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Washer Method

Washer Method

A technique for finding the volume of a solid of revolution. The washer method is a generalized version of the disk method. Both the washer and disk methods are specific cases of volume by parallel cross-sections.

Graph showing washer method: Volume = integral from a to b of π([f(x)]²−[g(x)]²)dx, with y=f(x) outer curve, y=g(x) inner curve.

 

 

See also

Shell method, axis of rotation

Key Formula

V=πab[R(x)2r(x)2]dxV = \pi \int_a^b \left[ R(x)^2 - r(x)^2 \right] \, dx
Where:
  • VV = Volume of the solid of revolution
  • R(x)R(x) = Outer radius — the distance from the axis of rotation to the farther curve
  • r(x)r(x) = Inner radius — the distance from the axis of rotation to the closer curve
  • a,ba, b = Bounds of integration along the axis of rotation
  • dxdx = Indicates integration with respect to x (use dy when rotating around a horizontal line and slicing horizontally)

Worked Example

Problem: Find the volume of the solid formed by rotating the region between y = x² and y = x about the x-axis.
Step 1: Find where the curves intersect by setting them equal. This gives the bounds of integration.
x2=x    x2x=0    x(x1)=0    x=0,  x=1x^2 = x \implies x^2 - x = 0 \implies x(x - 1) = 0 \implies x = 0, \; x = 1
Step 2: Identify the outer and inner radii. On [0, 1], the line y = x is above y = x², so it is farther from the x-axis. The outer radius is R(x) = x and the inner radius is r(x) = x².
R(x)=x,r(x)=x2R(x) = x, \quad r(x) = x^2
Step 3: Set up the washer method integral.
V=π01[x2(x2)2]dx=π01(x2x4)dxV = \pi \int_0^1 \left[ x^2 - (x^2)^2 \right] dx = \pi \int_0^1 \left( x^2 - x^4 \right) dx
Step 4: Evaluate the integral by finding the antiderivative.
V=π[x33x55]01=π(1315)=π215V = \pi \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_0^1 = \pi \left( \frac{1}{3} - \frac{1}{5} \right) = \pi \cdot \frac{2}{15}
Step 5: Simplify to get the final volume.
V=2π15V = \frac{2\pi}{15}
Answer: The volume of the solid is 2π/15 cubic units.

Another Example

This example differs because the axis of rotation (y = 3) is not the x-axis itself. It shows how to compute radii as distances from a shifted axis, which is a common source of confusion.

Problem: Find the volume of the solid formed by rotating the region between y = √x and y = 0, from x = 0 to x = 4, about the line y = 3.
Step 1: Sketch the setup. The region sits below y = √x and above y = 0, but the axis of rotation is y = 3, which is above the entire region. Each cross-section is a washer.
Step 2: Determine the outer and inner radii. The outer radius is the distance from y = 3 down to the farther boundary y = 0. The inner radius is the distance from y = 3 down to the closer boundary y = √x.
R(x)=30=3,r(x)=3xR(x) = 3 - 0 = 3, \quad r(x) = 3 - \sqrt{x}
Step 3: Set up the integral with bounds x = 0 to x = 4.
V=π04[32(3x)2]dxV = \pi \int_0^4 \left[ 3^2 - (3 - \sqrt{x})^2 \right] dx
Step 4: Expand the squared term and simplify the integrand.
9(96x+x)=6xx9 - (9 - 6\sqrt{x} + x) = 6\sqrt{x} - x
Step 5: Evaluate the integral.
V=π04(6xx)dx=π[623x3/2x22]04=π[4x3/2x22]04V = \pi \int_0^4 \left(6\sqrt{x} - x\right) dx = \pi \left[ 6 \cdot \frac{2}{3} x^{3/2} - \frac{x^2}{2} \right]_0^4 = \pi \left[ 4x^{3/2} - \frac{x^2}{2} \right]_0^4
Step 6: Substitute the bounds and simplify.
V=π(4(8)162)=π(328)=24πV = \pi \left( 4(8) - \frac{16}{2} \right) = \pi(32 - 8) = 24\pi
Answer: The volume of the solid is 24π cubic units.

Frequently Asked Questions

What is the difference between the disk method and the washer method?
The disk method is a special case of the washer method where the inner radius is zero — meaning the region touches the axis of rotation. The washer method handles the general case where there is a gap between the region and the axis, producing a hollow center (like a washer or ring). If r(x) = 0, the washer formula reduces to the disk formula V = π∫R(x)² dx.
When should you use the washer method instead of the shell method?
Use the washer method when you can easily express both radii as functions of the variable along the axis of rotation. If the axis of rotation is horizontal (like the x-axis), washers slice vertically and you integrate with respect to x. If setting up the radii becomes difficult — for example, if you'd need to solve complicated equations for x in terms of y — the cylindrical shell method may be simpler. Neither method is universally easier; the best choice depends on the problem.
How do you set up the washer method around a vertical line?
When rotating around a vertical line such as x = c, you integrate with respect to y. The outer radius R(y) and inner radius r(y) are the horizontal distances from the axis x = c to each bounding curve, expressed as functions of y. The formula becomes V = π∫[R(y)² − r(y)²] dy.

Washer Method vs. Cylindrical Shell Method

Washer MethodCylindrical Shell Method
Cross-section shapeRing (washer): a disk with a concentric holeThin cylindrical shell (hollow tube)
FormulaV = π∫[R(x)² − r(x)²] dxV = 2π∫x · f(x) dx
Slice directionPerpendicular to the axis of rotationParallel to the axis of rotation
Best used whenBoth radii are easily expressed as functions of the integration variableExpressing the height of the shell is easier than solving for radii
Axis of rotationTypically horizontal axis → integrate with respect to x; vertical axis → with respect to yTypically vertical axis → integrate with respect to x; horizontal axis → with respect to y

Why It Matters

The Washer Method appears throughout AP Calculus AB/BC and college Calculus II courses as a core technique for computing volumes of solids of revolution. Many real-world objects — pipes, rings, bowls, and bottles — have hollow interiors that the disk method alone cannot model. Mastering the washer method also deepens your understanding of how integration builds 3D volumes from 2D cross-sections.

Common Mistakes

Mistake: Subtracting the radii before squaring, writing π∫(R − r)² dx instead of π∫(R² − r²) dx.
Correction: You must square each radius separately and then subtract. The area of a washer is πR² − πr², not π(R − r)². Squaring the difference gives a completely different (and incorrect) result because (R − r)² = R² − 2Rr + r² ≠ R² − r².
Mistake: Mixing up which curve gives the outer radius and which gives the inner radius, especially when the axis of rotation is not y = 0 or x = 0.
Correction: Always measure each radius as a positive distance from the axis of rotation. The outer radius is the distance to the curve farther from the axis; the inner radius is the distance to the curve closer to the axis. A quick sketch helps you identify which is which.

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