Washer method around other axes — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Rotating bounded regions around horizontal non-x axes and vertical non-y axes, identifying outer and inner radii for the washer method, switching integration variables, setting up and evaluating volume integrals for solid volumes of revolution.

You should already know: Disk/washer method for rotation around the x-axis and y-axis, how to find intersection points of two curves, how to compute definite integrals of common functions.

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 BC 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 around other axes is an extension of the basic washer method for calculating the volume of a solid of revolution, where the axis of rotation is a horizontal or vertical line that is not one of the coordinate axes ($x=0$ or $y=0$). This topic is explicitly listed in the AP Calculus CED, accounting for approximately 2-4% of the total exam score, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections on every AP Calculus BC exam. Synonyms for this technique include "washer method for axes parallel to coordinate axes" or "volume of revolution around arbitrary horizontal/vertical lines."

The core logic matches the basic washer method: when you rotate a region between two curves around an axis, each cross-section perpendicular to the axis forms a washer (a disk with a concentric hole), with area $\pi (R^2-r^2)$ where $R$ is the outer radius and $r$ the inner radius. The total volume is the integral of this cross-sectional area along the axis. The key difference from rotation around coordinate axes is that radii are calculated as the distance between the curve and the shifted axis, not just the function value itself.

2. Rotation around a horizontal axis $y=k$, $k\ne 0$

When rotating around a horizontal line $y=k$, the axis is parallel to the x-axis, so we integrate with respect to $x$ for the washer method. Radius is defined as the perpendicular distance between the curve and the axis of rotation, which is always non-negative. For a region bounded above by $y=f(x)$ and below by $y=g(x)$ on the interval $x\in [a,b]$, if the entire region lies above the axis $y=k$, the outer radius (distance from the upper curve to the axis) is $R=f(x)-k$, and the inner radius (distance from the lower curve to the axis) is $r=g(x)-k$. If the entire region lies below the axis, reverse the subtraction to get positive distance: $R=k-g(x)$ and $r=k-f(x)$.

The volume formula for this case is: $$V=\pi {\int }_a^b\left[R^2-r^2\right]dx$$ Intuition: Each thin vertical slice perpendicular to the horizontal axis becomes a washer when rotated, and we sum the volume of all slices to get the total volume of the solid.

Worked Example

Problem: Find the volume of the solid formed by rotating the region bounded by $y=x^2$, $y=0$, and $x=2$ around the horizontal line $y=-1$. Solution:

1. Confirm bounds: The region runs from $x=0$ (intersection of $y=x^2$ and $y=0$) to $x=2$. The axis $y=-1$ is below the entire region, which spans $y\in [0,4]$.
2. Calculate radii: Upper curve is $y=x^2$, lower curve is $y=0$. Distance from upper curve to axis: $R=x^2-(-1)=x^2+1$. Distance from lower curve to axis: $r=0-(-1)=1$.
3. Substitute into the volume formula: $$V=\pi {\int }_0^2\left[(x^2+1{)}^2-1^2\right]dx$$
4. Simplify and integrate: The integrand simplifies to $x^4+2x^2$, so: $$V=\pi {\left[\frac{x^5}{5}+\frac{2x^3}{3}\right]}_0^2=\pi \left(\frac{32}{5}+\frac{16}{3}\right)=\frac{176}{15}\pi$$

Exam tip: Always calculate radius as the absolute value of $(y_{curve}-k)$. Writing the subtraction in the order that gives a positive value eliminates 90% of common sign errors on this topic.

3. Rotation around a vertical axis $x=k$, $k\ne 0$

When rotating around a vertical line $x=k$, the axis is parallel to the y-axis, so the washer method requires integrating with respect to $y$. This is because washers are perpendicular to the axis of rotation: a vertical axis means perpendicular slices are horizontal, with thickness $dy$. First, you must rewrite all boundary curves as functions of $y$ ($x=f(y)$) instead of functions of $x$.

Radius is again the perpendicular distance from the curve to the axis $x=k$, which is always non-negative. If the entire region lies to the right of $x=k$, the outer radius (distance from the rightmost curve to the axis) is $R=f(y)-k$, and the inner radius (distance from the leftmost curve to the axis) is $r=g(y)-k$. If the entire region lies to the left of $x=k$, reverse the subtraction to get positive distance: $R=k-g(y)$ and $r=k-f(y)$. The volume formula is: $$V=\pi {\int }_c^d\left[R^2-r^2\right]dy$$ where $c$ and $d$ are the bounds for $y$ across the region.

Worked Example

Problem: Find the volume of the solid formed by rotating the region bounded by $y=x^2$, $y=0$, and $x=2$ around the vertical line $x=3$. Solution:

1. Rewrite boundaries as functions of $y$: $y=x^2⟹x=\sqrt{y}$ (we take the positive root, since $x\in [0,2]$). Bounds for $y$ are $y=0$ to $y=4$ (at $x=2$).
2. The entire region is left of the axis $x=3$, so distance is calculated as $3-x$. The rightmost boundary is $x=2$, the leftmost boundary is $x=\sqrt{y}$. The outer radius (larger distance) is distance from the leftmost curve to the axis: $R=3-\sqrt{y}$. The inner radius (smaller distance) is distance from the rightmost curve to the axis: $r=3-2=1$.
3. Substitute into the volume formula: $$V=\pi {\int }_0^4\left[(3-\sqrt{y}{)}^2-1^2\right]dy$$
4. Simplify and integrate: The integrand simplifies to $8-6y^{1/2}+y$, so: $$V=\pi {\left[8y-4y^{3/2}+\frac{y^2}{2}\right]}_0^4=8\pi$$

Exam tip: When rotating around a vertical axis with the washer method, always confirm you are integrating with respect to $y$, not $x$. If you end up with an integral in $dx$, you are either using the shell method by mistake or set up the problem incorrectly.

4. Rotation when the axis cuts through the region

A common AP exam variation is when the axis of rotation passes through the interior of the bounded region, rather than lying entirely on one side. In this case, you cannot use a single outer and inner radius for the entire interval, because there is no hole in the cross-section: the entire region between the two boundary curves is rotated around the axis, so the cross-section is a full solid disk, not a washer.

To solve this, split the region into two sub-regions, one on each side of the axis. Each sub-region lies entirely on one side of the axis, so we can calculate the volume of each separately (each is a solid disk, so inner radius $r=0$ for each sub-region), then add the two volumes to get the total.

Worked Example

Problem: Set up (do not evaluate) the integral for the volume of the solid formed by rotating the region bounded by $y=x+2$ and $y=x^2$ around the line $y=1$. Solution:

1. Find intersection points to get bounds: $x+2=x^2⟹(x-2)(x+1)=0$, so $x\in [-1,2]$.
2. Check the position of the axis: The upper boundary is $y=x+2$, the lower boundary is $y=x^2$. For all $x$ in $[-1,2]$, $x^2<1<x+2$, so the axis $y=1$ cuts directly through the region between the two curves.
3. Split the region into two sub-regions: the lower sub-region between $y=x^2$ and $y=1$, and the upper sub-region between $y=1$ and $y=x+2$, both over $x\in [-1,2]$.
4. Calculate radii for each: Lower sub-region (entirely below $y=1$) has radius $R_1=1-x^2$, $r_1=0$. Upper sub-region (entirely above $y=1$) has radius $R_2=(x+2)-1=x+1$, $r_2=0$.
5. Total volume is the sum of the two volumes: $$V=\pi {\int }_{-1}^2(1-x^2{)}^{2}dx+\pi {\int }_{-1}^2(x+1{)}^{2}dx=\pi {\int }_{-1}^2\left[(1-x^2{)}^2+(x+1{)}^2\right]dx$$

Exam tip: If you end up with a negative radius when you calculate $R-r$, that is a clear sign the axis cuts through the region and you need to split your integral.

5. Common Pitfalls (and how to avoid them)

• Wrong move: For rotation around $y=k$, calculating radius as just $f(x)$ instead of $|f(x)-k|$. Why: Students get used to rotation around $y=0$, where radius equals $f(x)$, and forget to adjust for a shifted axis. Correct move: Always calculate radius as the perpendicular distance between the curve and the axis, every time, regardless of where the axis is.
• Wrong move: For rotation around a vertical axis $x=k$, setting up the integral with respect to $x$ instead of $y$ for the washer method. Why: Students confuse washer method with shell method, or default to integrating with respect to $x$ since most functions are given as $y=f(x)$. Correct move: For washer method, integrate with respect to the variable parallel to the axis: horizontal axis → $dx$, vertical axis → $dy$. Rewrite all functions in terms of the correct variable before setup.
• Wrong move: When the region is left of $x=k$, writing $R=k-x_{left}$ and $r=k-x_{right}$, swapping outer and inner radii. Why: Students assume "left = inner, right = outer" even when the axis is to the right of the entire region. Correct move: Calculate the distance from each curve to the axis first, then assign $R$ to the larger distance and $r$ to the smaller distance.
• Wrong move: When the axis cuts through the region, using a single washer term $R^2-r^2$ with $R=f(x)-k$ and $r=k-g(x)$. Why: Students incorrectly assume there is a hole between the lower curve and the axis, but the entire region between the curves is rotated, so there is no hole. Correct move: Split the region into two sub-regions on either side of the axis, add the volumes of the two solid disks.
• Wrong move: Expanding $(a+b{)}^2=a^2+b^2$, skipping the cross term. Why: Students rush to integrate after setup and skip simplifying the integrand. Correct move: Always expand $R^2-r^2$ fully and simplify term by term before integrating.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

The region bounded by $y=\sqrt{x}$, $y=0$, and $x=4$ is rotated around the line $y=3$. Which of the following gives the correct volume of the solid? A) $\pi {\int }_0^4\left((3-\sqrt{x}{)}^2\right)dx$ B) $\pi {\int }_0^4\left(3^2-(\sqrt{x}{)}^2\right)dx$ C) $\pi {\int }_0^4\left(3^2-(3-\sqrt{x}{)}^2\right)dx$ D) $\pi {\int }_0^2\left((4-y^2{)}^2\right)dy$

Worked Solution: The axis of rotation is the horizontal line $y=3$, which lies entirely above the region (the region spans $y$ from 0 to 2). The region is bounded above by $y=\sqrt{x}$ and below by $y=0$. The outer radius is the distance from the lower boundary ($y=0$) to the axis: $R=3-0=3$. The inner radius is the distance from the upper boundary ($y=\sqrt{x}$) to the axis: $r=3-\sqrt{x}$. Substituting into the washer volume formula gives $V=\pi {\int }_0^4\left(3^2-(3-\sqrt{x}{)}^2\right)dx$, which matches option C. Options A, B, and D all have incorrect radius calculations or bounds. Correct answer: $\boxed{C}$

Question 2 (Free Response)

Let $R$ be the region bounded by $y=e^x$, $y=1$, and $x=1$. (a) Find the area of $R$. (b) Write, but do not evaluate, an integral expression for the volume of the solid formed when $R$ is rotated around the horizontal line $y=-2$. (c) Write, but do not evaluate, an integral expression for the volume of the solid formed when $R$ is rotated around the vertical line $x=2$.

Worked Solution: (a) The region is bounded from $x=0$ (intersection of $y=e^x$ and $y=1$) to $x=1$. Area is: $$A={\int }_0^1(e^x-1)dx={\left[e^x-x\right]}_0^1=(e-1)-(1-0)=e-2$$

(b) For rotation around horizontal $y=-2$, integrate with respect to $x$. $R=e^x-(-2)=e^x+2$, $r=1-(-2)=3$. Volume: $$V=\pi {\int }_0^1\left[(e^x+2{)}^2-3^2\right]dx$$

(c) For rotation around vertical $x=2$, rewrite as $x=\ln y$, bounds for $y$ are $1$ to $e$. $R=2-\ln y$, $r=2-1=1$. Volume: $$V=\pi {\int }_1^e\left[(2-\ln y{)}^2-1^2\right]dy$$

Question 3 (Application / Real-World Style)

A machined metal cylinder has a recessed inner section cut out. The solid is formed by rotating the region between $y=2x+1$ (outer edge) and $y=2x$ (inner edge) from $x=0$ to $x=2$ centimeters around the horizontal axis $y=0$. Find the volume of metal in the finished part, in cubic centimeters.

Worked Solution: The axis of rotation is horizontal $y=0$, so we integrate with respect to $x$ from $0$ to $2$. The entire region is above $y=0$, so outer radius $R=2x+1$, inner radius $r=2x$. Volume is: $$V=\pi {\int }_0^2\left[(2x+1{)}^2-(2x{)}^2\right]dx=\pi {\int }_0^2(4x+1)dx$$ Integrate: $$V=\pi {\left[2x^2+x\right]}_0^2=\pi (8+2)=10\pi \approx 31.4{\text{ cm}}^3$$ The finished part has a volume of approximately 31.4 cubic centimeters of metal.

7. Quick Reference Cheatsheet

8. What's Next

The washer method around shifted axes is a core prerequisite for the cylindrical shell method (the other main method for volumes of revolution) and for more advanced AP topics including work done to pump fluid out of non-standard tanks, surface area of revolution, and volumes with arbitrary known cross-sections. Without mastering radius calculation for shifted axes, you will struggle to set up integrals correctly for these more complex multi-concept questions, which frequently appear on the AP Calculus BC FRQ section. The skills you practice here, especially correctly calculating distance between curves and axes and choosing the right integration variable, transfer directly to every other application of integration in the course. Follow-up topics to study next:

• Cylindrical shells method for volumes of revolution
• Volumes with known cross-sections
• Applications of integration: Work
• Arc length and surface area of revolution