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8.8 Volumes with Cross Sections: Triangles and Semicircles

Syllabus
2020
Topic
8.8
Level
—

Learning objectives

—CHA-5.B—Calculate volumes of solids with known cross sections using definite integrals—Topic 8.8• CHA-5.B Calculate volumes of solids with known cross sections using definite integrals.• CHA-5.B.2 Volumes of solids with triangular cross sections can be found using definite integrals and the area formulas for these shapes.• CHA-5.B.3 Volumes of solids with semicircular and other geometrically defined cross sections can be found using definite integrals and the area formulas for these shapes.• Enduring understanding CHA-5: Definite integrals allow us to solve problems involving the accumulation of change in area or volume over an interval.

Turn a Base Segment into a Geometric Cross Section

Let s(x)s(x) be the segment cut from the base region by a slice perpendicular to the xx-axis. The problem states what that segment represents—such as a triangle base, a triangle leg, or a semicircle diameter. First convert s(x)s(x) into cross-sectional area A(x)A(x); then accumulate those areas with V=∫abA(x) dxV=\int_a^b A(x)\,dx.

Cross section and meaning of ss Area function
Triangle with base ss and height h(x)h(x) A(x)=12s(x)h(x)A(x)=\frac12s(x)h(x)
Equilateral triangle with side ss A(x)=34[s(x)]2A(x)=\frac{\sqrt3}{4}[s(x)]^2
Semicircle with diameter ss A(x)=π8[s(x)]2A(x)=\frac{\pi}{8}[s(x)]^2
  1. Find s(x)s(x) from the base boundaries.
  2. Read exactly what geometric dimension s(x)s(x) represents.
  3. Substitute it into the stated shape's area formula.
  4. Integrate the resulting A(x)A(x) over the base interval.
  5. Check that the volume is nonnegative and has cubic units.

Example: the base lies between y=xy=x and y=0y=0 for 0≤x≤20\le x\le2, and each perpendicular cross section is a semicircle whose diameter is the vertical segment. Thus s(x)=xs(x)=x, so A(x)=π8x2A(x)=\frac{\pi}{8}x^2. Therefore V=∫02π8x2 dx=π8[x33]02=π3V=\int_0^2\frac{\pi}{8}x^2\,dx=\frac{\pi}{8}\left[\frac{x^3}{3}\right]_0^2=\frac{\pi}{3} cubic units.

For a semicircle, do not use the given diameter as the radius. If the base segment is the diameter ss, then r=s/2r=s/2 and A=12π(s/2)2=πs2/8A=\frac12\pi(s/2)^2=\pi s^2/8. Likewise, a triangle needs both base and height unless its type fixes their relationship.