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

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Washer method definition, volume formulas for revolution around x-axis, y-axis, and non-coordinate axes, identifying outer/inner radii, setting up and evaluating definite integrals for volumes of hollow solids of revolution.

You should already know: Definite integral evaluation, area between two curves, basics of solids of revolution.

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 the x- or y-axis?

The washer method is an integration technique used to calculate the volume of a solid of revolution that has a hollow core (a hole along the axis of rotation). It is an extension of the disk method: when you rotate a region between two curves around an axis, each cross-section perpendicular to the axis forms a "washer" (a flat disk with a smaller circular hole cut out of the center), hence the name. It is sometimes called the hollow disk method as a synonym.

Per the AP Calculus AB Course and Exam Description (CED), this topic is part of Unit 8: Applications of Integration, which accounts for 10–15% of the total AP exam score. Washer method problems appear on both multiple-choice (MCQ) and free-response (FRQ) sections of the exam. On FRQs, you will often be asked to either set up the integral (without evaluating) or calculate the exact volume, making this a high-weight topic for earning points.

2. Washer Method for Revolution Around the x-Axis (or Any Horizontal Axis)

When rotating around a horizontal axis (such as the x-axis, or any line of the form $y=k$), cross-sections perpendicular to the axis are vertical slices, so we integrate with respect to $x$. The area of one washer is the area of the outer circle minus the area of the inner circle: $$A=\pi R^2-\pi r^2=\pi \left(R^2-r^2\right)$$ where $R(x)$ is the outer radius: the distance from the axis of rotation to the farther curve from the axis, and $r(x)$ is the inner radius: the distance from the axis of rotation to the closer curve from the axis. To get total volume, we integrate the area of all washers between the bounds of the region, giving the formula: $$V=\pi {\int }_a^b\left({\left[R(x)\right]}^2-{\left[r(x)\right]}^2\right)dx$$ For rotation around the x-axis ($y=0$), this simplifies to $R(x)=y_{\text{upper}}(x)$ and $r(x)=y_{\text{lower}}(x)$, since the distance from $y=0$ to a curve is just the $y$-value of the curve.

Worked Example

Find the volume of the solid formed when the region bounded by $y=2x$, $y=x^2$, $0\le x\le 2$ is rotated around the x-axis.

1. Confirm the axis is horizontal ($y=0$), so we integrate with respect to $x$, with bounds $a=0$, $b=2$.
2. Identify radii: Between $0$ and $2$, $2x\ge x^2$, so the upper (farther from x-axis) curve is $y=2x$, lower (closer) is $y=x^2$. Thus $R(x)=2x$, $r(x)=x^2$.
3. Set up the volume integral: $$V=\pi {\int }_0^2\left((2x{)}^2-(x^2{)}^2\right)dx=\pi {\int }_0^2\left(4x^2-x^4\right)dx$$
4. Evaluate the definite integral: $$V=\pi {\left[\frac{4x^3}{3}-\frac{x^5}{5}\right]}_0^2=\pi \left(\left(\frac{32}{3}-\frac{32}{5}\right)-0\right)=\frac{64\pi }{15}$$

Exam tip: Always square each radius separately before subtracting. Never subtract first then square, as this will always give the wrong volume.

3. Washer Method for Revolution Around the y-Axis (or Any Vertical Axis)

When rotating around a vertical axis (such as the y-axis, or any line of the form $x=k$), cross-sections perpendicular to the axis are horizontal slices, so we integrate with respect to $y$. This means we must first rewrite all bounding curves as functions of $y$ ($x$ as a function of $y$, instead of $y$ as a function of $x$). The logic for radii and area is identical to the horizontal case, just with $y$ as the integration variable: $$V=\pi {\int }_c^d\left({\left[R(y)\right]}^2-{\left[r(y)\right]}^2\right)dy$$ Here, $R(y)$ is the distance from the axis of rotation to the farther (rightmost, if rotating around the y-axis) curve, and $r(y)$ is the distance to the closer (leftmost, if rotating around y-axis) curve. For rotation around the y-axis ($x=0$), this simplifies to $R(y)=x_{\text{right}}(y)$ and $r(y)=x_{\text{left}}(y)$.

Worked Example

Find the volume of the solid formed when the region bounded by $y=2x$, $y=x^2$, $0\le y\le 4$ is rotated around the y-axis.

1. Confirm the axis is vertical ($x=0$), so we integrate with respect to $y$, with bounds $c=0$, $d=4$. Rewrite both curves as $x$ in terms of $y$.
2. Rewrite: $y=2x⟹x=\frac{y}{2}$; $y=x^2⟹x=\sqrt{y}$ (we take the positive root for this region).
3. Identify radii: Between $0$ and $4$, $\sqrt{y}\ge \frac{y}{2}$, so $R(y)=\sqrt{y}$ (farther from y-axis), $r(y)=\frac{y}{2}$ (closer to y-axis).
4. Set up and evaluate the integral: $$V=\pi {\int }_0^4\left((\sqrt{y}{)}^2-{\left(\frac{y}{2}\right)}^2\right)dy=\pi {\int }_0^4\left(y-\frac{y^2}{4}\right)dy$$ $$V=\pi {\left[\frac{y^2}{2}-\frac{y^3}{12}\right]}_0^4=\pi \left(8-\frac{64}{12}\right)=\frac{8\pi }{3}$$

Exam tip: Always rewrite all functions in terms of the integration variable before calculating radii. Mixing $x$ and $y$ in the same integral is an immediate point loss on FRQs.

4. Washer Method for Revolution Around a Non-Coordinate Axis

AP Calculus AB exam questions very frequently ask for rotation around an axis that is not the x or y axis (e.g. $y=2$, $x=-1$). The core logic stays the same, but you must correctly calculate the radius as the distance between the curve and the shifted axis, not just the value of the curve. The key rule is: radius is the absolute difference between the curve and the axis, because distance is always positive. For a horizontal axis $y=k$, $R(x)=|y_{\text{farthest}}-k|$ and $r(x)=|y_{\text{closest}}-k|$. For a vertical axis $x=k$, $R(y)=|x_{\text{farthest}}-k|$ and $r(y)=|x_{\text{closest}}-k|$. The outer radius always belongs to the curve that is farther from the axis, not the curve with the largest $x$ or $y$ value.

Worked Example

Find the volume of the solid formed when the region bounded by $y=2x$, $y=x^2$, $0\le x\le 2$ is rotated around the horizontal line $y=5$.

1. Axis is horizontal ($y=5$), so integrate with respect to $x$, bounds $0$ to $2$. The entire region lies below $y=5$, so all distances are $5-y$.
2. Identify radii: The lower curve $y=x^2$ is farther from $y=5$, so it gives the outer radius: $R(x)=5-x^2$. The upper curve $y=2x$ is closer to $y=5$, so it gives the inner radius: $r(x)=5-2x$.
3. Set up the integral: $$V=\pi {\int }_0^2\left((5-x^2{)}^2-(5-2x{)}^2\right)dx$$
4. Simplify and evaluate: $$(25-10x^2+x^4)-(25-20x+4x^2)=x^4-14x^2+20x$$ $$V=\pi {\int }_0^2(x^4-14x^2+20x)dx=\pi {\left[\frac{x^5}{5}-\frac{14x^3}{3}+10x^2\right]}_0^2=\frac{136\pi }{15}$$

Exam tip: Test a sample point in your interval to confirm $R>r$: if your integrand is negative, you swapped outer and inner radii.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Writing $(R-r{)}^2$ instead of $R^2-r^2$, subtracting radii before squaring. Why: Students confuse factoring rules with the area formula for circles, or rush through setup. Correct move: Always square each radius separately first, then subtract the squared inner radius from the squared outer radius.
• Wrong move: Integrating with respect to $x$ when rotating around a vertical axis, or with respect to $y$ when rotating around a horizontal axis. Why: Students forget that cross-sections must be perpendicular to the axis of rotation, which dictates the integration variable. Correct move: For any horizontal axis, integrate with respect to $x$; for any vertical axis, integrate with respect to $y$, and rewrite functions first.
• Wrong move: Calculating radius for a non-coordinate axis as just the function value, e.g. for rotation around $y=2$, writing $R(x)=x^2$ instead of $2-x^2$. Why: Students get used to rotating around the x-axis ($y=0$) and forget to adjust radii for a shifted axis. Correct move: Always calculate radius as the absolute difference between the curve and the axis: $|curve-axis|$.
• Wrong move: Swapping outer and inner radii, resulting in a negative integrand. Why: Students mix up which curve is farther from the axis, especially when the axis is above or to the left of the region. Correct move: After identifying radii, test one point in the interval to confirm $R>r$ and the integrand is positive.
• Wrong move: Forgetting to multiply the entire integral by $\pi$, even when asked for a numerical volume. Why: Students remember the core integrand but omit the constant from the circle area formula. Correct move: Write $\pi$ outside the integral when setting up, before any other steps.
• Wrong move: Leaving functions in terms of $x$ when integrating with respect to $y$ for rotation around a vertical axis. Why: Students are more comfortable with $y$ as a function of $x$ and forget to invert. Correct move: As soon as you confirm you are integrating with respect to $y$, invert all functions to get $x$ as a function of $y$ before calculating radii.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

The region $R$ is bounded by $y=\sqrt{x}$, $y=2$, and $x=0$. Which of the following gives the volume of the solid formed when $R$ is rotated around the y-axis?

A) $\pi {\int }_0^4\left((y^2{)}^2-0^2\right)dy$ B) $\pi {\int }_0^2\left(2^2-(\sqrt{x}{)}^2\right)dx$ C) $\pi {\int }_0^2\left((y^2{)}^2-0^2\right)dy$ D) $\pi {\int }_0^4\left(y-4\right)dx$

Worked Solution: Rotation around the y-axis (a vertical axis) means we integrate with respect to $y$. First, rewrite $y=\sqrt{x}$ as $x=y^2$, with bounds for $y$ from $0$ to $2$. The rightmost curve is $x=y^2$, the left boundary is $x=0$, so $R(y)=y^2$ and $r(y)=0$. Substituting into the washer formula gives $V=\pi {\int }_0^2\left((y^2{)}^2-0^2\right)dy$, which matches option C. Option A has the wrong bounds, option B uses the wrong integration variable, option D has incorrect radii and variable mismatch. The correct answer is C.

Question 2 (Free Response)

Let $R$ be the region bounded by $y=x^2+1$, $y=x+3$, $x\ge 0$. (a) Set up, but do not evaluate, a definite integral for the volume of the solid formed when $R$ is rotated around the x-axis using the washer method. (b) Evaluate your integral from part (a) to find the exact volume. (c) Set up, but do not evaluate, a new integral for the volume if $R$ is rotated around the horizontal line $y=-1$ using the washer method.

Worked Solution: (a) First find intersection points: $x^2+1=x+3⟹x^2-x-2=0⟹(x-2)(x+1)=0$. For $x\ge 0$, bounds are $x=0$ to $x=2$. The upper curve is $y=x+3$, lower curve is $y=x^2+1$. Rotation around x-axis, so: $$V=\pi {\int }_0^2\left((x+3{)}^2-(x^2+1{)}^2\right)dx$$

(b) Expand the integrand: $(x^2+6x+9)-(x^4+2x^2+1)=-x^4-x^2+6x+8$. Evaluate: $$V=\pi {\int }_0^2(-x^4-x^2+6x+8)dx=\pi {\left[-\frac{x^5}{5}-\frac{x^3}{3}+3x^2+8x\right]}_0^2=\frac{304\pi }{15}$$

(c) Rotation around $y=-1$, so radius is distance from $y=-1$ to each curve. Outer radius (upper curve) is $(x+3)-(-1)=x+4$, inner radius (lower curve) is $(x^2+1)-(-1)=x^2+2$. New integral: $$V=\pi {\int }_0^2\left((x+4{)}^2-(x^2+2{)}^2\right)dx$$

Question 3 (Application / Real-World Style)

A water tank manufacturer makes a hollow spherical storage tank with an outer profile given by $x^2+y^2=16$ (radius 4 meters) and an inner wall given by $x^2+y^2=9$ (inner radius 3 meters). When rotated around the x-axis, the cross-section in the xy-plane forms the full 3D tank. Find the volume of plastic used to make the tank, in cubic meters.

Worked Solution: Rotation around the x-axis (horizontal), so integrate with respect to $x$ from $-4$ to $4$. Rewrite both curves as $y$ in terms of $x$: outer $y=\sqrt{16-x^2}$, inner $y=\sqrt{9-x^2}$. By symmetry, we can integrate from $0$ to $4$ and double, or integrate directly: $$V=\pi {\int }_{-4}^4\left((\sqrt{16-x^2}{)}^2-(\sqrt{9-x^2}{)}^2\right)dx=\pi {\int }_{-4}^{4}7dx=7\pi (8)=56\pi$$ Wait, correction: the inner hole extends from $-3$ to $3$, so split the integral: $$V=\pi {\int }_{-4}^{-3}(16-x^2)dx+\pi {\int }_{-3}^{3}7dx+\pi {\int }_3^4(16-x^2)dx=\frac{128\pi }{3}-18\pi +42\pi +\frac{128\pi }{3}-18\pi =\frac{224\pi }{3}\approx 234.5$$ Interpretation: The manufacturer needs approximately 235 cubic meters of plastic to build this hollow spherical storage tank.

7. Quick Reference Cheatsheet

8. What's Next

The washer method builds on your knowledge of area between two curves and extends integration from 2D area to 3D volume, a core application of integration that is heavily tested on the AP exam. This topic is a prerequisite for the cylindrical shells method, the other major volume technique for solids of revolution, and it also shares the same cross-section integration logic used for volumes with known cross-sections, which you will study next. Without mastering how to correctly identify radii, choose the right integration variable, and set up the washer integral, you will struggle to earn full points on all volume-based FRQ and MCQ problems, which make up a large portion of Unit 8's exam weight.

• Volumes with known cross-sections
• Disk method for solids of revolution
• Area between two curves
• Cylindrical shells method