}^{d} \pi \left[\right. \left(\right. u \left(\right. y \left.\right) \left.\right)^{2} - \left(\right. v \left(\right. y \left.\right) \left.\right)^{2} \left]\right. d y .\]

Rather than looking at an example of the washer method with the \(y -\text{axis}\) as the axis of revolution, we now consider an example in which the axis of revolution is a line other than one of the two coordinate axes. The same general method applies, but you may have to visualize just how to describe the cross-sectional area of the volume.

Example 6.11

The Washer Method with a Different Axis of Revolution

Find the volume of a solid of revolution formed by revolving the region bounded above by \(f \left(\right. x \left.\right) = 4 - x\) and below by the \(x -\text{axis}\) over the interval \(\left[\right. 0 , 4 \left]\right.\) around the line \(y = −2 .\)

Solution

The graph of the region and the solid of revolution are shown in the following figure.

This figure has two graphs. The first graph is labeled “a” and has the two curves f(x)=4-x and -2. There is a shaded region making a triangle bounded by the decreasing line f(x), the y-axis and the x-axis. The second graph is the same two curves. There is a solid formed by rotating the shaded region from the first graph around the line y=-2. There is a hollow cylinder inside of the solid represented by the lines y=-2 and y=-4.
Figure 6.24 (a) The region between the graph of the function \(f \left(\right. x \left.\right) = 4 - x\) and the \(x -\text{axis}\) over the interval \(\left[\right. 0 , 4 \left]\right. .\) (b) Revolving the region about the line \(y = −2\) generates a solid of revolution with a cylindrical hole through its middle.

We can’t apply the volume formula to this problem directly because the axis of revolution is not one of the coordinate axes. However, we still know that the area of the cross-section is the area of the outer circle less the area of the inner circle. Looking at the graph of the function, we see the radius of the outer circle is given by \(f \left(\right. x \left.\right) + 2 ,\) which simplifies to

\[f \left(\right. x \left.\right) + 2 = \left(\right. 4 - x \left.\right) + 2 = 6 - x .\]

The radius of the inner circle is \(g \left(\right. x \left.\right) = 2 .\) Therefore, we have

\[\begin{aligned} V & = \int_{0}^{4} \pi \left[\right. \left(\right. 6 - x \left.\right)^{2} - \left(\right. 2 \left.\right)^{2} \left]\right. d x \\ & = \pi \int_{0}^{4} \left(\right. x^{2} - 12 x + 32 \left.\right) d x = \left(\pi \left[\right. \frac{x^{3}}{3} - 6 x^{2} + 32 x \left]\right. \left|\right.\right)_{0}^{4} = \frac{160 \pi}{3} \text{units}^{3} . \end{aligned}\]

Checkpoint 6.11

Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of \(f \left(\right. x \left.\right) = x + 2\) and below by the \(x -\text{axis}\) over the interval \(\left[\right. 0 , 3 \left]\right.\) around the line \(y = −1 .\)

This lesson is part of:

Applications of Integration

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