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8.10 Volume with Disc Method: Revolving Around Other Axes

Syllabus
2020
Topic
8.10
Level
—

Learning objectives

—CHA-5.C—Calculate volumes of solids of revolution using definite integrals—Topic 8.10• CHA-5.C Calculate volumes of solids of revolution using definite integrals.• CHA-5.C.2 Volumes of solids of revolution around any horizontal or vertical line in the plane may be found by using definite integrals with the disc 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 Disc Radius from a Shifted Axis

For rotation around a line other than a coordinate axis, the disc radius is not usually the function value. It is the perpendicular distance from the axis of rotation to the outer boundary. The region must reach the axis so each perpendicular slice forms a solid disc.

y=k:\ R(x)=|f(x)-k|,\quad V=\pi\int_a^b[R(x)]^2,dx;\qquad x=h:\ R(y)=|g(y)-h|,\quad V=\pi\int_c^d[R(y)]^2,dy

  1. Mark the horizontal or vertical axis of rotation.
  2. Choose slices perpendicular to it.
  3. Write radius as outer boundary minus axis, or axis minus outer boundary, so it represents a nonnegative distance.
  4. Square the radius, multiply by π\pi, and integrate over the matching variable's bounds.

Example: rotate the region between y=−1y=-1 and y=x2y=x^2 for 0≤x≤10\le x\le1 about y=−1y=-1. Each vertical slice forms a disc with R(x)=x2−(−1)=x2+1R(x)=x^2-(-1)=x^2+1. Thus V=π∫01(x2+1)2 dx=π[x55+2x33+x]01=28π15V=\pi\int_0^1(x^2+1)^2\,dx=\pi\left[\frac{x^5}{5}+\frac{2x^3}{3}+x\right]_0^1=\frac{28\pi}{15} cubic units.

Do not use R=f(x)R=f(x) merely because the boundary is y=f(x)y=f(x); that is valid only when the axis is y=0y=0. For a shifted axis, compute the distance first. If the rotated region leaves a central gap, the cross section is not a disc and this single-radius formula does not apply.