The washer method around other axes finds the volume of a solid of revolution when there's a gap between the region and a horizontal or vertical line that isn't the x- or y-axis. You build each radius as a distance from the rotation line, square the outer and inner radii, subtract, and integrate. For AP Calculus, label the outer radius, inner radius, bounds, and axis of rotation before writing the integral.

Why This Matters for the AP Calculus Exam

Volume of revolution problems show up in both multiple-choice and free-response sections of the AP Calculus exam. This is the most flexible version of the washer method because the axis of rotation can be any horizontal or vertical line, so you have to reason carefully about distances instead of plugging into a memorized special case.

On free-response questions, you usually earn credit for a correct integral setup with proper notation before any answer is calculated, so writing the radii and bounds correctly matters even if you use a calculator to evaluate. A clean expression like

V = \pi \int_{a}^{b}\left[(R(x))^2 - (r(x))^2\right]\, dx

shows the reader exactly how you defined your outer and inner radii.

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Key Takeaways

The washer volume is $V = \pi \int_a^b \left[(R)^2 - (r)^2\right]\, dx$ , where $R$ is the outer radius and $r$ is the inner radius.
Each radius is the distance from the axis of rotation to a curve, not just the function value.
Rotating around $y = k$ uses radii like $|f(x) - k|$ and integrates with respect to $x$ . Rotating around $x = h$ uses radii like $|f(y) - h|$ and integrates with respect to $y$ .
Find your bounds by setting the two curves equal and solving for the intersection points.
The outer radius belongs to the curve farther from the axis. The inner radius belongs to the curve closer to the axis.
If the axis touches the region on part of the interval, the inner radius can become 0 and the washer turns into a disc there.

Revolving Around Other Axes

Revolving around another axis just means you pick a horizontal line $y = k$ or a vertical line $x = h$ to rotate around instead of the x- or y-axis.

The key idea is that a radius is always a distance from the axis to the curve. When you rotate around $y = k$ , the distance from the axis to a curve $y = f(x)$ is $|f(x) - k|$ . When you rotate around $x = h$ , the distance from the axis to a curve $x = f(y)$ is $|f(y) - h|$ .

For rotation around a horizontal line $y = k$ , integrate with respect to $x$ :

V = \pi \int_{a}^{b}\left[(f(x) - k)^2 - (g(x) - k)^2\right]\, dx

For rotation around a vertical line $x = h$ , solve for $x$ in terms of $y$ , then integrate with respect to $y$ :

V = \pi \int_{c}^{d}\left[(f(y) - h)^2 - (g(y) - h)^2\right]\, dy

Be careful with subtraction when the line value is negative. If $k = -2$ , then $f(x) - k$ becomes $f(x) + 2$ . Sign mistakes here are common.

The Washer Method

The washer method removes a small circle from a bigger circle, leaving a ring (washer) shape. The outer radius $R$ comes from the curve farther from the axis, and the inner radius $r$ comes from the curve closer to the axis.

The area of one washer cross section is $\pi R^2 - \pi r^2$ . Integrating that area across the interval gives the volume:

V = \pi \int_{a}^{b}\left[(R)^2 - (r)^2\right]\, dx

Always confirm which curve is the outer radius and which is the inner radius before you set up the integral. The bigger distance from the axis goes first.

Worked Example: Shifted Horizontal Axis

Find the volume of the solid formed by revolving the region between $f(x) = \sqrt{x}$ and $g(x) = x$ about the line $y = -1$ .

Step 1: Find the bounds. Set the curves equal:

\sqrt{x} = x

Squaring gives $x = x^2$ , so $x^2 - x = 0$ , which factors to $x(x - 1) = 0$ . The intersections are $x = 0$ and $x = 1$ . These are your bounds.

Step 2: Identify outer and inner radii. On $[0, 1]$ , the curve $\sqrt{x}$ sits above $x$ , so $\sqrt{x}$ is farther from the line $y = -1$ . Each radius is the distance from $y = -1$ :

Outer radius: $R = \sqrt{x} - (-1) = \sqrt{x} + 1$
Inner radius: $r = x - (-1) = x + 1$

Step 3: Set up the integral.

V = \pi \int_{0}^{1}\left[(\sqrt{x} + 1)^2 - (x + 1)^2\right]\, dx

Step 4: Expand the integrand.

(\sqrt{x} + 1)^2 = x + 2\sqrt{x} + 1

(x + 1)^2 = x^2 + 2x + 1

Subtracting:

(x + 2\sqrt{x} + 1) - (x^2 + 2x + 1) = 2\sqrt{x} - x^2 - x

So:

V = \pi \int_{0}^{1}\left(2\sqrt{x} - x^2 - x\right)\, dx

Step 5: Integrate.

V = \pi\left[\frac{4}{3}x^{3/2} - \frac{1}{3}x^3 - \frac{1}{2}x^2\right]_{0}^{1}

At $x = 1$ : $\frac{4}{3} - \frac{1}{3} - \frac{1}{2} = 1 - \frac{1}{2} = \frac{1}{2}$ . At $x = 0$ everything is 0.

V = \frac{\pi}{2}

How to Use This on the AP Calculus Exam

Free Response

Write the integral with $\pi$ outside and squared radii inside before evaluating. A clear setup like $V = \pi \int_a^b\left[(R)^2 - (r)^2\right] dx$ shows your reasoning.
Define each radius as a distance from the axis. Show the shift, for example $f(x) - (-1) = f(x) + 1$ .
If bounds come from solving $f(x) = g(x)$ , state the intersection values you use.

Problem Solving

Graph or sketch both curves and the axis of rotation first. This makes it obvious which curve is the outer radius.
Choose your variable by orientation: integrate with respect to $x$ for a horizontal axis, and with respect to $y$ for a vertical axis. For a vertical axis, you often need to rewrite functions as $x$ in terms of $y$ .
Check for places where the axis intersects the region. There, the inner radius is 0 and you may need more than one integral.

Common Trap

When the rotation line is negative, like $y = -2$ , the distance becomes $f(x) - (-2) = f(x) + 2$ . Forgetting the sign flip is one of the most frequent errors on these problems.

Common Misconceptions

Using function values instead of distances. The radius is the distance from the axis to the curve, not just $f(x)$ . When the axis is shifted, you must subtract the line value.
Squaring after subtracting incorrectly. You square each radius separately: $(R)^2 - (r)^2$ , which is not the same as $(R - r)^2$ . Never combine the radii before squaring.
Mixing up outer and inner radii. The outer radius is the curve farther from the axis, not always the "top" function. When the axis is below the region, the higher curve is usually outer, but always check against the axis location.
Forgetting to switch variables for a vertical axis. Rotating around $x = h$ means integrating with respect to $y$ , so you need each curve written as a function of $y$ .
Treating a shifted-axis problem like a disc problem. If there's a gap between the region and the axis, you have an inner radius and need the washer method, not the disc method.

When you use a calculator and approximate intersection points as your bounds, remember that your final decimal answer depends on those approximations. Carry enough digits to keep the volume accurate.

Vocabulary

The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.

Term	Definition
definite integral	The integral of a function over a specific interval [a, b], representing the net signed area between the curve and the x-axis.
ring-shaped cross sections	Annular (donut-shaped) slices of a solid of revolution formed when the region being rotated has a gap between the axis of rotation and the outer boundary.
solids of revolution	Three-dimensional solids formed by rotating a two-dimensional region around an axis.
washer method	A technique for finding the volume of a solid of revolution by integrating the areas of ring-shaped (washer-shaped) cross sections perpendicular to the axis of rotation.

Frequently Asked Questions

What is the washer method around another axis?

The washer method finds volume when a region is revolved around a horizontal or vertical line and the cross sections are rings. Around shifted axes, each radius must be measured as a distance from the rotation line.

What formula do I use for washer method volume?

Use V = pi integral of (R^2 - r^2) d(variable), where R is the outer radius and r is the inner radius. The radii are distances from the axis of rotation to the curves.

How do I set up washers around y = k?

For a horizontal axis y = k, use vertical slices and integrate with respect to x. Radii look like distances from the curve to the line, such as |f(x) - k| and |g(x) - k|.

How do I set up washers around x = h?

For a vertical axis x = h, use horizontal slices and integrate with respect to y. Rewrite the boundaries as x-functions of y, then measure each radius as a distance from x = h.

How do I know which radius is outer?

The outer radius is the curve farther from the axis of rotation, not always the top or rightmost curve. Sketch the region and the axis before deciding which radius goes first.

What mistakes should I avoid with shifted-axis washer problems?

Avoid using function values as radii without subtracting the axis value, forgetting sign changes like f(x) - (-2), mixing x and y variables, or squaring (R - r) instead of R^2 - r^2.