Volume of Rotation: Washer Method Simulator & Guide

The washer method finds the volume of a solid of revolution with a gap by subtracting the inner radius squared from the outer radius squared: volume equals pi times the integral from a to b of the outer radius squared minus the inner radius squared.

The Washer Method is just doing the disk method twice. You do it first for the outer radius, the larger circle, and then you do it for the smaller circle. Then you just subtract the volume of the smaller circle from the larger circle.

What is the washer method formula for volume?

\begin{gathered} \underbrace{V = \pi \int_{a}^{b} \left( [R(x)]^2 - [r(x)]^2 \right) dx}_{\text{Textbook Formula}} \\ V = \int_{a}^{b} \underbrace{\pi [\textcolor{#eab308}{R(x)}]^2}_{\text{Area of Big Circle}} dx - \int_{a}^{b} \underbrace{\pi [\textcolor{#60a5fa}{r(x)}]^2}_{\text{Area of Little Circle}} dx \end{gathered}

Washer method worked example: region between two curves

Find the volume of the solid formed by revolving the region bounded by y = x² and y = 2x about the x-axis.

1. Find the intersection points: x² = 2x ⟹ x(x - 2) = 0 ⟹ x = 0, 2.

2. The outer function is R(x) = 2x and the inner is r(x) = x².

3. Set up the integral:

V = \pi \int_{0}^{2} \left( (2x)^2 - (x^2)^2 \right) dx = \pi \int_{0}^{2} (4x^2 - x^4) dx

4. Integrate:

V = \pi \left[ \frac{4x^3}{3} - \frac{x^5}{5} \right]_{0}^{2} = \pi \left( \frac{32}{3} - \frac{32}{5} \right) = \frac{64\pi}{15}

Another worked example

Find the volume when the region between y = x and y = x squared, from x = 0 to x = 1, is revolved about the x-axis. Step 1: the outer radius is x and the inner radius is x squared. Step 2: integrate pi times (x squared minus x to the 4th) from 0 to 1. Step 3: that gives pi times (one-third minus one-fifth), which equals 2 pi over 15.

Aligned with the College Board AP Calculus CED — Unit 8 (Applications of Integration), Topic 8.10 (Volumes with Washers). College Board ↗

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