Volumes by Known Cross-Sections
Compute volumes of solids whose cross-sections perpendicular to an axis are known geometric shapes (squares, rectangles, semicircles, equilateral triangles). Set up the integral V = ∫A(x)dx.
A lesson of AP Calculus AB, in Applications of Integration, about 10 minutes.
Volumes with Known Cross Sections
By the end of this lesson, you will be able to set up and evaluate integrals to find volumes of solids whose cross sections are known geometric shapes.
The General Principle
A solid with known cross sections has cross sections (slices perpendicular to an axis) whose shapes are given. If each cross section has area $A(x)$ at position $x$, and we slice into thin slabs of thickness $dx$, then the volume of each slab is $A(x), dx$, and the total volume is:
$$V = \int_a^b A(x), dx$$
Common Cross Section Shapes
| Cross section | Area formula |
|---|---|
| Square with side $s$ | $A = s^2$ |
| Equilateral triangle with side $s$ | $A = \frac{\sqrt{3}}{4}s^2$ |
| Semicircle with diameter $d$ | $A = \frac{\pi}{8}d^2$ (or $\frac{\pi r^2}{2}$ where $r = d/2$) |
| Rectangle with width $w$ and fixed height $h$ | $A = wh$ |
Setting Up the Integral
Key: express the side length $s$ (or diameter $d$) in terms of $x$ using the given boundary functions.
If the cross sections are perpendicular to the x-axis: $s(x) = [\text{top function}] - [\text{bottom function}] = f(x) - g(x)$
If the cross sections are perpendicular to the y-axis: Work in terms of $y$.
Example 1: Square Cross Sections
The base of a solid is the region bounded by $y = \sqrt{x}$ and $y = x$ (from 0 to 1). Cross sections perpendicular to the x-axis are squares with sides from the bottom to top boundary.
Side length: $s(x) = \sqrt{x} - x$
$A(x) = s^2 = (\sqrt{x} - x)^2 = x - 2x^{3/2} + x^2$
$V = \int_0^1 (x - 2x^{3/2} + x^2), dx = \left[\frac{x^2}{2} - \frac{4x^{5/2}}{5} + \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{4}{5} + \frac{1}{3} = \frac{15-24+10}{30} = \frac{1}{30}$
Example 2: Semicircular Cross Sections
The base is the region bounded by $y = 4 - x^2$ and the x-axis (from $-2$ to 2). Cross sections perpendicular to the x-axis are semicircles with the diameter along the base.
Diameter: $d(x) = 4 - x^2$; radius: $r(x) = \frac{4-x^2}{2}$
$A(x) = \frac{\pi r^2}{2} = \frac{\pi}{2}\left(\frac{4-x^2}{2}\right)^2 = \frac{\pi(4-x^2)^2}{8}$