Washer method around other axes — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Washer method for calculating volumes of solid revolution around horizontal non-x-axes and vertical non-y-axes, how to correctly identify outer and inner radii, set up and evaluate definite integrals for volume, and adjust for shifted axes of rotation.

You should already know: 1. Disk/washer method for rotation around x and y coordinate axes. 2. Setting up definite integrals for area between two curves. 3. Evaluating definite integrals via the Fundamental Theorem of Calculus.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Washer method around other axes?

The washer method is a technique for finding the volume of a solid formed by rotating a bounded 2D region around a fixed straight axis. When the axis of rotation is not one of the two main coordinate axes (the x-axis $y=0$ or y-axis $x=0$), we need to adjust how we calculate the radius of each cross-sectional washer, hence the specific focus on "washer method around other axes". This topic is explicitly tested in AP Calculus AB, making up roughly 10-15% of the Unit 8 (Applications of Integration) exam weight, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often as a core step in a multi-part applications question. Unlike the disk method, the washer method is used when rotating a region between two curves, resulting in a cross-section with a hole (a "washer") rather than a solid disk. The core idea is identical to rotation around coordinate axes: find the area of each perpendicular cross-sectional washer, then integrate that area over the entire length of the solid to get total volume. The only critical difference from basic washer method is that radii are calculated relative to a shifted axis, not relative to a zero coordinate.

2. Rotation around horizontal axes parallel to the x-axis

A horizontal axis of rotation has the form $y=k$, where $k$ is a non-zero constant (if $k=0$, it reduces to the base case of rotation around the x-axis). For washer method with a horizontal axis, we take cross-sections perpendicular to the axis, which are vertical slices, so we always integrate with respect to $x$. The area of a single washer is $\pi (R^2-r^2)$, where $R$ (outer radius) is the distance from the axis to the farthest bounding curve, and $r$ (inner radius) is the distance from the axis to the closest bounding curve. Since distance is always non-negative, we always subtract the smaller coordinate from the larger coordinate to get a positive radius. 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 simplifies to: $$V=\pi {\int }_a^b\left[(f(x)-k{)}^2-(g(x)-k{)}^2\right]dx$$ If the entire region is below the axis, the formula becomes $V=\pi {\int }_a^b\left[(k-g(x){)}^2-(k-f(x){)}^2\right]dx$, since the lower bound is now farther from the axis.

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 $R=x-(-1)=x+1$. The closest curve is the lower boundary $y=0$, so $r=0-(-1)=1$.
3. Set up the volume integral with interval $x\in [0,4]$: $$V=\pi {\int }_0^4\left[(x+1{)}^2-1^2\right]dx$$
4. Simplify and evaluate: Expand $(x+1{)}^2=x^2+2x+1$, so the integrand reduces to $x^2+2x$. Integrate: $${\int }_0^4(x^2+2x)dx={\left[\frac{x^3}{3}+x^2\right]}_0^4=\frac{64}{3}+16=\frac{112}{3}$$ Multiply by $\pi$ to get $V=\frac{112}{3}\pi \approx 117.29$.

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 parallel to the y-axis

A vertical axis of rotation has the form $x=h$, where $h$ is a non-zero constant (if $h=0$, it reduces to the base case of rotation around the y-axis). For washer method with a vertical axis, cross-sections perpendicular to the axis are horizontal slices, so we always integrate with respect to $y$. As with 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 $x=h$, bounded on the right by $x=f(y)$ and on the 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 becomes $V=\pi {\int }_c^d\left[(h-g(y){)}^2-(h-f(y){)}^2\right]dy$. 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 $y$: $x=\sqrt{y}$.
3. 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 $r=3-\sqrt{y}$.
4. 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 $V=24\pi \approx 75.40$.

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. Adjusting for axis of rotation between two curves

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 of the axis. This means one curve is above/right of the axis, and the other is below/left of the axis, so radii are measured from the axis to each curve separately, rather than both being on the same side. For a horizontal axis $y=k$, with $f(x)$ above the axis and $g(x)$ below the axis over $[a,b]$, the outer radius is still the distance from the axis to the farthest curve, and the inner radius is the distance to the closest curve, so the formula becomes: $$V=\pi {\int }_a^b\left[(f(x)-k{)}^2-(k-g(x){)}^2\right]dx$$ The key rule here does not change: volume is always $\pi$ times the integral of (outer radius squared minus inner radius squared). The only change from the same-side case is how we calculate each radius relative to the axis.

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 $R=(x+2)-1=x+1$. The lower curve $y=x^2$ is below the axis, so inner radius $r=1-x^2$.
3. Set up the integral: $$V=\pi {\int }_{-1}^2\left[(x+1{)}^2-(1-x^2{)}^2\right]dx$$
4. Simplify and evaluate: Expand the integrand: $(x^2+2x+1)-(1-2x^2+x^4)=-x^4+3x^2+2x$. Integrate: $${\int }_{-1}^2(-x^4+3x^2+2x)dx={\left[-\frac{x^5}{5}+x^3+x^2\right]}_{-1}^2=\frac{28}{5}-\frac{1}{5}=\frac{27}{5}$$ Multiply by $\pi$ to get $V=\frac{27}{5}\pi =5.4\pi \approx 16.96$.

Exam tip: Do not simplify $(f(x)-k{)}^2-(k-g(x){)}^2$ to $(f(x)-g(x){)}^2$ — these are not equivalent, and this common mistake will always give the wrong volume.

5. Common Pitfalls (and how to avoid them)

• 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 absolute. Correct move: Always calculate radius as $\left|\text{curve coordinate}-\text{axis coordinate}\right|$; 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, ask "Is the axis horizontal or vertical?" If horizontal, integrate with respect to $x$; if vertical, integrate with respect to $y$.
• Wrong move: When the axis is between two curves, write $V=\pi \int (f(x)-g(x){)}^{2}dx$. Why: Students are used to same-side rotation where both distances share the axis as a reference, so they incorrectly subtract the curves first. Correct move: Always measure each radius from the axis individually, then square and subtract.
• Wrong move: Leaving curves as $y=f(x)$ when integrating with respect to $y$, resulting in mixed variables in the integral. Why: It is tempting to skip solving for $x$ when working quickly. Correct move: If integrating with respect to $y$, rewrite all boundary curves as 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 value in the interval to confirm $R>r$ before integrating.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

The region R is bounded by $y=2x$, $y=0$, and $x=3$. Which of the following gives the volume of the solid formed when R is rotated around the line $y=6$? A) $\pi {\int }_0^3\left[6^2-(6-2x{)}^2\right]dx$ B) $\pi {\int }_0^3\left[(2x-6{)}^2-0^2\right]dx$ C) $\pi {\int }_0^6\left[6^2-(6-2x{)}^2\right]dy$ D) $\pi {\int }_0^3\left[6^2-(2x{)}^2\right]dx$

Worked Solution: The axis of rotation is horizontal, so we integrate with respect to $x$ over the interval $x\in [0,3]$, eliminating option C (wrong variable and wrong interval). The entire region R is below the axis $y=6$, with lower bound $y=0$ and upper bound $y=2x$. The farthest curve from the axis is the lower bound $y=0$, so outer radius $R=6-0=6$. The closest curve is the upper bound $y=2x$, so inner radius $r=6-2x$. Substituting into the washer volume formula gives the expression in option A. Options B, C, and D all contain common mistakes: B swaps radii, C uses the wrong variable, D incorrectly calculates the inner radius as $2x$ instead of $6-2x$. Correct answer: A.

Question 2 (Free Response)

Let R be the region bounded by $y=e^x$, $y=1$, and $x=2$. (a) Set up, but do not evaluate, the integral for the volume of the solid formed when R is rotated around the horizontal axis $y=-2$. (b) Set up, but do not evaluate, the integral for the volume of the solid formed when R is rotated around the vertical axis $x=3$. (c) Explain why the washer method is required here instead of the disk method.

Worked Solution: (a) The axis is horizontal, so we integrate with respect to $x$. The region spans $x\in [0,2]$ (since $e^x=1$ at $x=0$), the entire region is above $y=-2$. Outer radius $=e^x-(-2)=e^x+2$, inner radius $=1-(-2)=3$. The integral is: $$V=\pi {\int }_0^2\left[(e^x+2{)}^2-3^2\right]dx$$

(b) The axis is vertical, so we integrate with respect to $y$. The region spans $y\in [1,e^2]$, rewrite $y=e^x$ as $x=\ln y$. The entire region is left of $x=3$, outer radius $R=3-0=3$, inner radius $r=3-\ln y$. The integral is: $$V=\pi {\int }_1^{e^2}\left[3^2-(3-\ln y{)}^2\right]dy$$

(c) The disk method only calculates volume for solids of revolution with no hole in the cross-section, formed by rotating a region bounded by one curve around the axis. Here, R is bounded by two curves, so every cross-section perpendicular to the axis has a hole, requiring the washer method's $R^2-r^2$ term to account for the hole.

Question 3 (Application / Real-World Style)

A machinist is creating a solid metal part by lathing a piece of aluminum around a lathe axis. The cross-section of the unlathed part is bounded by $y=\frac{1}{4}{x}^2$, $y=0$, and $x=4$ (in cm). The lathing process rotates this cross-section around the vertical axis $x=5$ to form the final solid part. What is the total volume of the final aluminum part? Give your answer in cm³, rounded to one decimal place.

Worked Solution: The axis of rotation is vertical, so we integrate with respect to $y$ over $y\in [0,4]$ (when $x=4$, $y=\frac{1}{4}(4^2)=4$). Rewrite the boundary as $x=2\sqrt{y}$. The entire region is left of $x=5$, so outer radius $R=5-0=5$, inner radius $r=5-2\sqrt{y}$. The volume integral is: $$V=\pi {\int }_0^4\left[5^2-(5-2\sqrt{y}{)}^2\right]dy$$ Expand the integrand: $25-(25-20\sqrt{y}+4y)=20y^{1/2}-4y$. Integrate: $${\int }_0^4(20y^{1/2}-4y)dy={\left[\frac{40}{3}{y}^{3/2}-2y^2\right]}_0^4=\frac{320}{3}-32=\frac{224}{3}\approx 74.667$$ Multiply by $\pi$: $V\approx 74.667\times 3.1416\approx 234.6$ cm³. The total volume of the final aluminum part is approximately 234.6 cubic centimeters, which the machinist can use to calculate the part's total weight for material ordering.

7. Quick Reference Cheatsheet

8. What's Next

After mastering the washer method around non-coordinate axes, you will next apply this core technique to the shell method for volumes of revolution, and also extend it to finding volumes with known cross-sections. Without correctly identifying radii relative to a shifted axis, both the shell method and cross-section volume problems will be impossible to set up correctly, as they rely on the same core skill of measuring distances between curves and a fixed axis. This topic is a critical foundation for any further calculus study of volumes and solids in 3D space, and it is heavily tested on the AP Calculus AB exam as part of the Applications of Integration unit. The skills you practice here (setting up integrals for geometric quantities, adjusting for shifted reference lines) transfer to nearly every other applied integration topic in AP Calculus.

• Area between two curves
• Disk and washer method around coordinate axes
• Shell method for volumes of revolution
• Volumes with known cross-sections