Volumes by Cross-Sections

Main idea

If you can find the area of every slice of a solid, you can find its volume: add up slice area times thickness, $\int A(x)\,dx$ . The slices don't have to be disks; squares, triangles and semicircles work just as well.

V = \int_a^b {\htmlData{sym=0}{A(x)}}\,dx

What the symbols mean

$A(x)$
Cross-section area
The area of the slice at position x: a square, triangle, semicircle, or disk.

Explanation

Volume by cross-sections is simply finding the volume of a solid by slicing it like bread and adding up the slices. All we're really saying is that a thin slice's volume is about its face area times its thickness, and the integral adds up all the slices. The formula has one new piece: $A(x)$ (the area of the slice at position $x$ , whatever shape it is). Everything else, the $\int$ and the $dx$ , is the same adding-up you've been doing all unit.

Notice that this is the idea behind the disk and washer methods, with the disk part removed. For a spun solid, every slice is a circle, so $A(x) = \pi R^2$ . But nothing about $\int A(x)\,dx$ needs the slice to be a circle. As long as you can write down the area of each slice, the same integral works.

Steps:

Find the slice at position $x$ : its shape and the length that determines its size (often the width of a base region).
Write the area $A(x)$ using the shape's area formula.
Integrate across the solid.

Shape formulas: square $s^2$ ; equilateral triangle $\frac{\sqrt3}{4}s^2$ ; semicircle with diameter $s$ : $\frac\pi8 s^2$ .

This is also how the classic formulas are proven: a pyramid's cross-sections shrink linearly, so its volume is $\frac13$ base × height.

See it

This solid sits on the disk x² + y² ≤ 4, and every slice across it is a square standing straight up out of the page (drawn tilted back so you can see it). The square's side is the width of the disk at that spot, 2√(4 − x²), so its area is 4(4 − x²). As you can see, the squares are biggest in the middle, area 16, and shrink to nothing at the edges. Adding them up gives V = ∫ 4(4 − x²) dx = 128/3 ≈ 42.7.

Worked examples

Example 1

Find the volume of a pyramid with a square base of side $s$ and height $h$ .

1
Slice horizontally. The square shrinks steadily from side $s$ at the base to $0$ at the top, so at height $y$ its side is $s\frac{h - y}{h}$ .

Example 2

A solid's base is the disk $x^2 + y^2 \le 1$ , and its cross-sections perpendicular to the x-axis are squares. Find its volume.

1
At position $x$ , the disk runs from $y = -\sqrt{1 - x^2}$ to $y = \sqrt{1 - x^2}$ , so the square's side is $2\sqrt{1 - x^2}$ and $A(x) = 4(1 - x^2)$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

A pyramid has a square base of side 6 and height 4. Find its volume.

Problem 2+5 XP

A solid's base is the region between $y = x^2$ and $y = 1$ . Cross-sections perpendicular to the y-axis are squares. Find the volume.

Problem 3+5 XP

A solid's base is the unit disk; cross-sections perpendicular to the x-axis are equilateral triangles. Find the volume.

Problem 4+5 XP

Use slices to find the volume of a cone with radius 3 and height 5.

Problem 5+5 XP

A solid's base is the triangle with vertices $(0,0)$ , $(1,0)$ , $(0,1)$ ; cross-sections perpendicular to the x-axis are semicircles. Find the volume.

Problem 6+5 XP

The Great Pyramid of Giza was about $146$ m tall, with a square base about $230$ m on a side. Its horizontal slice at height $y$ is a square of side $230\left(1 - \frac{y}{146}\right)$ . Find its volume.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Two pipes of radius $1$ cross at right angles: the cylinders $x^2 + y^2 \le 1$ and $x^2 + z^2 \le 1$ . Find the volume of the region inside both.

Where it's used

CT and MRI scans measure organ and tumor volumes exactly this way: the scanner produces cross-sectional slices, and software adds up area × thickness.

Volumes by Cylindrical Shells Work