What if the axis of revolution is not a coordinate axis?

Shift the radius function. For example, if revolving around the vertical line $x = c$, the radius is $r(x) = |x - c|$. Ensure you drop the absolute value based on whether the region lies to the left or right of $c$.

When should I use shells instead of washers?

Use shells when the axis of revolution is parallel to the variable you want to integrate with respect to. For example, if your functions are given as $y = f(x)$ and you are revolving around the y-axis, the shell method lets you integrate with respect to $x$. Washer method would force you to invert the function to $x = g(y)$.

Can the shell height be a difference of two functions?

Yes. If the region is bounded above by $f(x)$ and below by $g(x)$, the shell height is $h(x) = f(x) - g(x)$.

How to find the volume with the shell method

The shell method calculates the volume of a solid of revolution by integrating the surface areas of nested cylindrical shells. It applies when revolving a two-dimensional region about an axis parallel to the axis of integration, which is often preferable when solving the boundary equations for the inverse variable is algebraically difficult.

The setup

The general volume formula for shells is $V = \int_a^b 2\pi r h \, dx$ (for vertical shells) or $V = \int_c^d 2\pi r h \, dy$ (for horizontal shells). The variable $r$ represents the distance from the axis of revolution to the shell, and $h$ represents the height of the shell.

The steps

Identify the axis of revolution and determine the variable of integration. Integrate with respect to $x$ if revolving around a vertical line, and $y$ if revolving around a horizontal line. 2. Determine the limits of integration from the boundaries of the region. 3. Express the shell radius $r$ and the shell height $h$ strictly in terms of the integration variable. 4. Substitute $r$ and $h$ into the integral $V = \int 2\pi r h$ and evaluate.

Checking the result

Verify that the calculated volume is strictly positive. If the equations can be easily inverted, calculate the same volume using the disk or washer method to confirm the result matches.

Common errors

A frequent error is misidentifying the radius $r$ when the axis of revolution is not the y-axis or x-axis. Always verify whether $r = x - c$ or $r = c - x$ ensures a positive distance over the interval. Another common mistake is omitting the $2\pi$ constant outside the integral.

Worked example

Find the volume of the solid generated by revolving the region bounded by $y = 2x - x^2$ and $y = 0$ about the y-axis.

The region bounds are found by setting $2x - x^2 = 0$ , giving $x = 0$ and $x = 2$ . We are revolving around the y-axis (a vertical line), so we use vertical shells and integrate with respect to $x$ . The radius of a shell at position $x$ is $r(x) = x$ . The height of the shell is the curve itself, $h(x) = 2x - x^2$ . The volume integral is $V = \int_0^2 2\pi x (2x - x^2) \, dx$ . Distribute the $x$ to get $V = 2\pi \int_0^2 (2x^2 - x^3) \, dx$ . Integrate term by term: $V = 2\pi [ \frac{2}{3}x^3 - \frac{1}{4}x^4 ]_0^2$ . Evaluate at the upper limit: $V = 2\pi ( \frac{2}{3}(8) - \frac{1}{4}(16) ) = 2\pi ( \frac{16}{3} - 4 )$ . Simplify to get $V = 2\pi ( \frac{16}{3} - \frac{12}{3} ) = 2\pi ( \frac{4}{3} ) = \frac{8\pi}{3}$ .

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References: OpenStax Calculus Volume 1, Chapter 2.3 · Stewart Calculus, Chapter 6.3