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How do you find the volume of a solid of revolution with a hole about a line other than a coordinate axis using the washer method?

Topic 8.12 Volume with Washer Method: Revolving Around Other Axes: find the volume of a solid of revolution with a hole about a line other than a coordinate axis using the washer method.

A focused answer to AP Calculus AB Topic 8.12, finding volumes of solids of revolution about lines other than the coordinate axes with the washer method by shifting both radii, with worked examples.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
Combining the shift and the hole
A worked washer about a shifted line
Why the outer curve can switch
The full template for volumes of revolution
A sketch-first workflow

What this topic is asking

The College Board (Topic 8.12) is the most general volume-of-revolution case: a region with a hole revolved about a line other than a coordinate axis. You combine the washer method (outer squared minus inner squared) with the shifted-axis radius adjustment.

Combining the shift and the hole

A worked washer about a shifted line

Revolving a gap region about a horizontal line below it

The region between $y = \sqrt{x}$ (upper) and $y = \frac{x}{2}$ (lower) from $x = 0$ to $x = 4$ is revolved about the line $y = -1$ . Find the volume.

step 1 Determine which curve is farther from the axis

The axis $y = -1$ is below the region, so the upper curve $y = \sqrt{x}$ is farther (outer) and the lower curve $y = \frac{x}{2}$ is nearer (inner).

step 2 Write the shifted radii

$R = \sqrt{x} - (-1) = \sqrt{x} + 1$ ; $r = \frac{x}{2} - (-1) = \frac{x}{2} + 1$ .

step 3 Set up the washer integral

V = \pi\int_0^4 \Big[(\sqrt{x} + 1)^2 - \left(\tfrac{x}{2} + 1\right)^2\Big]\,dx.

step 4 Expand and evaluate

$(\sqrt{x}+1)^2 = x + 2\sqrt{x} + 1$ ; $\left(\frac{x}{2}+1\right)^2 = \frac{x^2}{4} + x + 1$ . The difference is $2\sqrt{x} - \frac{x^2}{4}$ . So $V = \pi\int_0^4\left(2\sqrt{x} - \frac{x^2}{4}\right)dx = \pi\left[\frac{4}{3}x^{3/2} - \frac{x^3}{12}\right]_0^4 = \pi\left(\frac{32}{3} - \frac{64}{12}\right) = \pi\cdot\frac{32 - 16}{3} = \frac{16\pi}{3}.$

Why the outer curve can switch

The trickiest part is that which curve is the outer radius depends on where the axis is. If the axis lies below the region, the curve higher up is farther from the axis and becomes the outer radius. If the axis lies above the region, the lower curve is farther and becomes the outer radius. So revolving the same region about $y = -1$ versus $y = 2$ swaps the roles of the two curves. Always orient yourself by the axis: measure the distance from the line to each curve and let the larger distance be $R$ . Drawing the axis and a representative washer makes this unambiguous.

The full template for volumes of revolution

This topic completes a four-case family: disc about a coordinate axis (8.9), disc about another line (8.10), washer about a coordinate axis (8.11), and washer about another line (8.12). The unifying template is: pick the integration variable from the axis orientation (perpendicular slices), write each radius as the distance from the axis of revolution to the relevant curve (adding the shift for a non-coordinate line), and integrate $\pi R^2$ for a disc or $\pi(R^2 - r^2)$ for a washer. Every volume-of-revolution question on the AB exam is one of these four cases, so mastering the radius bookkeeping, distance from the axis, with the gap and shift, handles them all.

A sketch-first workflow

For this most general case a deliberate, sketch-first workflow pays off. Draw the region, draw the axis of revolution as a dashed line, and draw one representative washer perpendicular to the axis. Mark the outer radius as the segment from the axis to the farther curve and the inner radius to the nearer curve, reading each length as a difference of coordinates so both come out positive. Then write $\pi\int (R^2 - r^2)$ with the shifted radii and integrate. The sketch settles the three things that most often go wrong, the integration variable, which curve is farther from the axis, and the sign of each shifted radius, before any algebra begins. With those fixed on paper, the integral is routine and the answer comes out in cubic units.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2021 (style)1 marksSection I (multiple choice). The region between

y = x

and

y = x^2

[0, 1]

is revolved about the line

y = 2

. The outer radius at

x

is the distance from

y = 2

to (A)

y = x

(B)

y = x^2

x

-axis (D)

y = 1

Show worked answer →

The correct answer is (B), $y = x^2$ .

Revolving about $y = 2$ (above the region), the farther curve from the axis is the lower one, $y = x^2$ . So the outer radius is $R = 2 - x^2$ and the inner is $r = 2 - x$ .

AP 2024 (style)4 marksSection II (free response). The region bounded by

y = x

and

y = x^2

[0, 1]

is revolved about the line

y = 2

. (a) Write the outer and inner radii. (b) Set up and evaluate the volume integral.

Show worked answer →

A 4-point washer-about-shifted-line volume.

(a) (2 points) Distances from $y = 2$ : outer (to the lower curve $y = x^2$ ) $R = 2 - x^2$ ; inner (to the upper curve $y = x$ ) $r = 2 - x$ .
(2 points) $V = \pi\int_0^1 \big((2 - x^2)^2 - (2 - x)^2\big)\,dx = \pi\int_0^1 \big((4 - 4x^2 + x^4) - (4 - 4x + x^2)\big)\,dx = \pi\int_0^1 (x^4 - 5x^2 + 4x)\,dx = \pi\left[\frac{x^5}{5} - \frac{5x^3}{3} + 2x^2\right]_0^1 = \pi\left(\frac{1}{5} - \frac{5}{3} + 2\right) = \frac{8\pi}{15}$ .

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Sources & how we know this

AP Calculus AB and BC Course and Exam Description — College Board (2020)