Volume and the Slicing Method

Just as area measures a two-dimensional region, volume measures a three-dimensional solid. Most of us have computed volumes of solids using basic geometric formulas. For example, the volume of a rectangular solid can be computed by multiplying length, width, and height:

[latex]V=lwh.[/latex]

Below are some other common volume formulas:

• Sphere: [latex](V=\frac{4}{3}\pi {r}^{3}),[/latex]
• Cone: [latex](V=\frac{1}{3}\pi {r}^{2}h),[/latex]
• Pyramid: [latex](V=\frac{1}{3}Ah)[/latex].

Although some of these formulas were derived using geometry alone, all these formulas can be obtained by using integration.

Notice we did not give the formula for calculating the volume of a cylinder. To explore cylinders in this broader sense, we first need to define some terminology.  Although we typically think of a cylinder as having a circular base (like a soup can or a metal rod), in mathematics, the word “cylinder” has a more general meaning. 

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 the figure below is an example of a cylinder with a non-circular base.

To calculate the volume of a cylinder, we multiply the area of the cross-section by the height of the cylinder:

[latex]V=A·h.[/latex]

In the case of a right circular cylinder (such as a soup can), this becomes:

[latex]V=\pi {r}^{2}h.[/latex]

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, the solid [latex]S[/latex] shown below, extending along the [latex]x\text{-axis}\text{.}[/latex]

We want to divide [latex]S[/latex] into slices perpendicular to the [latex]x\text{-axis}\text{.}[/latex]

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 [latex]y[/latex]-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 [latex]x\text{-axis}\text{.}[/latex]

Because the cross-sectional area is not constant, we let [latex]A(x)[/latex] represent the area of the cross-section at point [latex]x.[/latex]

Now let [latex]P=\left\{{x}_{0},{x}_{1}\text{…},{X}_{n}\right\}[/latex] be a regular partition of [latex]\left[a,b\right],[/latex] and for [latex]i=1,2\text{,…}n,[/latex] let [latex]{S}_{i}[/latex] represent the slice of [latex]S[/latex] stretching from [latex]{x}_{i-1}\text{ to }{x}_{i}.[/latex]

The following figure shows the sliced solid with [latex]n=3.[/latex]

Finally, for [latex]i=1,2\text{,…}n,[/latex] let [latex]{x}_{i}^{*}[/latex] be an arbitrary point in [latex]\left[{x}_{i-1},{x}_{i}\right].[/latex]

Then the volume of slice [latex]{S}_{i}[/latex] can be estimated by:

[latex]V({S}_{i})\approx A({x}_{i}^{*})\text{Δ}x.[/latex]

Adding these approximations together, we see the volume of the entire solid [latex]S[/latex] can be approximated by:

[latex]V(S)\approx \underset{i=1}{\overset{n}{\text{∑}}}A({x}_{i}^{*})\text{Δ}x[/latex]

By now, we can recognize this as a Riemann sum, and our next step is to take the limit as [latex]n\to \infty .[/latex]

Then we have:

[latex]V(S)=\underset{n\to \infty }{\text{lim}}\underset{i=1}{\overset{n}{\text{∑}}}A({x}_{i}^{*})\text{Δ}x=\underset{a}{\overset{b}{\displaystyle\int }}A(x)dx[/latex]

The technique we have just described is called the slicing method.

To apply it, we use the following strategy.

We know from geometry that the formula for the volume of a pyramid is [latex]V=\frac{1}{3}Ah.[/latex] If the pyramid has a square base, this becomes [latex]V=\frac{1}{3}{a}^{2}h,[/latex] where [latex]a[/latex] denotes the length of one side of the base. Use the slicing method to derive this formula.

Show Solution

We want to apply the slicing method to a pyramid with a square base. To set up the integral, consider the pyramid shown in Figure 4, oriented along the [latex]x\text{-axis}\text{.}[/latex]

We first want to determine the shape of a cross-section of the pyramid. We are know the base is a square, so the cross-sections are squares as well (step 1). Now we want to determine a formula for the area of one of these cross-sectional squares. Looking at Figure 4(b), and using a proportion, since these are similar triangles, we have

[latex]\frac{s}{a}=\frac{x}{h}[/latex]   or   [latex]s=\frac{ax}{h}[/latex]

Therefore, the area of one of the cross-sectional squares is

[latex]A(x)={s}^{2}={(\frac{ax}{h})}^{2}[/latex]  (step [latex]2[/latex])

Then we find the volume of the pyramid by integrating from [latex]0\text{ to }h[/latex] (step [latex]3)\text{:}[/latex]

[latex]\begin{array}{cc}\hfill V& =\underset{0}{\overset{h}{\displaystyle\int }}A(x)dx\hfill \\ & =\underset{0}{\overset{h}{\displaystyle\int }}{(\frac{ax}{h})}^{2}dx=\frac{{a}^{2}}{{h}^{2}}\underset{0}{\overset{h}{\displaystyle\int }}{x}^{2}dx\hfill \\ & ={\left[\frac{{a}^{2}}{{h}^{2}}(\frac{1}{3}{x}^{3})\right]|}_{0}^{h}=\frac{1}{3}{a}^{2}h.\hfill \end{array}[/latex]

This is the formula we were looking for.