6.2 Determining Volumes by Slicing

In the preceding section, we used definite integrals to find the area between two curves. In this section, we use definite integrals to find volumes of three-dimensional solids. We consider three approaches—slicing, disks, and washers—for finding these volumes, depending on the characteristics of the solid.

Volume and the Slicing Method

Just as area is the numerical measure of a two-dimensional region, volume is the numerical measure of a three-dimensional solid. Most of us have computed volumes of solids by using basic geometric formulas. The volume of a rectangular solid, for example, can be computed by multiplying length, width, and height: $V = l w h .$ The formulas for the volume of a sphere $(V = \frac{4}{3} π r^{3}),$ a cone $(V = \frac{1}{3} π r^{2} h),$ and a pyramid $(V = \frac{1}{3} A h)$ have also been introduced. Although some of these formulas were derived using geometry alone, all these formulas can be obtained by using integration.

We can also calculate the volume of a cylinder. Although most of us think of a cylinder as having a circular base, such as a soup can or a metal rod, in mathematics the word cylinder has a more general meaning. To discuss cylinders in this more general context, we first need to define some vocabulary.

We define the cross-section of a solid to be the intersection of a plane with the solid. A cylinder is defined as any solid that can be generated by translating a plane region along a line perpendicular to the region, called the axis of the cylinder. Thus, all cross-sections perpendicular to the axis of a cylinder are identical. The solid shown in Figure 6.11 is an example of a cylinder with a noncircular base. To calculate the volume of a cylinder, then, we simply multiply the area of the cross-section by the height of the cylinder: $V = A \cdot h .$ In the case of a right circular cylinder (soup can), this becomes $V = π r^{2} h .$

This graphic has two figures. The first figure is half of a cylinder, on the flat portion. The cylinder has a line through the center labeled “x”. Vertically cutting through the cylinder, perpendicular to the line is a plane. The second figure is a two dimensional cross section of the cylinder intersecting with the plane. It is a semi-circle. — Figure 6.11Each cross-section of a particular cylinder is identical to the others.

If a solid does not have a constant cross-section (and it is not one of the other basic solids), we may not have a formula for its volume. In this case, we can use a definite integral to calculate the volume of the solid. We do this by slicing the solid into pieces, estimating the volume of each slice, and then adding those estimated volumes together. The slices should all be parallel to one another, and when we put all the slices together, we should get the whole solid. Consider, for example, the solid S shown in Figure 6.12, extending along the $x -axis .$

This figure is a graph of a 3-dimensional solid. It has one edge along the x-axis. The x-axis is part of the 2-dimensional coordinate system with the y-axis labeled. The edge of the solid along the x-axis starts at a point labeled “a” and stops at a point labeled “b”. — Figure 6.12A solid with a varying cross-section.

We want to divide $S$ into slices perpendicular to the $x -axis .$ As we see later in the chapter, there may be times when we want to slice the solid in some other direction—say, with slices perpendicular to the y-axis. The decision of which way to slice the solid is very important. If we make the wrong choice, the computations can get quite messy. Later in the chapter, we examine some of these situations in detail and look at how to decide which way to slice the solid. For the purposes of this section, however, we use slices perpendicular to the $x -axis .$

Because the cross-sectional area is not constant, we let $A (x)$ represent the area of the cross-section at point $x .$ Now let $P = {x_{0}, x_{1} \dots, X_{n}}$ be a regular partition of and for let represent the slice of stretching from The following figure shows the sliced solid with

Finally, for $i = 1, 2, \dots n,$ let $x_{i}^{*}$ be an arbitrary point in $[x_{i - 1}, x_{i}] .$ Then the volume of slice can be estimated by Adding these approximations together, we see the volume of the entire solid can be approximated by

V (S) \approx i = 1 \sum n A (x_{i}^{*}) Δ x .

By now, we can recognize this as a Riemann sum, and our next step is to take the limit as $n \to \infty.$ Then we have

V (S) = n \to \infty lim i = 1 \sum n A (x_{i}^{*}) Δ x = \int_{a}^{b} A (x) d x .

The technique we have just described is called the slicing method. To apply it, we use the following strategy.

Recall that in this section, we assume the slices are perpendicular to the $x -axis .$ Therefore, the area formula is in terms of x and the limits of integration lie on the $x -axis .$ However, the problem-solving strategy shown here is valid regardless of how we choose to slice the solid.

Solids of Revolution

If a region in a plane is revolved around a line in that plane, the resulting solid is called a solid of revolution, as shown in the following figure.

This figure has three graphs. The first graph, labeled “a” is a region in the x y plane. The region is created by a curve above the x-axis and the x-axis. The second graph, labeled “b” is the same region as in “a”, but it shows the region beginning to rotate around the x-axis. The third graph, labeled “c” is the solid formed by rotating the region from “a” completely around the x-axis, forming a solid. — Figure 6.15(a) This is the region that is revolved around the x-axis. (b) As the region begins to revolve around the axis, it sweeps out a solid of revolution. (c) This is the solid that results when the revolution is complete.

Solids of revolution are common in mechanical applications, such as machine parts produced by a lathe. We spend the rest of this section looking at solids of this type. The next example uses the slicing method to calculate the volume of a solid of revolution.

The Disk Method

When we use the slicing method with solids of revolution, it is often called the disk method because, for solids of revolution, the slices used to over approximate the volume of the solid are disks. To see this, consider the solid of revolution generated by revolving the region between the graph of the function $f (x) = (x - 1)^{2} + 1$ and the $x -axis$ over the interval $[- 1, 3]$ around the $x -axis .$ The graph of the function and a representative disk are shown in Figure 6.18(a) and (b). The region of revolution and the resulting solid are shown in Figure 6.18(c) and (d).

This figure has four graphs. The first graph, labeled “a” is a parabola f(x)=(x-1)^2+1. The curve is above the x-axis and intersects the y-axis at y=2. Under the curve in the first quadrant is a vertical rectangle starting at the x-axis and stopping at the curve. The second graph, labeled “b” is the same parabola as in the first graph. The rectangle under the parabola from the first graph has been rotated around the x-axis forming a solid disk. The third graph labeled “c” is the same parabola as the first graph. There is a shaded region bounded above by the parabola, to the left by the line x=-1 and to the right by the line x=3, and below by the x-axis. The fourth graph labeled “d” is the same parabola as the first graph. The region from the third graph has been revolved around the x-axis to form a solid. — Figure 6.18(a) A thin rectangle for approximating the area under a curve. (b) A representative disk formed by revolving the rectangle about the $x -axis .$ (c) The region under the curve is revolved about the $x -axis,$ resulting in (d) the solid of revolution.

We already used the formal Riemann sum development of the volume formula when we developed the slicing method. We know that

V = \int_{a}^{b} A (x) d x .

The only difference with the disk method is that we know the formula for the cross-sectional area ahead of time; it is the area of a circle. This gives the following rule.

The volume of the solid we have been studying (Figure 6.18) is given by

V = \int_{a}^{b} π [f (x)]^{2} d x = \int_{- 1}^{3} π [(x - 1)^{2} + 1]^{2} d x = π \int_{- 1}^{3} [(x - 1)^{4} + 2 (x - 1)^{2} + 1] d x = π [\frac{1}{5} (x - 1)^{5} + \frac{2}{3} (x - 1)^{3} + x] ∣_{- 1}^{3} = π [(\frac{32}{5} + \frac{16}{3} + 3) - (- \frac{32}{5} - \frac{16}{3} - 1)] = \frac{412 π}{15} units^{3} .

Let’s look at some examples.

So far, our examples have all concerned regions revolved around the $x -axis,$ but we can generate a solid of revolution by revolving a plane region around any horizontal or vertical line. In the next example, we look at a solid of revolution that has been generated by revolving a region around the $y -axis .$ The mechanics of the disk method are nearly the same as when the $x -axis$ is the axis of revolution, but we express the function in terms of $y$ and we integrate with respect to y as well. This is summarized in the following rule.

The next example shows how this rule works in practice.

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 $x -axis$ or $y -axis$ 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 $f (x) = x$ and below by the graph of the function $g (x) = 1$ over the interval $[1, 4] .$ When this region is revolved around the 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 $x -axis .$ (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 (x) = π (x)^{2} - π (1)^{2} = π (x - 1) .

Then the volume of the solid is

V = \int_{a}^{b} A (x) d x = \int_{1}^{4} π (x - 1) d x = π [\frac{x ^{2}}{2} - x] ∣_{1}^{4} = \frac{9}{2} π units^{3} .

Generalizing this process gives the washer method.

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.

Rather than looking at an example of the washer method with the $y -axis$ 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.

Key Concepts

Definite integrals can be used to find the volumes of solids. Using the slicing method, we can find a volume by integrating the cross-sectional area.
For solids of revolution, the volume slices are often disks and the cross-sections are circles. The method of disks involves applying the method of slicing in the particular case in which the cross-sections are circles, and using the formula for the area of a circle.
If a solid of revolution has a cavity in the center, the volume slices are washers. With the method of washers, the area of the inner circle is subtracted from the area of the outer circle before integrating.

Key Equations

Disk Method along the x-axis	$V = \int_{a}^{b} π [f (x)]^{2} d x$
Disk Method along the y-axis	$V = \int_{c}^{d} π [g (y)]^{2} d y$