Washer method around the x- or y-axis — AP Calculus BC Study Guide

For: AP Calculus BC.

Covers: Derivation and application of the washer method for volumes of revolution around the x-axis and y-axis, including identifying outer/inner radii, setting up definite integrals, and solving for volumes of regions bounded between two curves.

You should already know: How to find points of intersection of two functions; How to evaluate definite integrals of elementary functions; How to describe bounded regions between two curves in the xy-plane.

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 the x- or y-axis?

The washer method is an integration technique used to calculate the volume of a solid of revolution formed when a region bounded between two curves is rotated around a horizontal or vertical axis of rotation. When the region rotated does not touch the axis of rotation, a cross-section cut perpendicular to the axis is a washer: a flat circular disk with a smaller concentric circular hole cut out of its center, giving the technique its name. Unlike the simpler disk method (which applies when the region touches the axis, leaving no hole), the washer method explicitly accounts for the empty volume between the inner boundary of the region and the axis of rotation. Per the AP Calculus BC Course and Exam Description (CED), this topic falls within Unit 8 (Applications of Integration), which accounts for 10-15% of the total AP exam score. Washer method problems regularly appear on both multiple-choice (MCQ) and free-response (FRQ) sections of the exam, often as part of multi-part FRQ questions testing volumes of revolution.

2. Washer Method for Rotation Around the x-Axis

When rotating a region bounded above by $y=R_{\text{curve}}(x)$ and below by $y=r_{\text{curve}}(x)$ between $x=a$ and $x=b$ around the x-axis, the axis of rotation is horizontal, so we cut cross-sections perpendicular to the x-axis. Each cross-section is a washer with outer radius equal to the distance from the x-axis to the farther (outer) curve, and inner radius equal to the distance from the x-axis to the closer (inner) curve. The area of a single washer at position $x$ is the area of the outer circle minus the area of the inner circle: $$A(x)=\pi [R(x){]}^2-\pi [r(x){]}^2=\pi \left([R(x){]}^2-[r(x){]}^2\right)$$ To get the total volume of the solid, we integrate the area function across the full interval from $a$ to $b$, since each thin washer has thickness $dx$, so its volume is $A(x)dx$. This gives the final volume formula: $$V=\pi {\int }_a^b\left([R(x){]}^2-[r(x){]}^2\right)dx$$ The core convention to remember: $R(x)$ is always the outer radius (farther from the axis of rotation) and $r(x)$ the inner radius (closer to the axis), so $[R(x){]}^2>[r(x){]}^2$ to avoid negative volume.

Worked Example

Find the volume of the solid formed when the region bounded by $y=2x$ and $y=x^2$ is rotated around the x-axis.

1. Find points of intersection to get integral bounds: set $2x=x^2⟹x^2-2x=0⟹x(x-2)=0$, so intersections at $x=0$ and $x=2$, so $a=0$, $b=2$.
2. Identify outer and inner radii: test a point in $(0,2)$, e.g. $x=1$: $2(1)=2>1^2=1$, so $y=2x$ is farther from the x-axis, so $R(x)=2x$, $r(x)=x^2$.
3. Substitute into the volume formula: $V=\pi {\int }_0^2\left((2x{)}^2-(x^2{)}^2\right)dx=\pi {\int }_0^2(4x^2-x^4)dx$.
4. Evaluate the definite integral: ${\int }_0^2(4x^2-x^4)dx={\left[\frac{4x^3}{3}-\frac{x^5}{5}\right]}_0^2=\left(\frac{32}{3}-\frac{32}{5}\right)-0=32\left(\frac{2}{15}\right)=\frac{64}{15}$. Multiply by $\pi$ to get $V=\frac{64\pi }{15}$.

Exam tip: Always confirm which function is farther from the axis of rotation by testing a value in the interval between the two intersection points; do not assume the higher-degree function is always inner or outer.

3. Washer Method for Rotation Around the y-Axis

When rotating a region around the y-axis (a vertical axis of rotation), we cut cross-sections perpendicular to the y-axis, so we must express all boundary functions as $x$ in terms of $y$, and integrate over the interval of $y$ values spanned by the region. For a region bounded on the right by $x=R(y)$ (farther from the y-axis) and on the left by $x=r(y)$ (closer to the y-axis) between $y=c$ and $y=d$, the area of each washer at fixed $y$ follows the same area rule as for rotation around the x-axis: $A(y)=\pi ([R(y){]}^2-[r(y){]}^2)$. Integrating across the interval of $y$ gives the volume formula: $$V=\pi {\int }_c^d\left([R(y){]}^2-[r(y){]}^2\right)dy$$ The only structural change from rotation around the x-axis is that we integrate with respect to $y$ instead of $x$, because our slices are perpendicular to the vertical y-axis. This method is an alternative to the cylindrical shells method for rotation around the y-axis; washer method is simpler when you can easily solve for $x$ as a function of $y$. If an AP question explicitly asks to use the washer method, you must use this form even if shells is computationally easier.

Worked Example

Find the volume of the solid formed when the region bounded by $y=x^3$, $y=8$, and $x=0$ is rotated around the y-axis.

1. Rewrite $y=x^3$ as $x$ in terms of $y$: $x=y^{1/3}$ (we use the positive root since $x\ge 0$ in the region).
2. Find bounds for $y$: the region spans from $y=0$ (at $x=0$) to $y=8$, so $c=0$, $d=8$.
3. Identify outer and inner radii: for any $y$ between $0$ and $8$, the rightmost boundary is $x=y^{1/3}$, which is farther from the y-axis than the left boundary $x=0$, so $R(y)=y^{1/3}$, $r(y)=0$.
4. Substitute into the formula: $V=\pi {\int }_0^8\left((y^{1/3}{)}^2-0^2\right)dy=\pi {\int }_0^8{y}^{2/3}dy$.
5. Evaluate: ${\int }_0^8{y}^{2/3}dy={\left[\frac{y^{5/3}}{5/3}\right]}_0^8=\frac{3}{5}(8^{5/3})=\frac{3}{5}(32)=\frac{96}{5}$. Multiply by $\pi$ to get $V=\frac{96\pi }{5}$.

Exam tip: If your original functions are given as $y=f(x)$, do not forget to solve explicitly for $x$ in terms of $y$ before setting up the integral for rotation around the y-axis; integrating with respect to $x$ for the washer method around a vertical axis is always incorrect.

4. Washer Method for Rotation Around a Shifted Axis (Not Through the Origin)

The washer method generalizes to any axis of rotation, not just the x-axis ($y=0$) or y-axis ($x=0$). The core rule for finding radii does not change: the radius for any boundary curve is the absolute distance between the curve and the axis of rotation, regardless of where the axis is located. For a horizontal axis of rotation at $y=k$ (still integrate with respect to $x$): $R(x)=|\text{outer boundary}-k|$, $r(x)=|\text{inner boundary}-k|$, so the volume formula remains $V=\pi {\int }_a^b([R(x){]}^2-[r(x){]}^2)dx$. For a vertical axis of rotation at $x=h$ (integrate with respect to $y$): $R(y)=|\text{outer boundary}-h|$, $r(y)=|\text{inner boundary}-h|$, so volume is $V=\pi {\int }_c^d([R(y){]}^2-[r(y){]}^2)dy$. The absolute value can be dropped after identifying which radius is larger, because squaring removes the sign: $(-a{)}^2=a^2$ for any real $a$. This shifted axis case is one of the most commonly tested variations of the washer method on AP FRQ.

Worked Example

Find the volume of the solid formed when the region bounded by $y=2x$ and $y=x^2$ is rotated around the horizontal line $y=-2$.

1. Bounds for $x$ are still the intersection points we found earlier: $a=0$, $b=2$.
2. Calculate distance from each boundary to the axis $y=-2$: for $y=2x$, distance is $2x-(-2)=2x+2$; for $y=x^2$, distance is $x^2-(-2)=x^2+2$. Both are positive because both boundaries are above the axis $y=-2$.
3. Identify outer and inner radii: between $0$ and $2$, $2x>x^2$, so $2x+2>x^2+2$, so $R(x)=2x+2$, $r(x)=x^2+2$.
4. Substitute into the formula: $V=\pi {\int }_0^2\left((2x+2{)}^2-(x^2+2{)}^2\right)dx$. Expand the integrand: $(4x^2+8x+4)-(x^4+4x^2+4)=8x-x^4$.
5. Evaluate: ${\int }_0^2(8x-x^4)dx={\left[4x^2-\frac{x^5}{5}\right]}_0^2=(16-\frac{32}{5})=\frac{80-32}{5}=\frac{48}{5}$. Multiply by $\pi$ to get $V=\frac{48\pi }{5}$.

Exam tip: If the axis of rotation cuts through the region between the two boundaries, you still calculate distance from the axis to each boundary separately; one distance is not automatically zero just because the axis passes through the region.

5. Common Pitfalls (and how to avoid them)

• Wrong move: For rotation around the y-axis, you leave functions as $y=f(x)$ and integrate with respect to $x$. Why: You confuse the washer method with the shell method, which integrates with respect to $x$ for rotation around the y-axis. Correct move: Always solve for $x$ as a function of $y$ and integrate with respect to $y$ when using the washer method for rotation around any vertical axis.
• Wrong move: You write the integrand as $\pi (R-r{)}^2$ instead of $\pi (R^2-r^2)$. Why: You incorrectly factor the square when subtracting the areas of the two circles. Correct move: Always expand the area subtraction first: $\pi R^2-\pi r^2=\pi (R^2-r^2)$, never $\pi (R-r{)}^2$.
• Wrong move: When rotating around a horizontal axis at $y=k$, you subtract $k$ only from the outer radius and forget to subtract it from the inner radius. Why: You assume the inner radius is still measured from the origin, not the new axis. Correct move: Calculate the distance from the axis of rotation to every boundary curve before identifying inner and outer radii.
• Wrong move: You do not split the integral when the inner/outer radius changes at an intermediate value of $x$ or $y$. Why: You do not check if the boundary of the region changes across the integration interval. Correct move: Test the boundary at multiple points across the interval; split the integral at any point where the identity of the inner or outer radius changes.
• Wrong move: You use the washer method when the region is bounded by one curve and touches the axis of rotation. Why: You confuse the washer and disk method; there is no hole so you do not need to subtract an inner area. Correct move: If the region touches the axis, use the disk method, which is just a special case of the washer method with inner radius $r=0$.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

The region bounded by $y=x+2$ and $y=x^2$ (with intersections at $x=-1$ and $x=2$) is rotated around the x-axis. Which of the following is the correct definite integral for the volume of the solid using the washer method?

A) $\pi {\int }_{-1}^2(x+2-x^2{)}^{2}dx$ B) $\pi {\int }_{-1}^2((x+2{)}^2-(x^2{)}^2)dx$ C) $\pi {\int }_{-1}^2(x+2{)}^{2}dx$ D) $\pi {\int }_{-1}^2((x^2{)}^2-(x+2{)}^2)dx$

Worked Solution: For rotation around the x-axis using the washer method, the formula is $V=\pi {\int }_a^b(R(x{)}^2-r(x{)}^2)dx$, with bounds $a=-1$ and $b=2$ given. Between $x=-1$ and $x=2$, $y=x+2$ is above $y=x^2$, so it is farther from the x-axis, meaning $R(x)=x+2$ and $r(x)=x^2$. Substituting into the formula gives exactly the integral in option B. Option A is the common error of squaring the difference between radii, option C forgets to subtract the inner area, and option D swaps the inner and outer radii. The correct answer is B.

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 x-axis, using the washer method. (b) Set up, but do not evaluate, the integral for the volume of the solid formed when R is rotated around the y-axis, using the washer method. (c) How would your answer to part (a) change if the rotation was around the horizontal line $y=3$? Give the new updated integral.

Worked Solution: (a) The region R starts at the intersection of $y=e^x$ and $y=1$, which is at $x=0$, and ends at $x=2$, so $a=0$, $b=2$. The outer radius is $R(x)=e^x$ (farther from x-axis) and inner radius is $r(x)=1$, so the integral is: $$V=\pi {\int }_0^2\left((e^x{)}^2-1^2\right)dx=\pi {\int }_0^2(e^{2x}-1)dx$$

(b) For rotation around the y-axis, rewrite $y=e^x$ as $x=\ln y$. Bounds for $y$ go from $y=1$ to $y=e^2$. The right boundary is $x=2$, so $R(y)=2$, and the left boundary is $x=\ln y$, so $r(y)=\ln y$. The integral is: $$V=\pi {\int }_1^{e^2}\left(2^2-(\ln y{)}^2\right)dy=\pi {\int }_1^{e^2}(4-(\ln y{)}^2)dy$$

(c) For rotation around $y=3$, we split the integral at $x=\ln 3$, where the axis cuts the region. For $0\le x\le \ln 3$, the farthest distance from $y=3$ is $|1-3|=2$, and the closest distance is $|e^x-3|=3-e^x$. For $\ln 3\le x\le 2$, the farthest distance is $|e^x-3|=e^x-3$, and the closest distance is $2$. The updated integral is: $$V=\pi {\int }_0^{\ln 3}\left(2^2-(3-e^x{)}^2\right)dx+\pi {\int }_{\ln 3}^2\left((e^x-3{)}^2-2^2\right)dx$$

Question 3 (Application / Real-World Style)

A manufacturing company produces hollow propeller shafts for marine vessels, with a parabolic fillet cut into the end of the shaft to reduce stress. The cross-section of the fillet (in the xy-plane) is bounded by $y=0.5x^2$ and $y=2$, between $x=-2$ and $x=2$ (all coordinates in centimeters). The finished end is formed by rotating this region around the y-axis (the central axis of the shaft). What is the volume of material removed to create this fillet, in cubic centimeters?

Worked Solution: We use the washer method for rotation around the y-axis. First, rewrite $y=0.5x^2$ as $x$ in terms of $y$: $x^2=2y⟹x=\sqrt{2y}$ for positive $x$. Bounds for $y$ go from $y=0$ to $y=2$. The outer radius is the distance from the y-axis to the outer edge of the shaft at $x=2$, so $R(y)=2$, and the inner radius is the distance from the y-axis to the edge of the fillet at $x=\sqrt{2y}$, so $r(y)=\sqrt{2y}$. Substitute into the volume formula: $$V=\pi {\int }_0^2\left(2^2-(\sqrt{2y}{)}^2\right)dy=\pi {\int }_0^2(4-2y)dy$$ Evaluate the integral: ${\int }_0^2(4-2y)dy={\left[4y-y^2\right]}_0^2=(8-4)=4$. Multiply by $\pi$ to get $V=4\pi \approx 12.57$ cubic centimeters. This means approximately 12.57 cubic centimeters of material are removed from the end of the shaft to create the stress-reducing parabolic fillet.

7. Quick Reference Cheatsheet

8. What's Next

Mastery of the washer method is required for all further topics involving volumes of revolution, which are a regular component of both MCQ and FRQ on the AP Calculus BC exam. Next, you will learn the cylindrical shells method, an alternative method for volumes of revolution that is often easier for rotation around the y-axis when you cannot easily solve for $x$ as a function of $y$. You will also apply washer method concepts to find volumes of solids with known cross-sections, another core topic in Unit 8. Without mastering the process of identifying inner/outer radii and setting up the correct definite integral, both of these topics will be much harder to master. The washer method also reinforces the core conceptual idea of integrating cross-sectional area to find total volume, which unifies all applications of integration across the course.

• Cylindrical shells method around x- or y-axis
• Volumes of solids with known cross-sections