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8.12 Volume with Washer Method: Revolving Around Other Axes

Syllabus
2020
Topic
8.12
Level
—

Learning objectives

—CHA-5.C—Calculate volumes of solids of revolution using definite integrals—Topic 8.12• CHA-5.C Calculate volumes of solids of revolution using definite integrals.• CHA-5.C.4 Volumes of solids of revolution around any horizontal or vertical line 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.

Measure Both Washer Radii from the Rotation Line

A region rotated around a horizontal line y=ky=k or vertical line x=hx=h forms washers when neither boundary reaches the axis. Measure both radii perpendicular to the axis. The outer radius RR is the farther distance and the inner radius rr is the nearer distance, regardless of which function is named first.

V=\pi\int\big(R^2-r^2\big),d(\text{perpendicular variable}),\qquad R=\text{farther distance to the axis},\quad r=\text{nearer distance}

  1. Use slices perpendicular to the rotation line: dxdx for a horizontal axis, dydy for a vertical axis.\n2. Write each boundary's distance from the line.\n3. Compare those distances and label the larger one RR.\n4. Integrate π(R2−r2)\pi(R^2-r^2) over the matching bounds.\n5. Split the interval if the farther boundary changes.

Example: rotate the region between y=x2y=x^2 and y=xy=x on 0≤x≤10\le x\le1 around y=2y=2. The axis lies above both curves. The lower curve y=x2y=x^2 is farther away, so R(x)=2−x2R(x)=2-x^2; the upper curve is nearer, so r(x)=2−xr(x)=2-x. Thus V=π∫01[(2−x2)2−(2−x)2]dx=π∫01(4x−5x2+x4) dx=8π15V=\pi\int_0^1\left[(2-x^2)^2-(2-x)^2\right]dx=\pi\int_0^1(4x-5x^2+x^4)\,dx=\frac{8\pi}{15} cubic units.

“Upper” does not always mean “outer.” With an axis above the region, the lower boundary is farther from the axis; with an axis below, the upper boundary may be farther. Determine distance first, then square, and keep the area as π(R2−r2)\pi(R^2-r^2) rather than π(R−r)2\pi(R-r)^2.