AP Calculus AB and BC

Disk vs Washer vs Shell: Choosing the Volume Method

Q: How do I know whether to integrate with respect to x or y?

Look at the axis of revolution. A horizontal axis (the x-axis or a line y = k) gives vertical slices of thickness dx, so you integrate in x. A vertical axis (the y-axis or a line x = k) gives horizontal slices of thickness dy, so you integrate in y and rewrite every boundary as a function of y.

Slice the region perpendicular to the axis of revolution: no gap between the slice and the axis means the disk method; a gap means the washer method. On the AP exam only disk and washer are tested (CED 8.9-8.12); the shell method is a college topic, not on the exam.

Draw the axis, then draw one slice

The fastest way to choose a volume method is to draw two things and read the picture. First draw the axis of revolution. Then draw one representative slice cut perpendicular to that axis. When you spin that perpendicular slice around the axis it sweeps out a flat circular piece, and whether that piece is a solid disk or a ring with a hole punched out is the whole decision. This is what Topics 8.9 through 8.12 are testing.

If the slice runs from the axis straight out to the boundary with nothing in between, revolving it produces a solid disk. If there is a gap between the axis and the region, so the slice starts out away from the axis, revolving it produces a washer: a disk with a smaller disk removed from the center. Disk or washer comes down to one question: is there a gap between the region and the axis?

Decision rule

Slice perpendicular to the axis of revolution. No gap between the region and the axis means the disk method. A gap means the washer method. That single test decides every AP volume-of-revolution problem.

Disk or washer: run the gap test

Three situations cover almost every exam problem, and the gap test sorts them instantly. When one boundary of the region is the axis itself, say the region sits between a curve and the $x$ -axis and you revolve around the $x$ -axis, the slice starts right at the axis. No gap, so it is a disk.

When the region is trapped between two curves that both stay off the axis and you revolve around that axis, every slice starts away from the axis at the nearer curve and ends at the farther curve. That is a gap, so it is a washer, and the two curves supply the inner and outer radii.

The third case is the one students miss: the region touches the axis in its original position, but the problem revolves it around a different line. Moving the axis can open a gap where there was none. The same region that gives a disk around the $x$ -axis gives a washer around the line $y = 2$ , because the slice no longer starts at the axis. Always run the gap test against the axis the problem actually names, which is the whole point of Topics 8.10 and 8.12.

A quick check before you commit: label the nearer and farther boundaries as seen from the axis. If the nearer boundary is the axis line itself, the inner radius is zero and the washer collapses to a disk, which confirms the no-gap case. If the nearer boundary is a curve, that curve supplies the inner radius (measure the distance from the axis to it) and you are in washer territory.

The formulas, and why the radius is always a distance

Both methods integrate the area of the circular slice. A disk of radius $R$ has area $\pi R^2$ . A washer with outer radius $R_{\text{out}}$ and inner radius $R_{\text{in}}$ has area $\pi\left(R_{\text{out}}^2 - R_{\text{in}}^2\right)$ . Multiply by the slice thickness and add up with an integral.

V = \pi \int_a^b \left[R(x)\right]^2 \,dx

V = \pi \int_a^b \left(\left[R_{\text{out}}(x)\right]^2 - \left[R_{\text{in}}(x)\right]^2\right) \,dx

Here is the idea that prevents most setup errors: the radius is a distance from the axis to a boundary, not simply a function value. When you revolve around the $x$ -axis the radius equals the function value, which hides the distinction. The instant the axis moves to a line such as $y = 2$ or $x = 3$ , you must write the radius as the axis position minus the curve, or the curve minus the axis position, whichever comes out positive. For a washer the outer radius reaches the boundary farther from the axis and the inner radius reaches the nearer one.

The slice thickness tells you the variable of integration. A horizontal axis, the $x$ -axis or any line $y = k$ , produces vertical slices of thickness $dx$ , so you integrate in $x$ . A vertical axis, the $y$ -axis or any line $x = k$ , produces horizontal slices of thickness $dy$ , so you integrate in $y$ and every boundary must be rewritten as a function of $y$ . Match the variable to the direction of the axis and the rest of the setup follows. Finding the limits is the same intersection work you did for area between curves in Topics 8.4 through 8.6.

Axis of revolution	Region touches the axis?	Method	Integrate in
Horizontal: $x$ -axis or $y = k$	Yes (no gap)	Disk	$x$ , thickness $dx$
Horizontal: $x$ -axis or $y = k$	No (a gap)	Washer	$x$ , thickness $dx$
Vertical: $y$ -axis or $x = k$	Yes (no gap)	Disk	$y$ , thickness $dy$
Vertical: $y$ -axis or $x = k$	No (a gap)	Washer	$y$ , thickness $dy$

Where the shell method fits (and why the AP exam skips it)

The shell method slices parallel to the axis instead of perpendicular to it. Revolving a parallel slice sweeps out a thin cylindrical shell, like the paper label on a soup can, with volume $2\pi$ times radius times height times thickness. It is a genuine and useful technique, and you will most likely meet it in a college calculus course.

V = 2\pi \int_a^b (\text{radius})(\text{height}) \,dx

Not on the AP exam

The shell method is not part of AP Calculus AB or BC. The CED covers only the disk method (Topics 8.9 and 8.10) and the washer method (Topics 8.11 and 8.12). Every exam volume problem is solvable with disk or washer, so you never need shells on a free-response question. Learn it if a college course or your teacher asks for it, but do not reach for it on the AP exam.

Shells exist because a perpendicular slice sometimes forces you to solve a curve for the awkward variable or to split the region into two separate integrals, while a parallel slice keeps everything in a single clean integral. AP problems are written so that disk or washer always stays workable, which is exactly why the exam can leave shells out without limiting what it can ask.

Mistakes that cost points

Forgetting to square before integrating. The integrand is the radius squared, not the radius. Square first, then antidifferentiate.
Subtracting the radii instead of their squares. A washer is $\pi\left(R_{\text{out}}^2 - R_{\text{in}}^2\right)$ , never $\pi\left(R_{\text{out}} - R_{\text{in}}\right)^2$ . Those two expressions are not equal.
Using the bare function value as the radius after the axis has moved. Around $y = 2$ the radius is $2 - (\text{curve})$ or $(\text{curve}) - 2$ , not the curve by itself.
Mismatching the variable and the axis. Around a vertical line you must integrate in $y$ , with every boundary rewritten as a function of $y$ .
Skipping the intersection step. Find where the boundaries meet first; those points are your limits of integration.

Worked examples

Worked example

Disk method around the x-axis

The region bounded by $y = \sqrt{x}$ , the $x$ -axis, and the line $x = 4$ is revolved around the $x$ -axis. Find the volume.

Draw the axis of revolution (the $x$ -axis) and a representative vertical slice, which is perpendicular to that axis. The slice runs from the $x$ -axis up to the curve $y = \sqrt{x}$ , so it touches the axis with no gap. A perpendicular slice that touches the axis sweeps out a solid disk, so this is the disk method.
The radius of each disk is the distance from the axis to the curve: $R(x) = \sqrt{x}$ . Because the axis is horizontal, the slices have thickness $dx$ and you integrate in $x$ , from $x = 0$ to $x = 4$ .
Set up the disk integral: $V = \pi \int_0^4 \left(\sqrt{x}\right)^2 \,dx$ .
Simplify the integrand: $\left(\sqrt{x}\right)^2 = x$ , so $V = \pi \int_0^4 x \,dx$ .
Antidifferentiate: $\int x \,dx = \frac{x^2}{2}$ , giving $V = \pi \left[\frac{x^2}{2}\right]_0^4$ .
Evaluate: $V = \pi \left(\frac{16}{2} - 0\right) = 8\pi$ .

$V = 8\pi$ cubic units.

Worked example

Washer method around the x-axis

The region enclosed by $y = 2x$ and $y = x^2$ is revolved around the $x$ -axis. Find the volume.

Find where the curves meet: set $2x = x^2$ , so $x^2 - 2x = 0$ and $x(x - 2) = 0$ , giving $x = 0$ and $x = 2$ . The region lives on $[0, 2]$ .
Decide which curve is on top. At $x = 1$ , $2x = 2$ and $x^2 = 1$ , so $y = 2x$ is the upper boundary and $y = x^2$ is the lower boundary.
Draw the $x$ -axis and a vertical slice. The lower boundary $y = x^2$ sits above the axis (except at the endpoint $x = 0$ ), so there is a gap between the region and the axis. A perpendicular slice with a gap sweeps out a washer, so this is the washer method.
Identify the radii as distances from the $x$ -axis. The outer radius reaches the farther curve: $R_{\text{out}} = 2x$ . The inner radius reaches the nearer curve: $R_{\text{in}} = x^2$ .
Set up the washer integral: $V = \pi \int_0^2 \left[(2x)^2 - (x^2)^2\right] \,dx$ .
Expand each square: $(2x)^2 = 4x^2$ and $(x^2)^2 = x^4$ , so $V = \pi \int_0^2 \left(4x^2 - x^4\right) \,dx$ .
Antidifferentiate: $V = \pi \left[\frac{4x^3}{3} - \frac{x^5}{5}\right]_0^2$ .
Evaluate at $x = 2$ : $\frac{4(8)}{3} - \frac{32}{5} = \frac{32}{3} - \frac{32}{5}$ .
Use a common denominator of $15$ : $\frac{32}{3} = \frac{160}{15}$ and $\frac{32}{5} = \frac{96}{15}$ , so the difference is $\frac{160 - 96}{15} = \frac{64}{15}$ .

$V = \frac{64\pi}{15}$ cubic units.

Worked example

Same region, new axis: a disk becomes a washer

Take the same region as Example 1 (bounded by $y = \sqrt{x}$ , the $x$ -axis, and $x = 4$ ), but revolve it around the horizontal line $y = 2$ . Find the volume.

Nothing about the region changed, only the axis moved. Draw the line $y = 2$ and a vertical slice. Since $\sqrt{x} \le 2$ on $[0, 4]$ , the whole region sits below the axis, and the slice no longer touches $y = 2$ except at the single point $x = 4$ . The gap means this is now a washer, not a disk.
Measure each radius as a distance from the axis $y = 2$ . The boundary farther from the axis is the $x$ -axis, where $y = 0$ : $R_{\text{out}} = 2 - 0 = 2$ . The nearer boundary is the curve $y = \sqrt{x}$ : $R_{\text{in}} = 2 - \sqrt{x}$ .
Set up the washer integral (horizontal axis, so thickness $dx$ ): $V = \pi \int_0^4 \left[2^2 - \left(2 - \sqrt{x}\right)^2\right] \,dx$ .
Expand the inner square: $\left(2 - \sqrt{x}\right)^2 = 4 - 4\sqrt{x} + x$ .
Subtract inside the brackets: $4 - \left(4 - 4\sqrt{x} + x\right) = 4\sqrt{x} - x$ , so $V = \pi \int_0^4 \left(4\sqrt{x} - x\right) \,dx$ .
Antidifferentiate. Write $\sqrt{x} = x^{1/2}$ , so $\int 4x^{1/2} \,dx = 4 \cdot \frac{x^{3/2}}{3/2} = \frac{8}{3}x^{3/2}$ and $\int x \,dx = \frac{x^2}{2}$ . Then $V = \pi \left[\frac{8}{3}x^{3/2} - \frac{x^2}{2}\right]_0^4$ .
Evaluate at $x = 4$ : $x^{3/2} = 4^{3/2} = 8$ , so $\frac{8}{3}(8) = \frac{64}{3}$ , and $\frac{x^2}{2} = \frac{16}{2} = 8 = \frac{24}{3}$ .
Combine: $\frac{64}{3} - \frac{24}{3} = \frac{40}{3}$ .

$V = \frac{40\pi}{3}$ cubic units.

Frequently asked questions

Is the shell method on the AP Calculus exam?

No. AP Calculus AB and BC test only the disk method (Topics 8.9 and 8.10) and the washer method (Topics 8.11 and 8.12). The shell method is common in college calculus but is never required on the AP exam, and every exam volume-of-revolution problem can be done with disk or washer.

How do I know whether to integrate with respect to x or y?

Look at the axis of revolution. A horizontal axis (the $x$ -axis or a line $y = k$ ) gives vertical slices of thickness $dx$ , so you integrate in $x$ . A vertical axis (the $y$ -axis or a line $x = k$ ) gives horizontal slices of thickness $dy$ , so you integrate in $y$ and rewrite every boundary as a function of $y$ .

What is the difference between the disk and washer method?

Both revolve a perpendicular slice around the axis. The disk method applies when the slice touches the axis with no gap, giving a solid disk of area $\pi R^2$ . The washer method applies when there is a gap between the region and the axis, giving a ring of area $\pi\left(R_{\text{out}}^2 - R_{\text{in}}^2\right)$ . The gap is the only thing that decides between them.