6.1 Disk Method

Volume

A geometric problem where Riemann sums play a key role, besides computing area, is the problem of finding the volume of a given solid. The purpose of this chapter is to present approximation processes to to compute volume.

To be more precise, let \(\mathcal{R}\) be a region enclosed by the graph of a continuous function \(f\) with \(f(x) \geq 0\) for all \(x\in [a,b]\).

1. Let \(\mathcal{S}\) be the solid obtained by revolving such a region \(\mathcal{R}\) around the \(x\)-axis. We will use the disk method to approximate the volume of such a solid.
2. Let \(\mathcal{S}\) be the solid whose base is the region \(\mathcal{R}\) with given cross sections. To compute the volume of such a solid, we will use the slicing method.
3. Let \(\mathcal{S}\) be the solid obtained by revolving such a region \(\mathcal{R}\) around the \(y\)-axis, we will use the cylindrical shells method to approximate the volume of such a solid.
4. The disk method can be extended to the deal with regions enclosed by the graphs of two functions, via the washer method. This is the content of 6.4.
5. The slicing method can be extended to include other solids. This is the content of 6.5.

6.1 Disk Method

6.1.1 Visualization of the Disk Method

Consider the solid \(\mathcal{S}\) obtained by revolving the region \(\mathcal{R}\) enclosed by the graph of
\(f(x)=\sin x\), \(x \in [0,\pi]\) and the \(x\)-axis.
Figure 6.1 shows the solid together with a rectangular approximation, and a disk obtained by revolving one of the rectangles in the rectangular approximation.

Figure 6.1a

Figure 6.1b

Figure 6.1c

The disk method consists on adding the volumes of the disks thus obtained, and then taking the limit as the number of disks goes to infinity. To accomplish this, we first put the formula to compute the volume of a disk in this context.

6.1.2 Prior Knowledge

• The volume of a right circular cylinder of radius \(r\) and height \(h\) is given by
  \[V=\pi r^2h.\]
• Let \(\mathcal{R}\) be a rectangular region with small width \(w\) and length \(l\), and let
  \(\mathcal{C}\) be the thin disk obtained by revolving \(\mathcal{R}\) around one edge corresponding to a width (see Figure2}). Then, the volume of the disk, which is actually a thin cylinder, is obtained by noticing that
  its thickness, which is the height \(h\) (as a cylinder) is given by \(w\), and the
  radius \(r\) is actually \(l\), so that the volume of the disk is (see Figure 6.2)
  \begin{equation}\label{ch06-01-eq01}\tag{6.1}
  V=\pi r^2h=\pi l^2w
  \end{equation}

Figure 6.2a

Figure 6.2b

6.1.3 Reformulation of Prior Knowledge in the Context of a Solid of Revolution

Let \(f\) be a function with \(f(x)\geq 0\) for all \(x\in [a,b]\), and let $R(f,n)$ be the right-endpoint rectangular approximation with $n$ rectangular regions. Let \(x_i\) be the right endpoint of the \(i\)-th interval in a partition of \([a,b]\). Then the \(i\)-th rectangle has width \(w_i=\Delta x\) and height \(h_i=f(x_i)\). It follows from
(\ref{ch06-01-eq01}) that the volume \(V_i\) of the \(i\)-th disk generated by revolving the \(i\)-th rectangular region is
\begin{equation}\label{ch06-01-eq02} \tag{6.2}
V_i=\pi \left(f\left(x_i\right)\right)^2\Delta x,
\end{equation}
and the corresponding Riemann sum obtained by adding the volumes of the cylinders is
\begin{equation}\label{ch06-01-eq03}\tag{6.3}
\sum _{i=1}^n V_i=\sum _{i=1}^n \pi \left(f\left(x_i\right)\right)^2\Delta x.
\end{equation}
Observation. The sum in (\ref{ch06-01-eq03}) shows
that the approximation to the volume by means of \(n\) disks (using right endpoints) is actually \(R\left(\pi f^2,n\right)\).

6.1.4 Disk Method and Examples

The following example, serves two purposes: on one hand, it shows in an explicit fashion the steps to execute the disk method. On the other hand, it provides all details of the method when applied to find the volume of a right circular cone.

Example 1 [and Method]
A right circular cone of height \(h\) and radius \(r\) is obtained by revolving, around the \(x\)-axis, the region \(\mathcal{R}\) enclosed by the graph of \(\displaystyle f(x)=\frac{r}{h}x\vphantom{x^{2^2}}\), \(x\in [0,h]\) and the \(x\)-axis (see Figure 6.2 ). Using the disk method, show that its volume is given by \( V=\frac{1}{3}\pi r^2h\vphantom{x^{2^2}}\).

Figure 6.3a

Figure 6.3b

Figure 6.3c

Solution.

Step 1. Rectangular approximation to the region enclosed by the graph of $f$.

– Partition of the interval:
\[ \Delta x=\frac{h-0}{n}=\frac{h}{n}, \;\;\; x_i=x_0+i \Delta x=\frac{hi}{n}.\]
– Width and height of the \(i\)-th rectangular region:
\[ w_i=\Delta x=\frac{h}{n}, \;\;\; h_i= f\left(x_i\right)=\frac{r}{h}x_i=\frac{r}{h}\frac{hi}{n}=\frac{ri}{n}, \;\;\; i=1, 2, \dots , n.
\]

Step 2. Volume of the disks generated by revolving the rectangular regions.

In the context of the cone, where \(\displaystyle f\left(x_i\right)=\frac{r}{h}x_i=\frac{ri}{n}\), the volume of the \(i\)-th disk is given by (see (\ref{ch06-01-eq02}))
\[V_i =\pi \left(f\left(x_i\right)\right)^2\Delta x = \pi \left(\frac{ri}{n}\right)^2\frac{h}{n}= \frac{\pi r^2 h}{n^3} i^2 .\]

Step 3. Riemann sum to approximate the volume.
\begin{align*}
\sum _{i=1}^n V_i
&=\sum _{i=1}^n \pi \left(f\left(x_i\right)\right)^2\Delta x
=\sum _{i=1}^n \frac{\pi r^2 h}{n^3} i^2
=\frac{\pi h r^2}{n^3} \sum _{i=1}^n i^2
=\frac{\pi h r^2}{n^3}\frac{n(n+1)(2n+1)}{6}.
\end{align*}

Step 4. Evaluation of the definite integral to compute the exact volume.
\begin{align*}
\int_{0}^{h}\pi (f(x))^2\,dx
&=\lim_{n\to \infty} \sum _{i=1}^n \pi \left(f\left(x_i\right)\right)^2\Delta x
=\lim_{n\to \infty} \pi \frac{r^2h}{n^3}\frac{n(n+1)(2n+1)}{6} =\\
&\underset{?}{=}\lim_{n\to \infty}\pi \frac{r^2h}{6}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)=\pi \frac{r^2h}{6}(1+0)(2+0)= \frac{1}{3}\pi r^2h.
\end{align*}

On Your Own 1.
Use the disk method to show that for a sphere of radius \(r\), \(\displaystyle V=\frac{4}{3}\pi r^3\).

Hint: naturally, a sphere is obtained by revolving a semicircle. Thus, you need to show that
\[
\int_{-r}^r \pi \left(\sqrt{r^2-x^2}\right)^2\,dx
= \lim_{n\to \infty}\sum _{i=1}^n \pi \left(\sqrt{r^2-x_i^2}\right)^2\,\Delta x = \frac{4}{3}\pi r^2h.
\]

Example 2
Compute the volume of the solid \(\mathcal{S}\) obtained by revolving the region \(\mathcal{R}\) enclosed by the graph of \(f(x)=\sin x +2\), \(\displaystyle x\in \left [-\frac{\pi}{2}, \frac{\pi}{2}\right ]\) around the \(x\)-axis.

Solution.

Step 1. Rectangular approximation to the region enclosed by the graph of $f$.

– Partition of the interval:
\[ \Delta x=\frac{\frac{\pi}{2}-\left (-\frac{\pi}{2}\right )}{2}=\frac{\pi}{n}, \;\;\; x_i=x_0+i \Delta x= -\frac{\pi}{2}+\frac{\pi i}{n}.\]
– Width and height of the \(i\)-th rectangular region:
\[ w_i=\Delta x=\frac{\pi}{n}, \;\;\;
h_i= f\left(x_i\right)=f\left(-\frac{\pi}{2}+\frac{\pi i}{n}\right)= \sin \left (-\frac{\pi}{2}+\frac{\pi i}{n}\right )+2, \;\;\; i=1, 2, \dots , n.
\]

Step 2. Volume of the disks generated by revolving the rectangular regions.

The volume of the \(i\)-th disk is given by (see (\ref{ch06-01-eq02}))
\[V_i =\pi \left(f\left(x_i\right)\right)^2\Delta x = \pi\left( \sin x_i + 2\right )^2\Delta x.\]

Step 3. Riemann sum to approximate the volume.
\begin{align*}
\sum _{i=1}^n V_i
&=\sum _{i=1}^n \pi \left( f\left(x_i\right)\right)^2\Delta x
= \sum _{i=1}^n \pi \left( \sin x_i + 2 \right )^2\Delta x.
\end{align*}

Step 4. Evaluation of the definite integral to compute the exact volume.
\[
\int_{-\pi/2}^{\pi/2}\pi (f(x))^2\,dx
=\lim_{n\to \infty} \sum _{i=1}^n \pi \left(f\left(x_i\right)\right)^2\Delta x,
\]
that is,
\[
\int_{-\pi/2}^{\pi/2}\pi (\sin x + 2)^2\,dx
=\int_{-\pi/2}^{\pi/2}\pi \left (\sin^2 x + 4\sin x + 4\right )\,dx.\]
Using a double-angle identity, the last integral can be rewritten as
\[
\int_{-\pi/2}^{\pi/2}\pi \left (\frac{1-\cos 2 x}{2} + 4\sin x + 4\right )\,dx =\frac{\pi}{2}\int_{-\pi/2}^{\pi/2} \left (9-\cos 2 x + 8\sin x\right )\,dx.
\]
To evaluate the definite integral, we use the integration formulas
\[ \int_a^b\sin k x\,dx = \frac{1}{k}\left(-\cos(kb)+\cos(ka)\right) \;\;\; \mathrm{and} \;\;\;
\int_a^b\cos k x\,dx =\frac{1}{k}\left(\sin(kb)-\sin(ka)\right).\]

\[
\frac{\pi}{2}\int_{-\pi/2}^{\pi/2} \left (9-\cos 2 x + 8\sin x\right )\,dx
=
\frac{\pi}{2}\left [9\int_{-\pi/2}^{\pi/2} \,dx
– \int_{-\pi/2}^{\pi/2} \cos 2 x \,dx
+8\int_{-\pi/2}^{\pi/2} \sin x \,dx \right ]= \]
\[\underset{?}{=}\frac{\pi}{2}\left [
9\left (\frac{\pi}{2} -\left (-\frac{\pi}{2}\right )\right )
-\frac{1}{2}\left (\sin \left (2\cdot\frac{\pi}{2}\right)- \sin \left(-2\cdot\frac{\pi}{2}\right)\right)
+8\left ( -\cos \frac{\pi}{2} +\cos \left (-\frac{\pi}{2}\right ) \right )
\right ]
\underset{?}{=}\frac{9\pi^2}{2}
\]

On Your Own 2.
Do as in Example 2 for \(f(x)=2\cos x +3\), \(\displaystyle x\in\left [-\frac{\pi}{3},\frac{2\pi}{3}\right ]\). Answer. \(V(\mathcal{S})=\pi(12\sqrt{3}+11\pi)\).