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Calculus Honors · AP-aligned study guide

5.4 · Volumes with known cross sections

Convert the width of a base region into a cross-sectional area, then integrate that area through the solid.

AP 8.7–8.8 · Reading: Larson & Edwards §7.2

Learning objectives

  • Write a cross-sectional area function from a base width.
  • Distinguish square, semicircle, and triangular cross sections.
  • Interpret a volume integral as area times thickness.

The base width is not the cross-sectional area

For a solid with square sections perpendicular to the x-axis, the side length is the width of the base region at x\displaystyle x. If the upper boundary is f\displaystyle f and lower boundary is g\displaystyle g, the width is w(x)=f(x)−g(x)\displaystyle w(x)=f(x)-g(x) and area is w(x)2\displaystyle w(x)^2.

The volume is the integral of the cross-sectional area, not the integral of the base width. The solid does not need to be generated by rotation.

V=∫abA(x) dxV=\int_a^b A(x)\,dx

The section shape determines the coefficient

A square of side w\displaystyle w has area w2\displaystyle w^2. A semicircle with diameter w\displaystyle w has radius w/2\displaystyle w/2 and area πw2/8\displaystyle \pi w^2/8. An equilateral triangle of side w\displaystyle w has area 3 w2/4\displaystyle \sqrt3\,w^2/4.

If triangles have a specified height rather than an equilateral shape, use A=bh/2\displaystyle A=bh/2. Read whether the base width represents a radius, diameter, side, or another dimension.

Match section orientation to the variable

Sections perpendicular to the x-axis produce dx\displaystyle dx slices; sections perpendicular to the y-axis produce dy\displaystyle dy slices. Rewrite the base boundaries in terms of the slice variable.

Check units and endpoint behavior. If a base region narrows to width zero, the cross-sectional area should also approach zero for these similar shapes.

Worked examples

Square sections

The base lies between y=0\displaystyle y=0 and y=x\displaystyle y=x on [0,2]\displaystyle [0,2]. Sections perpendicular to the x-axis are squares. Find the volume.

  1. The side length is x\displaystyle x.
  2. Area is x2\displaystyle x^2, so V=∫02x2 dx\displaystyle V=\int_0^2x^2\,dx.
  3. The volume is 8/3\displaystyle 8/3.

Semicircular sections

Use the same base with semicircles whose diameters equal the base width.

  1. The diameter is x\displaystyle x, so the radius is x/2\displaystyle x/2.
  2. Cross-sectional area is πx2/8\displaystyle \pi x^2/8.
  3. The volume is (π/8)(8/3)=π/3\displaystyle (\pi/8)(8/3)=\pi/3.

Common mistakes

  • Integrating base width instead of cross-sectional area.
  • Treating a semicircle’s diameter as its radius.
  • Adding a rotation factor when the solid is defined by noncircular cross sections.

Try it yourself

No calculator required.

1. The base is between y=x\displaystyle y=x and y=x2\displaystyle y=x^2 on [0,1]\displaystyle [0,1]. Perpendicular square sections have a side equal to the base width. Find the volume.

The region between y = x and y = x squared on [0, 1], with a vertical width indicated.
Each vertical width becomes the side of a square section.
Show worked solution

The width is x−x2\displaystyle x-x^2. Thus V=∫01(x−x2)2 dx=1/3−1/2+1/5=1/30\displaystyle V=\int_0^1(x-x^2)^2\,dx=1/3-1/2+1/5=1/30.

The region between y = x and y = x squared on [0, 1], with a vertical width indicated. The side is x minus x squared; cross-sectional area is its square.
The side is x minus x squared; cross-sectional area is its square.

2. For the same base, use semicircular sections with diameter equal to the width. Find the volume.

Show worked solution

The section area is π(x−x2)2/8\displaystyle \pi(x-x^2)^2/8. Therefore V=(π/8)(1/30)=π/240\displaystyle V=(\pi/8)(1/30)=\pi/240.

3. A solid has triangular sections with base x\displaystyle x and height 2x\displaystyle 2x for 0≤x≤3\displaystyle 0\leq x\leq3. Find its volume.

Show worked solution

The section area is (1/2)x(2x)=x2\displaystyle (1/2)x(2x)=x^2. Hence V=∫03x2 dx=9\displaystyle V=\int_0^3x^2\,dx=9.

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