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8.11 Volume with Washer Method: Revolving Around the x- or y-Axis

Syllabus
2020
Topic
8.11
Level
—

Learning objectives

—CHA-5.C—Calculate volumes of solids of revolution using definite integrals—Topic 8.11• CHA-5.C Calculate volumes of solids of revolution using definite integrals.• CHA-5.C.3 Volumes of solids of revolution around the x- or y-axis whose cross sections are ring shaped may be found using definite integrals with the washer method.• Enduring understanding CHA-5: Definite integrals allow us to solve problems involving the accumulation of change in area or volume over an interval.

Subtract the Hole with the Washer Method

If a region does not reach the axis of rotation, a perpendicular slice produces a washer rather than a solid disc. Let RR be the farther distance from the axis and rr the nearer distance, with R≥r≥0R\ge r\ge0. The washer area is the area of the outer circle minus the inner circular hole.

A=\pi(R^2-r^2);\qquad \text{about the }x\text{-axis: }V=\pi\int_a^b\big(R(x)^2-r(x)^2\big),dx;\qquad \text{about the }y\text{-axis: }V=\pi\int_c^d\big(R(y)^2-r(y)^2\big),dy

  1. Use slices perpendicular to the rotation axis.\n2. Measure both boundary distances from that axis.\n3. Label the farther distance RR and the nearer distance rr.\n4. Integrate π(R2−r2)\pi(R^2-r^2) over the appropriate bounds.\n5. Check that R≥rR\ge r throughout each interval.

Example: rotate the region between y=xy=\sqrt{x} and y=x2y=x^2 on 0≤x≤10\le x\le1 around the xx-axis. Since x≥x2\sqrt{x}\ge x^2, R(x)=xR(x)=\sqrt{x} and r(x)=x2r(x)=x^2. Thus V=π∫01[(x)2−(x2)2]dx=π∫01(x−x4) dx=π[x22−x55]01=3π10V=\pi\int_0^1\left[(\sqrt{x})^2-(x^2)^2\right]dx=\pi\int_0^1(x-x^4)\,dx=\pi\left[\frac{x^2}{2}-\frac{x^5}{5}\right]_0^1=\frac{3\pi}{10} cubic units.

A washer is a difference of two circular areas: πR2−πr2=π(R2−r2)\pi R^2-\pi r^2=\pi(R^2-r^2). It is not π(R−r)2\pi(R-r)^2. If the farther and nearer boundaries exchange, split the integral where their roles change so the cross-sectional area stays nonnegative.