Module 5 · Lesson 5.5

Volumes of Revolution — Slicing 3D Solids

Spin a flat 2D region around an axis to sweep out a 3D solid, then find its exact volume by stacking infinitely thin disks.

Intuition

Take a flat shape, like the region under a curve, and spin it a full turn around a line. Every point traces out a circle, and the whole flat region sweeps through space to become a solid — a vase, a cone, a lightbulb shape, depending on the curve. To find how much space it fills, slice it back into thin coins.

Core concept

Each thin slice of the solid is a disk: a flat circle with radius $f(x)$ and a tiny thickness $dx$. A disk's volume is $\pi [f(x)]^2\,dx$ — circle area times thickness. Stack these disks along the axis from $a$ to $b$, and their total volume is the volume of the whole solid. It's the exact same accumulation idea as area under a curve, just measuring circles instead of rectangles.

f(x) = x, spun about the axis
Curve & intervalf(x) = x, from 0 to 2
Disk radius at xf(x) = x
Disk volume elementπ x² dx
Antiderivativeπ x³/3
Evaluate at 2 minus at 0π(8/3) − 0
AnswerV = 8π/3 ≈ 8.38

Interactive Lab

Slide to extend how far along the axis the solid has been swept. Each disk stamped onto the axis represents one thin slice, and the running volume total updates as more disks stack up.

axis of revolution
x = 2.0 radius f(x) = 2.0 volume so far = 8.38

Key Idea

$$ V = \pi \int_a^b \big[f(x)\big]^2 \, dx $$

This is the disk method: square the radius function to get each disk's circular cross-sectional area, multiply by $\pi$, and accumulate from $a$ to $b$ exactly like any other definite integral. The only new ingredient is that each thin slice is a circle instead of a rectangle.

$f(x)$The radius of the disk at position $x$, taken from the generating curve
$\pi[f(x)]^2$The cross-sectional area of one circular disk
$a, b$The start and end of the axis segment being swept
$V$The total volume of the solid formed by revolution

Think Further

  1. Why does the disk method square $f(x)$ instead of just using $f(x)$ directly, the way the area-under-a-curve formula does?
  2. A 3D printer builds an object layer by layer, each layer a thin disk of measured radius. How is that physical process the same accumulation idea as the disk method?
  3. Predict what happens to the volume formula if the generating curve is doubled to $2f(x)$ instead of $f(x)$. Does the volume double, or change by some other factor?
Show suggested answers
  1. Because a disk's cross section is a circle, and a circle's area depends on the square of its radius ($\pi r^2$) — squaring $f(x)$ turns the radius into the correct cross-sectional area before accumulating.
  2. Each printed layer is a physical disk of known radius and small thickness, and stacking their volumes is exactly the same disk-accumulation idea the integral formalizes — the printer is running the Riemann sum in real material.
  3. The volume scales by a factor of 4, not 2. Since the formula squares the radius, doubling $f(x)$ makes $[2f(x)]^2 = 4[f(x)]^2$, quadrupling each disk's area and therefore the total volume.