Mathematics › 🔬 Grade 12 › The Potter's Wheel: Volumes of Revolution

The Potter's Wheel: Volumes of Revolution

Spin a flat region about an axis and it sweeps out a solid — stack up its circular cross-sections and calculus gives the exact volume.

Grade 12Calculus / AP level

The Potter's Wheel: Volumes of Revolution — illustration

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The big idea: Spin a region about an axis, the way a potter shapes clay on a wheel, and the flat shape sweeps out a three-dimensional solid. Every cross-section perpendicular to the axis is a circle, so the solid is a stack of infinitely thin disks. Add up the disk volumes with an integral — π times radius squared times thickness — and you get the exact volume of revolution.

🎯 By the end, you'll be able to

Describe how rotating a region produces a solid with circular cross-sections
Set up the disk-method integral for a volume of revolution
Compute volumes of revolution about the x-axis
Extend the idea to the washer method when the solid has a hole

📎 You should already know

Evaluating definite integrals
Area under a curve
Area of a circle

Clay on the wheel

Watch a potter: a flat profile pressed against spinning clay traces out a bowl, a vase, a goblet. Mathematically, take a region bounded by a curve y = f(x) and the x-axis, and rotate it about that axis. The region sweeps through space and encloses a solid of revolution.

Because every point travels in a circle around the axis, the solid is perfectly round — each slice across it is a disk.

Slice it into disks

Cut the solid with planes perpendicular to the axis, spacing them a tiny width dx apart. Each slice is a thin coin — a disk — whose radius is the height of the curve, f(x), at that position. A disk of radius R and thickness dx has volume πR² dx. Summing all the disks with an integral gives the total volume.

🔑 The disk method

Rotating y = f(x) about the x-axis, the radius of each disk is R(x) = f(x). The volume is the integral of the disk areas: π times the radius squared, integrated across the interval.

\[ V = \pi \int_a^b \left[R(x)\right]^2 \, dx = \pi \int_a^b \left[f(x)\right]^2 \, dx \]

The disk method: add up pi-r-squared-thickness for every slice from x = a to x = b.

🎮 Potter's Wheel LIVE

Spin a region about an axis to sweep out a solid of revolution.

Radius squared, then integrate

The workflow: identify the radius as a function of x (the distance from the axis to the curve), square it, integrate over the interval, and multiply by π. The squaring is essential — the cross-section is a circle, whose area grows with the square of the radius.

📝 Worked example: The region under y = √x from x = 0 to x = 4 is rotated about the x-axis. Find the volume.

Radius of each disk: \( R(x) = \sqrt{x} \), so \( [R(x)]^2 = x \).
Set up the integral: \( V = \pi \displaystyle\int_0^4 x \, dx \).
Integrate: \( \pi\left[\dfrac{x^2}{2}\right]_0^4 = \pi\left(\dfrac{16}{2} - 0\right) = 8\pi \).

✓ The volume is <strong>8π</strong> cubic units.

📝 Worked example: The region under y = x from x = 0 to x = 1 is rotated about the x-axis. Find the volume (this is a cone).

Radius: \( R(x) = x \), so \( [R(x)]^2 = x^2 \).
Volume: \( V = \pi \displaystyle\int_0^1 x^2 \, dx = \pi\left[\dfrac{x^3}{3}\right]_0^1 = \dfrac{\pi}{3} \).
Check with the cone formula \( \tfrac13 \pi r^2 h \) with r = 1, h = 1: \( \tfrac13\pi(1)^2(1) = \dfrac{\pi}{3} \). ✓

✓ The volume is <strong>π/3</strong> — matching the classic cone formula, as it should.

When the solid has a hole: washers

If the region does not touch the axis — say it lies between an outer curve and an inner curve — then rotating it produces a solid with a hollow core. Each slice is a washer (a disk with a circular bite removed). Its area is the big circle minus the small circle: π(R² − r²).

\[ V = \pi \int_a^b \left(\left[R_{\text{outer}}(x)\right]^2 - \left[r_{\text{inner}}(x)\right]^2\right) dx \]

The washer method: subtract the inner disk from the outer disk before integrating.

⚠️ Square the radius before integrating

You must square the radius inside the integral — the cross-sectional area is πR², not πR. And for a washer you cannot square the difference: (R − r)² is not R² − r². Subtract the two separate squares, π(R² − r²).

✨ Revolution makes every slice a circle

The reason volumes of revolution are so tractable is that rotation ensures circular cross-sections. Whatever the profile curve, the slices are disks (or washers), and a circle’s area is the simple πR². That is why a single integral captures the whole 3-D solid.

Check your understanding

1. When a region is rotated about the x-axis (with no gap to the axis), each cross-section perpendicular to the axis is a:

Every point circles the axis, so each perpendicular slice is a circular disk of radius f(x).

2. The volume of one thin disk of radius R and thickness dx is:

A disk's area is πR², so its volume is area times thickness: πR² dx.

3. Rotating y = √x on [0, 4] about the x-axis gives a volume of:

Here [R]² = (√x)² = x, so V = π∫₀⁴ x dx = π[x²/2]₀⁴ = π(8) = 8π.

4. Rotating y = x on [0, 1] about the x-axis gives a volume of:

[R]² = x², so V = π∫₀¹ x² dx = π[x³/3]₀¹ = π/3 — the cone formula (1/3)πr²h with r = h = 1.

5. When the rotated region leaves a gap from the axis, you should use the washer method with cross-sectional area:

A washer is the outer disk minus the inner disk: area = πR² − πr² = π(R² − r²). You cannot combine into (R − r)².

✅ Key takeaways

Rotating a flat region about an axis sweeps out a solid of revolution with circular cross-sections.
Each slice perpendicular to the axis is a disk of radius R(x) = f(x) and thickness dx, with volume πR² dx.
Disk method: V = π∫ₐᵇ [f(x)]² dx — square the radius, integrate, multiply by π.
Example: y = √x on [0,4] about the x-axis gives 8π; y = x on [0,1] gives π/3, matching the cone formula.
When the solid has a hollow core, each slice is a washer with area π(R² − r²).
Always square the radius inside the integral, and never replace R² − r² with (R − r)².