Washer method around other axes

AP Calculus AB· AP Calculus AB CED — Applications of Integration· 14 min read

1. Core Concept: Washer Method for Non-Coordinate Axes★★☆☆☆⏱ 3 min

The washer method calculates the volume of a solid formed by rotating a bounded 2D region around a fixed straight axis. When the axis of rotation is not the x-axis ( $y = 0$ ) or y-axis ( $x = 0$ ), we only need to adjust how we calculate radii — the core logic of the method stays the same. Like with coordinate axes, we find the area of each perpendicular cross-sectional washer, then integrate the area over the length of the solid to get total volume.

📘 Definition

Washer Cross-Section

A perpendicular cross-section of a solid of revolution formed by rotating a region between two curves, with an outer circular edge and an inner circular hole. The area is always given by $e x t A r e a = π (R^{2} - r^{2})$ , where $R$ = outer radius (distance from axis to farthest curve) and $r$ = inner radius (distance from axis to closest curve), both always positive.

2. Rotation around Horizontal Axes ($y=k$)★★☆☆☆⏱ 4 min

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A horizontal axis of rotation has the form $y = k$ , where $k$ is a non-zero constant. Cross-sections perpendicular to a horizontal axis are vertical slices, so we always integrate with respect to $x$ . Radius is the positive distance between the curve and the axis, so we always subtract the smaller coordinate from the larger to get a positive value.

If the entire region is above the axis $y = k$ , bounded above by $y = f (x)$ and below by $y = g (x)$ over $x \in [a, b]$ , the volume formula is:

V = \pi \int_a^b \left[ (f(x) - k)^2 - (g(x) - k)^2 \right] dx

If the entire region is below the axis, the farthest curve from the axis is the lower bound, so the formula adjusts accordingly.

📐 Worked Example

Find the volume of the solid formed by rotating the region bounded by $y = x$ , $y = 0$ , and $x = 4$ around the horizontal axis $y = - 1$ .

1
Sketch the region: it is a right triangle with vertices at $(0, 0)$ , $(4, 0)$ , and $(4, 4)$ . The entire region lies above the axis of rotation $y = - 1$ .
2
Calculate radii for a vertical slice at position $x$ : The farthest curve from $y = - 1$ is the upper boundary $y = x$ , so:
3
$R = x - (-1) = x+1$
4
The closest curve is the lower boundary $y = 0$ , so:
5
$r = 0 - (-1) = 1$
6
Set up the volume integral with interval $x \in [0, 4]$ :
7
$V = \pi \int_0^4 \left[ (x+1)^2 - 1^2 \right] dx$
8
Simplify and evaluate: Expand $(x + 1)^{2} = x^{2} + 2 x + 1$ , so the integrand reduces to $x^{2} + 2 x$ . Integrate using the power rule:
9
$\int_0^4 (x^2 + 2x) dx = \left[ \frac{x^3}{3} + x^2 \right]_0^4 = \frac{64}{3} + 16 = \frac{112}{3}$
10
Multiply by $π$ to get the final volume:
11
$V = \frac{112}{3}\pi \approx 117.29$

Exam tip:

Always draw a quick sketch of the region and axis to avoid misordering subtraction for radii. Even a 1-minute sketch will catch most sign errors.

3. Rotation around Vertical Axes ($x=h$)★★★☆☆⏱ 4 min

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A vertical axis of rotation has the form $x = h$ , where $h$ is a non-zero constant. Cross-sections perpendicular to a vertical axis are horizontal slices, so we always integrate with respect to $y$ . Like horizontal rotation, radius is the positive distance from the axis to the bounding curve.

If the entire region is to the right of the axis, bounded right by $x = f (y)$ and left by $x = g (y)$ over $y \in [c, d]$ , the volume formula is:

V = \pi \int_c^d \left[ (f(y) - h)^2 - (g(y) - h)^2 \right] dy

If the entire region is to the left of the axis, the farthest curve from the axis is the left boundary, so the formula adjusts. The most common mistake here is failing to rewrite curves given as $y = f (x)$ as functions of $y$ before setting up the integral.

📐 Worked Example

Find the volume of the solid formed by rotating the region bounded by $y = x^{2}$ , $y = 0$ , and $x = 2$ around the vertical axis $x = 3$ .

1
Sketch the region: it is the area under the parabola from $x = 0$ to $x = 2$ , spanning $y \in [0, 4]$ . The entire region lies to the left of the axis $x = 3$ .
2
Rewrite the boundary curve as a function of $y$ :
3
$x = \sqrt{y}$
4
Calculate radii for a horizontal slice at position $y$ : The farthest curve from $x = 3$ is the left boundary $x = 0$ , so:
5
$R = 3 - 0 = 3$
6
The closest curve is the right boundary $x = y$ , so:
7
$r = 3 - \sqrt{y}$
8
Set up and evaluate the integral over $y \in [0, 4]$ :
9
$V = \pi \int_0^4 \left[ 3^2 - (3 - \sqrt{y})^2 \right] dy$
10
Expand the integrand: $9 - (9 - 6 y + y) = 6 y^{1/2} - y$ . Integrate:
11
$\int_0^4 (6y^{1/2} - y) dy = \left[ 4y^{3/2} - \frac{y^2}{2} \right]_0^4 = 4(8) - 8 = 24$
12
Multiply by $π$ to get the final volume:
13
$V = 24\pi \approx 75.40$

Exam tip:

Never leave $x$ terms in a $d y$ integral (or $y$ terms in a $d x$ integral). Always solve for the correct variable before setting up the volume expression.

4. Axis of Rotation Between Two Curves★★★★☆⏱ 4 min

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In many AP problems, the axis of rotation lies between the two bounding curves of the region, rather than having the entire region on one side. One curve is above/right of the axis, and the other is below/left of the axis, so we measure each radius from the axis separately. The core rule $V = π \int (R^{2} - r^{2})$ does not change, only the calculation of each radius.

For a horizontal axis $y = k$ , with $f (x)$ above the axis and $g (x)$ below the axis over $[a, b]$ , the volume formula is:

V = \pi \int_a^b \left[ (f(x) - k)^2 - (k - g(x))^2 \right] dx

📐 Worked Example

The region R is bounded by $y = x + 2$ and $y = x^{2}$ , for $x \in [- 1, 2]$ . Find the volume when R is rotated around the horizontal axis $y = 1$ .

1
Confirm the axis position: For all $x \in [- 1, 2]$ , $x^{2} < 1 < x + 2$ , so the axis $y = 1$ lies between the two bounding curves.
2
Calculate radii: The upper curve $y = x + 2$ is above the axis, so outer radius:
3
$R = (x+2) - 1 = x+1$
4
The lower curve $y = x^{2}$ is below the axis, so inner radius:
5
$r = 1 - x^2$
6
Set up the integral:
7
$V = \pi \int_{-1}^2 \left[ (x+1)^2 - (1 - x^2)^2 \right] dx$
8
Simplify and evaluate: Expand the integrand: $(x^{2} + 2 x + 1) - (1 - 2 x^{2} + x^{4}) = - x^{4} + 3 x^{2} + 2 x$ . Integrate:
9
$\int_{-1}^2 (-x^4 + 3x^2 + 2x) dx = \left[ -\frac{x^5}{5} + x^3 + x^2 \right]_{-1}^2 = \frac{27}{5}$
10
Multiply by $π$ to get the final volume:
11
$V = \frac{27}{5}\pi = 5.4\pi \approx 16.96$

5. AP-Style Practice Check★★★☆☆⏱ 3 min

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✓ Quick check

Test your understanding of integral setup for shifted axes:

The region R is bounded by $y = 2 x$ , $y = 0$ , and $x = 3$ . Which of the following gives the volume of the solid formed when R is rotated around the line $y = 6$ ?
- A) $π \int_{0}^{3} [6^{2} - (6 - 2 x)^{2}] d x$
- B) $π \int_{0}^{3} [(2 x - 6)^{2} - 0^{2}] d x$
- C) $π \int_{0}^{6} [6^{2} - (6 - 2 x)^{2}] d y$
- D) $π \int_{0}^{3} [6^{2} - (2 x)^{2}] d x$
Reveal answer
A) $\pi \int_0^3 \left[ 6^2 - (6 - 2x)^2 \right] dx$ —
The axis is horizontal, so we integrate with respect to $x$ over $[0, 3]$ . The entire region is below $y = 6$ , so outer radius $R = 6 - 0 = 6$ , inner radius $r = 6 - 2 x$ , giving the correct expression. Other options: B swaps radii, C uses wrong variable, D miscalculates inner radius.

📐 Worked Example

A machinist creates a solid metal part by rotating a cross-section bounded by $y = \frac{1}{4} x^{2}$ , $y = 0$ , and $x = 4$ (in cm) around the vertical axis $x = 5$ . What is the total volume of the final part, rounded to one decimal place in cm³?

1
Axis of rotation is vertical, so integrate with respect to $y$ . The region spans $y \in [0, 4]$ (when $x = 4$ , $y = 4$ ).
2
Rewrite the boundary as a function of $y$ :
3
$x = 2\sqrt{y}$
4
The entire region is left of $x = 5$ , so outer radius $R = 5 - 0 = 5$ , inner radius $r = 5 - 2 y$ .
5
Set up the volume integral:
6
$V = \pi \int_0^4 \left[ 5^2 - (5 - 2\sqrt{y})^2 \right] dy$
7
Expand the integrand: $25 - (25 - 20 y + 4 y) = 20 y^{1/2} - 4 y$ . Integrate:
8
$\int_0^4 (20y^{1/2} - 4y) dy = \left[ \frac{40}{3}y^{3/2} - 2y^2 \right]_0^4 = \frac{224}{3} \approx 74.667$
9
Multiply by $π$ to get the final volume:
10
$V \approx 74.667 \times 3.1416 \approx 234.6$
11
The total volume of the part is approximately 234.6 cm³.

6. Common Pitfalls

Wrong move:

Subtracting the axis from the curve when the curve is below/left of the axis, leaving a negative sign before squaring.

Why:

Students memorize 'radius = curve minus axis' without remembering distance is always non-negative.

Correct move:

Always calculate radius as $∣ curve coordinate - axis coordinate ∣$ ; squaring removes negatives anyway, so absolute value guarantees a positive radius.

Wrong move:

Integrating with respect to $x$ for rotation around a vertical axis, or $y$ for rotation around a horizontal axis.

Why:

Students default to the variable the curves are given in, rather than matching the variable to the axis orientation.

Correct move:

Before setting up the integral, confirm: if axis is horizontal, integrate with respect to $x$ ; if vertical, integrate with respect to $y$ .

Wrong move:

When the axis is between two curves, write $V = π \int (f (x) - g (x))^{2} d x$ .

Why:

Students incorrectly subtract the curves first, assuming they share the same side of the axis.

Correct move:

Always measure each radius from the axis individually, then square and subtract per the washer area formula.

Wrong move:

Leaving curves as $y = f (x)$ when integrating with respect to $y$ , resulting in mixed variables in the integral.

Why:

Students skip solving for the correct variable to save time when working quickly.

Correct move:

If integrating with respect to $y$ , rewrite all boundary curves as explicit functions of $y$ before calculating radii.

Wrong move:

Swapping outer and inner radii, resulting in a negative integrand.

Why:

Students guess which curve is farther from the axis without checking.

Correct move:

After identifying radii, pick a test point in the interval to confirm $R > r$ before setting up the integral.

7. Quick Reference Cheatsheet

Category	Formula	Notes
General washer volume	$V = π \int_{a}^{b} (R^{2} - r^{2}) d u$	$u$ = variable parallel to axis; $R$ = outer radius, $r$ = inner radius (positive distances)
Radius calculation	$r = curve coordinate - axis coordinate$	Works for any axis position, avoids negative radius errors
Horizontal $y = k$ (same side)	$V = π \int_{a}^{b} [(f (x) - k)^{2} - (g (x) - k)^{2}] d x$	$f (x)$ upper, $g (x)$ lower; flip subtraction if region is all below $y = k$
Horizontal $y = k$ (axis between curves)	$V = π \int_{a}^{b} [(f (x) - k)^{2} - (k - g (x))^{2}] d x$	$f (x)$ above axis, $g (x)$ below axis
Vertical $x = h$ (same side)	$V = π \int_{c}^{d} [(f (y) - h)^{2} - (g (y) - h)^{2}] d y$	$f (y)$ right, $g (y)$ left; flip subtraction if region is all left of $x = h$
Vertical $x = h$ (axis between curves)	$V = π \int_{c}^{d} [(f (y) - h)^{2} - (h - g (y))^{2}] d y$	$f (y)$ right of axis, $g (y)$ left of axis

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Volume around horizontal shifted axis
2022 · FRQ
Multi-part volume question

What's Next

After mastering the washer method around non-coordinate axes, you will next apply this core distance-measurement skill to the shell method for volumes of revolution, and extend it to finding volumes with known cross-sections. Correctly identifying radii relative to a shifted axis is a critical foundation for all further volume problems, as they rely on the same skill of measuring distances between curves and a fixed reference axis. This topic is heavily tested on the AP Calculus AB exam, and the skills you build here transfer to nearly every other applied integration topic.