6.4 Washer Method

6.4.1 Prior Knowledge

 
Let \(\mathcal{R}\) be a rectangular region with small width \(w\) and length \(l\), and let \(\mathcal{C}\) be the washer obtained by revolving \(\mathcal{R}\) around an axis parallel to the width-edges (see Figure 6.14). Then, the volume of the washer, which is actually a thin cylinder with a cylindrical hole at its center, is obtained by subtracting the volume of the cylindrical hole to the volume of the outer cylinder that encloses the washer.
Let \(V_O\) and \(V_I\) be the volume of the outer and inner cylinders, respectively, and let \(r_O\) and \(r_I\) be their corresponding radii. The thickness of the washer, which corresponds to the height \(h\) of the cylinders is given by \(w\) (see Figure 6.14). Thus, we have
\begin{equation}\label{ch06-04-eq01}\tag{6.10}
V= V_O-V_I = \pi r_O^2 h – \pi r_I^2 h = \pi (r_O^2 – r_I^2 )h,
\end{equation}

Figure 6.13a

Figure 6.13b

6.4.2 Examples and the Washer Method

 
Example 1. Consider the solid \(\mathcal{S}\) obtained by revolving the region \(\mathcal{R}\) enclosed by the graphs of \(f(x)=4-x^2\) and \(g(x)= x^2+2\), \(x \in [-1,1]\) (see Figure 6.14(a)), around the line \(y=-1\) (see Figure 6.14 (b) and (c)). (a) Set up a Riemann sum to approximate the volume of the solid by means of volumes of washers; (b) compute the exact volume.

Figure 6.14a

Figure 6.14b

Figure 6.14c

Outline of the Washer Method. The method consists on producing a rectangular approximation to the region \(\mathcal{R}\) enclosed by the graphs of the functions, and then revolving each rectangle to generate a washer (see Figure 6.14(d)). The sum of the volumes of the washers, a Riemann sum, provides an approximation to the volume of the solid. The limit of the Riemann sum, a definite integral, yields the exact value of the volume.

Solution.

Step 1. Rectangular approximation to the region enclosed by the graph of \(f\).

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

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

Since \(f(x) \geq g(x) \geq -1\) for all \(x\in [a,b]\), then (see Figure 6.15)

\(r_O = \) distance from the axis to the point \((x_i, f(x_))\), that is \(|f(x_i)-(-1)|= f(x_i)+1\);

\(r_I = \) distance from the axis to the point \((x_i, g(x_))\), that is \(|g(x_i)-(-1)|= g(x_i)+1\).

Figure 6.15

It follows from (\ref{ch06-04-eq01}) that the volume of the \(i\)-th washer, \(W_i\), generated by revolving the \(i\)-th rectangular region is given by
\begin{equation}\label{ch06-04-eq02}\tag{6.11}
W_i = V_O-V_I = \pi \left (r_O ^2- r_I^2\right )\Delta x = \pi \left (\left (f(x_i)+1\right ) ^2-\left (g(x_i)+1\right )^2\right )\Delta x.
\end{equation}

Step 3. Riemann sum to approximate the volume.

The corresponding Riemann sum obtained by adding the volumes of the washers is
\begin{equation}\label{ch06-04-eq03}\tag{6.12}
\sum _{i=1}^n V_i=\sum _{i=1}^n \pi \left (\left (f(x_i)+1\right ) ^2-\left (g(x_i)+1\right )^2\right )\Delta x.
\end{equation}

Step 4. Evaluation of the definite integral to compute the exact volume.
\[
V=\int_{-1}^{1}\pi \left ((f(x)+1)^2-(g(x)+1)^2\right )\,dx
=\lim_{n\to \infty} \sum _{i=1}^n \pi \left (\left (f(x_i)+1\right ) ^2-\left (g(x_i)+1\right )^2\right )\Delta x .
\]
Substituting the expressions for \(f\) and \(g\), a direct integration yields \(\displaystyle V= \frac{64\pi}{3}\).

The following example, differs from the previous one in one respect: the axis of revolution is above the region. As it turns out, the computations are essentially the same. The only slight change occurs in the expressions for \(r_O\) and \(r_I\).

Example 2. (and Method) Consider the solid \(\mathcal{S}\) obtained by revolving the region \(\mathcal{R}\) enclosed by the graphs of \(\displaystyle f(x)=\sin \frac{\pi x}{2}\) and \(g(x)= (x-1)^2\), \(x \in [0,2]\) (see Figure 6.14(a)), around the line \(y=3\) (see Figure 6.14 (b)). (a) Set up a Riemann sum to approximate the volume of the solid by means of volumes of washers; (b) compute the exact volume.

Figure 6.16a

Figure 6.16b

Figure 6.16c

Solution.

Step 1. Rectangular approximation to the region enclosed by the graph of \(f\).

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

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

Note here that since \(3> f(x) \geq g(x)\) for all \(x\in [a,b]\), then (see Figure 6.16(c) )

\(r_O = \) distance from the axis to the point \((x_i, g(x_))\), that is \(|g(x_i)-(3)|=3- g(x_i)\);

\(r_I = \) distance from the axis to the point \((x_i, f(x_))\), that is \(|f(x_i)-3|= 3 – f(x_i)\).

It follows that the volume of the \(i\)-th washer is given by
\[W_i=\pi \left (\left(3-g\left(x_i\right)\right)^2- \left(3-f\left(x_i\right)\right)^2\right )\Delta x.\]

This is point where the location of the axis of revolution with respect to the region \(\mathcal{R}\) makes a difference.

Step 3. Riemann sum to approximate the volume.
\[\sum _{i=1}^n W_i
=\sum _{i=1}^n \pi \left (\left(3-g\left(x_i\right)\right)^2- \left(3-f\left(x_i\right)\right)^2\right )\Delta x
\]

Step 4. Evaluation of the definite integral to compute the exact volume.
\[
\int_{0}^{2}\pi \left (\left(3-g\left(x\right)\right)^2- \left(3-f\left(x\right)\right)^2\right )\,dx
=\lim_{n\to \infty} \sum _{i=1}^n \pi \left (\left(3-g\left(x_i\right)\right)^2- \left(3-f\left(x_i\right)\right)^2\right )\Delta x.
\]
To evaluate the definite integral, we proceed in three steps:

(1)
\begin{align*}
\int_{0}^{2}\pi \left(3-g\left(x\right)\right)^2 \,dx
&= \int_{0}^{2}\pi \left( x^4-4 x^3+8 x+4 \right) \,dx
=\frac{72\pi}{5}.
\end{align*}

(2)
\begin{align*}
\int_{0}^{2}\pi \left(3-f\left(x\right)\right)^2 \,dx
&= \int_{0}^{2}\pi \left( \sin^2 \frac{\pi x}{2}- 4\sin \frac{\pi x}{2}+ 4 \right) \,dx \\
&= \int_{0}^{2}\pi \left( \frac{1}{2}\left (1-\cos \pi x \right ) – 4\sin \frac{\pi x}{2}+ 4 \right) \,dx = 9\pi -16.
\end{align*}

(3) Then \(\displaystyle V= \frac{72\pi}{5} – \left (9\pi -16\right ) = 16 + \frac{27\pi}{5} \).