Calculus AB · Applications of Integration · 14 min read · Updated 2026-05-10

Washer method around other axes — AP Calculus AB

AP Calculus AB · 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.

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

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 $y = -1$.

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

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

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 $x=3$.

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

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

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

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 $V = \pi \int (R^2 - r^2)$ 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 $y=1$.

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.
Calculate radii: The upper curve $y = x+2$ is above the axis, so outer radius:
$R = (x+2) - 1 = x+1$
The lower curve $y=x^2$ is below the axis, so inner radius:
$r = 1 - x^2$
Set up the integral:
$V = \pi \int_{-1}^2 \left[ (x+1)^2 - (1 - x^2)^2 \right] dx$
Simplify and evaluate: Expand the integrand: $(x^2 + 2x +1) - (1 - 2x^2 + x^4) = -x^4 + 3x^2 + 2x$. Integrate:
$\int_{-1}^2 (-x^4 + 3x^2 + 2x) dx = \left[ -\frac{x^5}{5} + x^3 + x^2 \right]_{-1}^2 = \frac{27}{5}$
Multiply by $\pi$ to get the final volume:
$V = \frac{27}{5}\pi = 5.4\pi \approx 16.96$

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

📐 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?

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

Common Pitfalls

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

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

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

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

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

Quick Reference Cheatsheet

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Washer method around other axes — AP Calculus AB

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

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

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

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

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

Common Pitfalls

Quick Reference Cheatsheet

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