Volumes by cylindrical shells
Some solids of revolution are hard to find with disks or washers. For example, rotate the region under
$y=2x^2-x^3$, for $0\le x\le 2$,
about the $y$-axis. Washers would be horizontal, so we would need to solve $y=2x^2-x^3$ for $x$. That is very hard. The shell method avoids this.
The shell method
Instead of slicing across the axis, cut the region into thin vertical strips, parallel to the axis. When a strip at distance $x$ from the $y$-axis is rotated, it sweeps out a thin cylindrical shell, like a tin can with no top or bottom.
Cut the shell down one side and unroll it. It becomes a thin, flat sheet:
- Its length is the circumference of the shell, $2\pi x$.
- Its height is $f(x)$.
- Its thickness is $\Delta x$.
So the shell has volume about $2\pi x\,f(x)\,\Delta x$. Adding the shells and taking the limit gives the volume of the solid.
The shell method: Suppose the region under the curve
$y=f(x)$, for $a\le x\le b$ (with $a\ge 0$),
is rotated about the $y$-axis. Then
$\displaystyle V=\int_a^b 2\pi x\,f(x)\,dx$.
In words: integrate circumference times height. The radius is $x$, the distance from the axis.
Example 1: Find the volume when the region under
$y=2x^2-x^3$, for $0\le x\le 2$,
is rotated about the $y$-axis.
Solution:
Radius: $x$. Height: $2x^2-x^3$.
$\begin{align*}&V\\&=\int_0^2 2\pi x(2x^2-x^3)\,dx\\&=2\pi\int_0^2(2x^3-x^4)\,dx\\&=2\pi\left[\dfrac{x^4}{2}-\dfrac{x^5}{5}\right]_0^2\\&=2\pi\left(8-\dfrac{32}{5}\right)\\&=\dfrac{16\pi}{5}\end{align*}$
Example 2: The region under $y=\sqrt{x}$, from $x=0$ to $x=1$, is rotated about the $y$-axis. Find the volume in two ways: with shells, and with washers.
Solution:
Shells: The radius is $x$, and the height is $\sqrt{x}$:
$\begin{align*}&V\\&=\int_0^1 2\pi x\cdot x^{1/2}\,dx\\&=2\pi\int_0^1 x^{3/2}\,dx\\&=2\pi\cdot\dfrac{2}{5}\\&=\dfrac{4\pi}{5}\end{align*}$
Washers: Use horizontal slices. At height $y$, the region goes from $x=y^2$ to $x=1$. So the outer radius is $1$, and the inner radius is $y^2$:
$\begin{align*}&V\\&=\int_0^1\pi\left(1^2-(y^2)^2\right)dy\\&=\pi\left[y-\dfrac{y^5}{5}\right]_0^1\\&=\dfrac{4\pi}{5}\end{align*}$
Both methods give the same answer, as they must.
Example 3: The region between $y=x$ and $y=x^2$ is rotated about the $y$-axis. Find the volume.
Solution:
On $[0,1]$, $x\ge x^2$. So a strip at $x$ has height $x-x^2$ (top minus bottom):
$\begin{align*}&V\\&=\int_0^1 2\pi x(x-x^2)\,dx\\&=2\pi\int_0^1(x^2-x^3)\,dx\\&=2\pi\left(\dfrac{1}{3}-\dfrac{1}{4}\right)\\&=\dfrac{\pi}{6}\end{align*}$
Rotating about another vertical line
If the axis is the line $x=c$ instead of the $y$-axis, the radius is the distance from the strip to that line. For example, for strips to the left of $x=c$, the radius is $c-x$.
Example 4: The region between the curve
$y=x-x^2$
and the $x$-axis is rotated about the line
$x=2$.
Find the volume.
Solution:
The curve meets the $x$-axis at $x=0$ and $x=1$. All the strips are to the left of the axis of rotation.
Radius: $2-x$. Height: $x-x^2$.
Multiply out the product inside the integral:
$(2-x)(x-x^2)=2x-3x^2+x^3$
So
$\begin{align*}&V\\&=\int_0^1 2\pi(2-x)(x-x^2)\,dx\\&=2\pi\int_0^1(2x-3x^2+x^3)\,dx\end{align*}$
Find an antiderivative, and evaluate it from $0$ to $1$:
$\begin{align*}&=2\pi\left[x^2-x^3+\dfrac{x^4}{4}\right]_0^1\\&=2\pi\cdot\dfrac{1}{4}\end{align*}$
Simplify:
$\begin{align*}&V=\dfrac{\pi}{2}\end{align*}$
Which method should I use?
- Disks and washers: the slices are perpendicular to the axis of rotation.
- Shells: the strips are parallel to the axis of rotation.
Both give the same volume. Choose the one that is easier: the one where you do not have to solve for $x$ (or $y$), or where you need only one integral.
Summary
- A thin shell has volume about circumference $\times$ height $\times$ thickness.
- About the $y$-axis: $\displaystyle V=\int_a^b 2\pi x\,f(x)\,dx$.
- About the line $x=c$, use the distance to that line as the radius.
- Shells and washers give the same volume. Pick the easier one.