6.3 Cylindrical Shells Method | Math 140 - Calculus and Analytic Geometry 1

6.3.1 Visualization of the Cylindrical Shells Method

Example 1.

Consider the solid $\mathcal{S}$ obtained by revolving the region $\mathcal{R}$ enclosed by the graphs of
$f(x)= 4x-x^2$ and $g(x)=x$, $x \in [0,3]$, around the vertical line $x=4$. Here we set about to compute the volume of $\mathcal{S}$.

Figure 6.9 shows the following:

the region $\mathcal{R}$ and the solid of revolution $\mathcal{S}$;
a portion of the solid $\mathcal{S}$ as the region $\mathcal{R}$ is being revolved;
a midpoint rectangular approximation to $\mathcal{R}$ together with a cylindrical shell obtained by revolving one of the rectangles.

Solid obtained by revolving the region enclosed by y = 4x-x^2 and y = x around the line x=4

Figure 6.9a

The plane x,y cross section of solid obtained by revolving the region enclosed by y = 4x-x^2 and y = x around the line x=4.

Figure 6.9b

Solid obtained by revolving the region enclosed by y = 4x-x^2 and y = x around the line x=4. An approximating cylindrical shell.

Figure 6.9(c)

The cylindrical shells method consists in adding the volumes of the shells to set up a Riemann sum, and then taking the limit as the number of shells goes to infinity to compute a definite integral. To accomplish this, we need first to put in this context the formula to compute the volume of a cylindrical shell.

6.3.2 Prior Knowledge

A cylindrical shell is obtained by revolving a thin rectangular region around an axis parallel to its height. It is clear that such a shape consists of the solid region enclosed by two cylinders, a large outer cylinder and a smaller inner cylinder (see Figure 6.10).

Here we are interested in computing the volume $V$ of such a shell. It is immediate that $V$ is the difference of the volumes of the two cylinders, that is,
\[V=V_O-V_I =\pi r_O^2 h -\pi r_I^2h
=\pi \left(r_O^2- r_I^2\right)h.\]

Washer obtained by revolving a thin vertical rectangle around a vertical line together with rectangle of width w (small) and height h

Figure 6.10

Factoring the difference of squares, we obtain
\begin{equation}\label{ch06-03-eq01}\tag{6.7}
V=\pi \left(r_O- r_I\right)\left(r_O+r_I\right)h=\pi \left(r_O+r_I\right)h \Delta x,
\end{equation}
where we have used the following notation:

$V=$ volume of the cylindrical shell,
$V_O=$ volume of the outer cylinder,
$V_I=$ volume of the inner cylinder.

$r_O=$ radius of the outer cylinder, $r_I=$ radius of the inner cylinder, $h=$ height of the cylinders/rectangle,

$\Delta x =r_O- r_I =\text{thickness}=$ difference between the radii $=w=$ width of the rectangle.

In our context, $\Delta x$ will be very small.

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

Continuing with Example 1, Figure 6.11 shows the following:

the region $\mathcal{R}$ enclosed by the graphs of
$f(x)= 4x-x^2$ and $g(x)=x$, $x \in [0,3]$, and the axis of revolution $x=4$;
a midpoint rectangular approximation, and the outer and inner radii of the cylinders.
the cylindrical shell obtained by revolving a rectangle around the axis of revolution.

Objective. Here, we want to compute the volume of the cylindrical shells generated by revolving the rectangles in a midpoint rectangular approximation to the region $\mathcal{R}$. To accomplish this, we need to reformulate (\ref{ch06-03-eq02}) and (\ref{ch06-03-eq03}) in this context,

The region enclosed by y = 4x-x^2 and y = x, vertical axis y=4, outer and inner radii of a cylindrical shell, and shell

Figure 6.11a

Rectangular approximation to the region enclosed by y = 4x-x^2 and y = x, vertical axis y=4, outer.

Figure 6.11b

Rectangular approximation to the region enclosed by y = 4x-x^2 and y = x, vertical axis y=4, outer and inner radii of a cylindrical shell, and shell

Figure 6.11c

Key Points

The approximation is a midpoint rectangular approximation. Denote by $\overline{x}_i$ the midpoint of the $i$-th interval in a partition of the interval $[0,3]$.
The width of the rectangular regions is the same for all of them and is given by $\Delta x$.

In out present situation $\Delta x =\frac{3}{n}$.

The height of the $i$-th rectangular region is given by $h_i=|f(\overline{x}_i)-g(\overline{x}_i)|$, the distance between the $y$-coordinates of points on the graphs of $f$ and $g$ evaluated at the midpoint of the $i$-th interval.

In our present case, $h_i=f(\overline{x}_i)-g(\overline{x}_i)$ because $g(x)\leq f(x)$, $x\in [0,3]$).

It follows from (\ref{ch06-03-eq01}) that the volume $V_i$ of the $i$-th cylindrical shell generated by revolving the $i$-th rectangular region is
\begin{equation} \label{ch06-03-eq02}\tag{6.8}
V_i=\pi \left(r_O+r_I\right)h_i \Delta x = \pi (r_O+r_I)|f(\overline{x}_i)-g(\overline{x}_i)|\Delta x,
\end{equation}
and the sum of the volumes of the cylinders is
\begin{equation}\label{ch06-03-eq03}\tag{6.9}
\sum _{i=1}^n V_i=\sum _{i=1}^n \pi (r_O+r_I) |f(\overline{x}_i)-g(\overline{x}_i)|\Delta x.
\end{equation}
It remains to determine the values of $r_O$ and $r_I$. This part depends on the location of the axis with respect to the region $\mathcal{R}$.

[$r_O$] is given by the distance to the axis of revolution from the vertical side of the $i$-th rectangle which is furthest away from the axis of revolution.

In the present case this is given by $r_O = 4-x_{i-1}$ (see Figure 6.11(b)).

[$r_I$] is given by the distance to the axis of revolution from the vertical side of the $i$–th rectangle which is closest to the axis of revolution.

In the present case this is given by $r_I= 4-x_{i}$ (see Figure 6.11(b)).

6.3.4 Cylindrical Shells Method and Examples

Here we conclude Example 1.

Example 1. (Continuation.)
Consider the solid $\mathcal{S}$ obtained by revolving the region $\mathcal{R}$ enclosed by the graphs of
$f(x)= 4x-x^2$ and $g(x)=x$, $x \in [0,3]$, around the vertical line $x=4$. Compute the volume of $\mathcal{S}$.

Solution. As Figure 6.11 shows, revolving each of the rectangles in $R(f,n)$ leads to a corresponding disk. The method consists on adding the volumes of the disks to approximate the volume of the solid of revolution.

Step 1. Midpoint rectangular approximation to the region enclosed by the graph of $f$.}

– Partition of the interval:
\[ \Delta x=\frac{3-0}{n}=\frac{3}{n}, \;\;\; x_i=x_0+i \Delta x=\frac{3i}{n},\;\;\;
\overline{x}_i=\frac{x_{i-1}+x_i}{2} = \mathrm{midpoint\, of\, the\, }i\mathrm{th\, interval}.\]
– Width and height of the $i$-th rectangular region:
\[ w_i=\Delta x=\frac{3}{n}, \;\;\; h_i= f\left(\overline{x}_i\right)-g(\overline{x}_i), \;\;\; i=1, 2, \dots , n.
\]
Note that the height is obtained by evaluating $f$ and $g$ at the midpoint of the interval, as we are using the midpoint rectangular approximation.

In our present case, $h_i=f(\overline{x}_i)-g(\overline{x}_i)$ because $g(x)\leq f(x)$, $x\in [0,3]$.

Step 2. Volume of the cylindrical shells generated by revolving the rectangular regions.}

Formula (\ref{ch06-03-eq02}) yields
\[V_i=\pi \left(r_O+r_I\right)h_i \Delta x = \pi (r_O+r_I)\left|f(\overline{x}_i)-g(\overline{x}_i)\right|\Delta x
= \pi (r_O+r_I)\left(f(\overline{x}_i)-g(\overline{x}_i)\right)\Delta x.\]
And, as show in the last key point in 6.3.3, $r_O=4-x_{i-1}$ and $r_I=4-x_i$. Thus,
\begin{align*}
V_i &= \pi (4-x_{i-1}+4-x_{i})\left(f(\overline{x}_i)-g(\overline{x}_i)\right)\Delta x
= \pmb{2}\pi \left(\frac{4-x_{i-1}+4-x_{i}}{\pmb{2}}\right )\left(f(\overline{x}_i)-g(\overline{x}_i)\right)\Delta x =\\
&\underset{?}{=} \pmb{2}\pi \left (4-\overline{x}_i \right )\left(f(\overline{x}_i)-g(\overline{x}_i)\right)\Delta x
.
\end{align*}

Step 3. Riemann sum to approximate the volume.}

\begin{align*}
\sum _{i=1}^n V_i
&=\sum _{i=1}^n \pi (r_O+r_I)\left|f(\overline{x}_i)-g(\overline{x}_i)\right|\Delta x
=\sum _{i=1}^n 2\pi \left (4-\overline{x}_i \right )\left(f(\overline{x}_i)-g(\overline{x}_i)\right)\Delta x.
\end{align*}

Step 4. Evaluation of the definite integral to compute the volume.}

\[
\int_{0}^3 2\pi(4-x) \left(f(x)-g(x)\right)\,dx
= \lim_{n\to \infty}\sum _{i=1}^n 2\pi \left (4-\overline{x}_i \right )\left(f(\overline{x}_i)-g(\overline{x}_i)\right)\Delta x.
\]
To find the exact value of the definite integral, we apply TFTCP2:
# 2 in 2.4.3.
\begin{align*}
&\int_{0}^3 2\pi(4-x) \left(f(x)-g(x)\right)\,dx = \int_{0}^3 2\pi(4-x) \left(4x-x^2-x\right)\,dx
\underset{?}{=} \frac{45\pi}{2}
\end{align*}
On Your Own 1.
Use cylindrical shells to show that the volume of a if the solid $\mathcal{S}$ obtained by revolving the region $\mathcal{R}$ in Example 1 around the line $x=-1$
(see Figure 6.12) is given by
\[
\int_{0}^3 2\pi(x+1) \left(f(x)-g(x)\right)\,dx
= \lim_{n\to \infty}\sum _{i=1}^n 2\pi \left (\overline{x}_i +1\right )\left(f(\overline{x}_i)-g(\overline{x}_i)\right)\Delta x =\frac{45\pi}{2}.
\]

Solid obtained by revolving the region enclosed by y = 4x-x^2 and y = x around the line x=-1.

Figure 6.12a

An approximating cylindrical shell. Solid obtained by revolving the region enclosed by y = 4x-x^2 and y = x around the line x=-1.

Figure 6.12b

6.3.5 Exercises

In problems 1—6, for the given functions $f$ and $g$, let $\mathcal{R}$ be the region enclosed by their graphs, and let $\mathcal{S}$ be the solid obtained by revolving $\mathcal{R}$ around the given axis. Do all of the following by explicitly showing all details for all four steps in the cylindrical shells method as in
Example 1.

Sketch $\mathcal{R}$ and set up a Riemann sum to approximate the volume $V(\mathcal{S})$.
Evaluate the definite integral to compute the exact value of $V(\mathcal{S})$ via TFTCP2.

$\displaystyle f(x)=\frac{1}{\sqrt{x}}$ and $g(x)=0$, $ x\in [1,4]$; $x=0$.
$f(x)=x^2$ and $g(x)= x$; (i) $x=2$; (ii) $x=-2$.
Hint: find the points of intersection of the two graphs.
$f(x)=x^2-2x$ and $g(x)= x^3-4x$, $x\geq 0$; (i) $x=2$; (ii) $x=-2$.
$f(x)=x^2-2x$ and $g(x)= -x$; (i) $x=1$; (ii) $x=-1$.
$f(x)=\cos x +2$ and $y=0$, $x\in [-\pi,\pi]$; $y=0$.
$f(x)=x^2+2x$ and $y=0$; (i) $x=0$; (ii) $y=0$.
$f(x)=\sin x+1$ and $y=0$, $x\in [0,\pi ]$; $y=0$.
Show that the volume of the solid of revolution obtained by revolving the circle with center at $(R,0)$ and radius $r$ with (0 < r < R) around the $x$-axis (the solid is called a torus) is $V=2\pi R\cdot \pi r^2 $. Do this in two different ways (a) via cylindrical shells; (b) via slicing

Hint: for (b) interpret the integral geometrically as area.

On Your Own 1.

Step 1. Midpoint rectangular approximation to the region enclosed by the graph of $f$.}

Partition of the interval:
\[ \Delta x=\frac{3-0}{n}=\frac{3}{n}, \;\;\; x_i=x_0+i \Delta x=\frac{3i}{n},\;\;\;
\overline{x}_i=\frac{x_{i-1}+x_i}{2} = \mathrm{midpoint\, of\, the\, }i\mathrm{th\, interval}.\]

Width and height of the $i$-th rectangular region:
\[ w_i=\Delta x=\frac{3}{n}, \;\;\; h_i= |f\left(\overline{x}_i\right)-g(\overline{x}_i)|
\underset{?}{=}f(\overline{x}_i)-g(\overline{x}_i), \;\;\; i=1, 2, \dots , n.
\]

Step 2. Volume of the cylindrical shells generated by revolving the rectangular regions.}

Formula (\ref{ch06-03-eq02}) yields
\[V_i=\pi \left(r_O+r_I\right)h_i \Delta x = \pi (r_O+r_I)\left|f(\overline{x}_i)-g(\overline{x}_i)\right|\Delta x
= \pi (r_O+r_I)\left(f(\overline{x}_i)-g(\overline{x}_i)\right)\Delta x.\]
$r_O=x_{i}-(-1)$ and $r_I=4-x_{i-1}-(-1)$. Thus,
\begin{align*}
V_i &= \pi (x_{i}+1+x_{i-1}+1)\left(f(\overline{x}_i)-g(\overline{x}_i)\right)\Delta x
= \pmb{2}\pi \left(\frac{x_{i}+1+x_{i-1}+1}{\pmb{2}}\right )\left(f(\overline{x}_i)-g(\overline{x}_i)\right)\Delta x =\\
&\underset{?}{=} \pmb{2}\pi \left (\overline{x}_i+1 \right )\left(f(\overline{x}_i)-g(\overline{x}_i)\right)\Delta x
,
\end{align*}

Step 3. Riemann sum to approximate the volume.}

\begin{align*}
\sum _{i=1}^n V_i
&=\sum _{i=1}^n \pi (r_O+r_I)\left|f(\overline{x}_i)-g(\overline{x}_i)\right|\Delta x
=\sum _{i=1}^n 2\pi \left (\overline{x}_i + 1 \right )\left(f(\overline{x}_i)-g(\overline{x}_i)\right)\Delta x.
\end{align*}

Step 4. Evaluation of the definite integral to compute the volume.}

\[
\int_{0}^3 2\pi(x+1) \left(f(x)-g(x)\right)\,dx
= \lim_{n\to \infty}\sum _{i=1}^n 2\pi \left (\overline{x}_i + 1 \right )\left(f(\overline{x}_i)-g(\overline{x}_i)\right)\Delta x =\frac{45\pi}{2}.
\]

Previous Section

Next Section