Volume with Washer Method: Revolving Around Other Axes: Study Guide (Calculus AB 8.12)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

Integrals are written in a compact form: π ∫ (a to b) [R(x)² − r(x)²] dx means π times the definite integral of R(x)² − r(x)² from x = a to x = b. On paper, write a at the bottom of the integral sign and b at the top. R is the outer radius and r the inner radius. Keep the brackets and the dx or dy.

The same washers, a different axis

In Topic 8.11 the axis was the x-axis or the y-axis. A radius was then just a height (a y-value) or an x-value. Now the axis can be any horizontal line y = k or vertical line x = h. Nothing about the slices changes:

• slices are still perpendicular to the axis;
• each slice still spins into a washer with face area π(R² − r²);
• the volume is still π times the integral of R² − r².

What changes is how you measure the radii. A radius is the distance from the axis to an edge of the slice. Distances along a vertical line are differences of y-values; distances along a horizontal line are differences of x-values.

Radius rule. Radius = larger coordinate − smaller coordinate. This is always positive, so you never need to guess whether to add or subtract k.

Axis	Slice	Boundary curve	Radius if axis is on this side
y = k, above the region	vertical, dx	y = f(x)	k − f(x)
y = k, below the region	vertical, dx	y = f(x)	f(x) − k
x = h, right of the region	horizontal, dy	x = g(y)	h − g(y)
x = h, left of the region	horizontal, dy	x = g(y)	g(y) − h

Which boundary is the outer one? The one farther from the axis. If the axis is above the region, the lower curve is farther away, so the lower curve gives R. This surprises many students, and it is a common error in this topic.

The method, step by step

1. Sketch the region and draw the axis as a dashed line.
2. Slice perpendicular to the axis: vertical slices (dx) for a horizontal axis, horizontal slices (dy) for a vertical axis.
3. Find the limits in the slice variable.
4. Write each radius as a distance: larger coordinate minus smaller coordinate.
5. Choose R and r with a test point: R belongs to the boundary farther from the axis.
6. Write and evaluate π ∫ [R² − r²], with the right differential.
7. Check the answer is positive, and compare it with a simpler shape if you can.

A quick test before you integrate. Put your test point into both radius formulas and compare the numbers with your sketch. In Figure 1 at x = 1, your formulas should give R = 4 and r = 1, and on the picture the dashed measurement to the parabola is visibly about four times the one to the line. If a formula gives a negative number, or the larger number belongs to the curve that looks closer to the axis, fix it now. This one check catches most sign and order errors in this topic.

Worked example 1: axis above the region (no calculator)

Question. The region between y = x² and y = 4 (Figure 1) is revolved around the line y = 5. Find the exact volume.

1. Limits. x² = 4 gives x = −2 and x = 2.
2. Slices. The axis is horizontal, so slices are vertical and everything is in x.
3. Distances from y = 5. The axis is above both boundaries, so each distance is 5 minus the curve’s y-value. To the line: 5 − 4 = 1. To the parabola: 5 − x².
4. R and r. Test x = 0: the parabola is 5 units from the axis and the line is 1 unit away. So R(x) = 5 − x² and r(x) = 1.
5. Integral. V = π ∫ (−2 to 2) [(5 − x²)² − 1²] dx
6. Expand. (5 − x²)² − 1 = 25 − 10x² + x⁴ − 1 = x⁴ − 10x² + 24.
7. Evaluate. The integrand is even: ∫ (0 to 2) (x⁴ − 10x² + 24) dx = 32/5 − 80/3 + 48 = (96 − 400 + 720)/15 = 416/15. Double it: V = 832π/15.

Answer. V = 832π/15 cubic units (about 174.254).

What goes wrong. Taking R from the line and r from the parabola gives −832π/15, a negative volume. Ignoring the axis and revolving around the x-axis gives 256π/5, a different solid.

The same region, axis below

Revolve the same region around y = −1 instead. Now the axis is below the region, so each distance is the curve’s y-value minus (−1), which is the y-value plus 1. The line y = 4 is now farther away:

• R(x) = 4 − (−1) = 5
• r(x) = x² − (−1) = x² + 1

V = π ∫ (−2 to 2) [25 − (x² + 1)²] dx = 1088π/15 (about 227.870).

The region is the same, but the volume is larger because the region now sits farther from the axis on average. Also, the line that gave the inner radius before now gives the outer radius. Subtracting 1 instead of adding it (writing x² − 1 and 3) gives 448π/15, which is wrong: the distance from y = −1 up to y = 4 is 5, not 3.

Worked example 2: a vertical axis (no calculator)

Question. The region bounded by x = y² and x = 4 is revolved around the line x = 6. Find the volume.

1. Limits. The axis is vertical, so slices are horizontal and everything is in y. y² = 4 gives y = −2 and y = 2.
2. Distances from x = 6. The axis is to the right of the region, so each distance is 6 minus the curve’s x-value. To the parabola x = y²: 6 − y². To the line x = 4: 6 − 4 = 2.
3. R and r. Test y = 0: the parabola is at x = 0, which is 6 units from the axis; the line is 2 units away. So R(y) = 6 − y² and r(y) = 2.
4. Integral. V = π ∫ (−2 to 2) [(6 − y²)² − 2²] dy
5. Expand. (6 − y²)² − 4 = y⁴ − 12y² + 32.
6. Evaluate. ∫ (−2 to 2) 32 dy = 128. ∫ (−2 to 2) 12y² dy = 64. ∫ (−2 to 2) y⁴ dy = 64/5. So the integral is 128 − 64 + 64/5 = 384/5.

Answer. V = 384π/5 cubic units (about 241.274).

Check. The solid fits inside a thick tube with outer radius 6, inner radius 2 and height 4, whose volume is π(36 − 4)(4) = 128π ≈ 402.1. The answer is smaller, as it should be, because the parabola is closer to the axis than x = 0 for every y except y = 0. Revolving around the y-axis by mistake (R = 4, r = y²) gives 256π/5.

Worked example 3: intersections from a calculator

Question. The region enclosed by y = 4 − x² and y = 2ˣ is revolved around the line y = −1. Find the volume. (Calculator allowed.)

1. Intersections. Solve 4 − x² = 2ˣ on the calculator: x = A ≈ −1.933442 and x = B ≈ 1.264166. Store A and B.
2. Distances from y = −1. The axis is below the region, so add 1 to each y-value: (4 − x²) + 1 = 5 − x² and 2ˣ + 1.
3. R and r. Test x = 0: 5 − 0 = 5 and 2⁰ + 1 = 2. So R(x) = 5 − x² and r(x) = 2ˣ + 1.
4. Integral. V = π ∫ (A to B) [(5 − x²)² − (2ˣ + 1)²] dx
5. Evaluate with the stored limits: V ≈ 130.938 cubic units. On paper: “π ∫ (−1.933 to 1.264) [(5 − x²)² − (2ˣ + 1)²] dx = 130.938”.

Sense check. The same region revolved around the x-axis gives π ∫ (A to B) [(4 − x²)² − (2ˣ)²] dx ≈ 89.341. Moving the axis 1 unit further from the region should make every washer larger, and 130.938 is indeed larger. In fact (R + 1)² − (r + 1)² = R² − r² + 2(R − r), so each face grows by an amount that depends on the slice. That is why you must square the shifted radii rather than adjust the final answer. Subtracting 1 instead of adding it would measure from the line y = 1, which even cuts through the region, and gives about 47.744; rounding the limits to −1.9 and 1.3 first gives 130.863. Both are wrong.

Common misconceptions

• “The top curve is always R.” Only when the axis is below the region. R comes from the boundary farther from the axis.
• Adding k when you should subtract (or the reverse). Use larger coordinate minus smaller coordinate. For the axis y = −1, that means adding 1.
• Ignoring the axis. Using the plain curve values, as if the axis were y = 0 or x = 0, gives a different solid.
• Shifting the answer instead of the radii. Changing the axis changes each radius before squaring; you cannot adjust the final volume by a simple amount.
• Wrong variable for a vertical axis. Around x = h, slices are horizontal: write the curves as x in terms of y and use y-limits.
• (R − r)² instead of R² − r². The face of a washer is a big circle minus a small circle.
• Rounding early on a calculator question. Store the intersection values and round only the final volume.

Where this leads

Topic 8.12 completes the Calculus AB course content: you can now find volumes with known cross sections, discs and washers around any horizontal or vertical axis. For AB students, the next step is mixed revision across Unit 8 and the whole course, including recognising which method a question needs. Calculus BC students continue with arc length (Topic 8.13) and then Units 9 and 10. Look back at Volume with Washer Method: Revolving Around the x- or y-Axis if the radii here felt unfamiliar. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.