Calculus

Volumes of Revolution

You know how to find the area under a curve. Now imagine spinning that curve around the x-axis, like a potter shaping clay on a wheel. The result is a 3D solid, and calculus can tell you its exact volume.

1. From Area to Volume: The Disk Method

When you rotate the curve y = f(x) around the x-axis, each thin vertical strip sweeps out a disk — a flat cylinder. The radius of the disk at position x is f(x), and its thickness is dx.

Volume of one disk=π[f(x)]2 dx\text{Volume of one disk} = \pi [f(x)]^2 \, dx
Total volume=∫abπ[f(x)]2 dx\text{Total volume} = \int_a^b \pi [f(x)]^2 \, dx

Think of it as stacking infinitely many coins of varying size.

2. Rotate y = sqrt(x)

Let’s rotate the curve y = sqrt(x) from x = 0 to x = b. Each disk has radius sqrt(x), so its area is pi * x.

b (right boundary)4
0.56
V=∫04π(0)2 dx=∫04πx dx=π⋅420V = \int_0^{{4}} \pi (\sqrt0)^2 \, dx = \int_0^{{4}} \pi x \, dx = \frac{\pi \cdot 4^2}0
246-4-22468101214161820y = sqrt(x) (top half)y = -sqrt(x) (bottom half)pi x (disk area at x)
Try This

Try this: The blue curves show the top and bottom halves of the shape you would see if you sliced the solid along the x-axis. The red curve shows how the cross-sectional area (of each disk) grows as you move right. The total volume is the area under the red curve from 0 to b.

3. Cross-Section Radius = f(x)

The key to the disk method is recognizing that the radius of each cross-section equals the function value. Let’s explore this with a different function.

k (controls the curve shape)1
0.53
f(x)=1⋅sin⁡(x),0=1sin⁡(x)f(x) = 1 \cdot \sin(x), \quad \text0 = 1 \sin(x)
V=∫0ππ⋅(1sin⁡x)2 dx=12π20V = \int_0^\pi \pi \cdot (1 \sin x)^2 \, dx = \frac{{1}^2 \pi^2}0
24-4-2246810f(x) = k sin(x) (top)-k sin(x) (bottom)pi [f(x)]² (disk area)
Connection

The pair of curves (top and bottom) shows the profile of the solid of revolution. Rotating k*sin(x) from 0 to pi creates a shape like a football. Increasing k makes the football wider, and the volume scales with k^2 — because the disk area depends on the square of the radius.

4. Comparing Different Solids

Different functions produce dramatically different solids. Let’s compare three rotations on the interval [0, 2]:

51015202530pi x² (rotate y=x)pi x⁴ (rotate y=x²)pi x (rotate y=sqrt(x))
Rotate y=x:V=πb30\text{Rotate } y = x: \quad V = \frac{\pi b^3}0
Rotate y=x2:V=πb50\text{Rotate } y = x^2: \quad V = \frac{\pi b^5}0
Rotate y=0:V=πb20\text{Rotate } y = \sqrt0: \quad V = \frac{\pi b^2}0

The steeper the function grows, the faster the disk areas increase, and the larger the volume.

5. Adjusting the Bounds

The limits of integration control which part of the curve gets rotated. Slide the bounds to see how the volume changes.

a (left bound)0
02
b (right bound)3
15
V=∫00πx2 dx=π0(33−03)V = \int_{{0}}^{0} \pi x^2 \, dx = \frac{\pi}0\left(3^3 - 0^3\right)
5-55101520253035404550f(x) = x-f(x) = -xpi x² (disk area)
Try This

Try this: With a = 0 and b = 3, you get a cone (volume = 9pi). Move a to 1: now you have a frustum (a cone with the tip sliced off). The volume is pi/3 * (27 - 1) = 26pi/3. The disk method handles any bounds naturally.
Challenge

Challenge: What solid do you get when you rotate the line y = 3 (a horizontal line) from x = 0 to x = h around the x-axis? What is its volume? You should recognize a familiar shape.

Hint: every disk has the same radius.

The Big Idea

To find the volume of a solid of revolution, slice it into infinitely thin disks, find each disk’s area (pi times the radius squared), and integrate.

The disk method transforms a 3D volume problem into a 1D integral. The function value f(x) gives you the radius, squaring it gives the disk area, and integrating adds up all the disks. It is the same slice-and-sum philosophy as Riemann sums, but now in three dimensions.

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