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Volume of Rotation: Washer Method Simulator & Guide

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.

Formula

V=π∫ab([R(x)]2−[r(x)]2)dx⏟Textbook FormulaV=∫abπ[R(x)]2⏟Area of Big Circledx−∫abπ[r(x)]2⏟Area of Little Circledx\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}

Easy Example Problem

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=π∫02((2x)2−(x2)2)dx=π∫02(4x2−x4)dxV = \pi \int_{0}^{2} \left( (2x)^2 - (x^2)^2 \right) dx = \pi \int_{0}^{2} (4x^2 - x^4) dx

4. Integrate:

V=π[4x33−x55]02=π(323−325)=64π15V = \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}
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