Pappus's Theorems
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 2.6: “Moments and Centers of Mass” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/2-6-moments-and-centers-of-mass |
| Textbook used in class | Stewart, Calculus, Section 8.3: “Applications to Physics and Engineering” |
Opening Scenario
A donut (torus) is formed by rotating a disk of radius $r$ around an axis that is $R$ units away from the disk’s center. What is the volume of the donut? What is its surface area?
You could compute these directly with integrals. But Pappus’s theorems say: volume equals the area of the cross-section times the distance traveled by the centroid. Distance traveled by a centroid at radius $R$ in one full revolution: $2\pi R$. So $V = 2\pi R \cdot \pi r^2$ and $S = 2\pi R \cdot 2\pi r$. No integration needed beyond knowing the centroid.
Quick Reference
Pappus’s Theorem (Volume). Let $R$ be a plane region of area $A$ with centroid at distance $\bar{d}$ from an axis that does not intersect $R$. The volume of the solid obtained by rotating $R$ about that axis is: $$V = 2\pi\bar{d}\cdot A.$$
Pappus’s Theorem (Surface Area). Let $C$ be a plane curve of length $L$ with centroid at distance $\bar{d}$ from an axis that does not intersect $C$. The surface area of the surface obtained by rotating $C$ about that axis is: $$S = 2\pi\bar{d}\cdot L.$$
In both formulas, $2\pi\bar{d}$ is the circumference of the circle traced by the centroid.
Key Concepts
1. Intuition: Average Radius Times Arc Length
When a region rotates about an axis, every point in the region sweeps a circle. The circles have different radii (from the inner edge to the outer edge of the region). The average radius is the centroid’s distance $\bar{d}$. The total volume swept is as if the entire area $A$ traveled the average circumference $2\pi\bar{d}$: $$V = A \cdot 2\pi\bar{d}.$$
This is not an obvious formula -- it requires that the centroid (not the midpoint of the region’s width) is the right average.
2. The Volume Theorem in Detail
If the region $R$ lies to one side of a vertical axis (say the $y$-axis, so $x \geq 0$ throughout), then: $$V = 2\pi\bar{x}\cdot A,$$ where $\bar{x}$ is the $x$-coordinate of the centroid (horizontal distance from the $y$-axis) and $A$ is the area of $R$.
For rotation about a horizontal axis ($x$-axis): $$V = 2\pi\bar{y}\cdot A.$$
3. Reversing: Finding Centroids from Known Volumes
If the volume of revolution is known independently (e.g., from geometry), Pappus’s theorem can be used to find the centroid: $$\bar{d} = \frac{V}{2\pi A}.$$
Example: The centroid of a semicircle of radius $r$ above the $x$-axis. Rotating the semicircle about the $x$-axis gives a sphere of volume $\dfrac{4}{3}\pi r^3$. Area of semicircle $= \dfrac{\pi r^2}{2}$.
$$\bar{y} = \frac{V}{2\pi A} = \frac{\frac{4}{3}\pi r^3}{2\pi \cdot \frac{\pi r^2}{2}} = \frac{\frac{4}{3}r}{\pi} = \frac{4r}{3\pi}.$$
This is the well-known centroid height of a semicircle.
Worked Example
Find the volume and surface area of the torus generated by rotating the disk $(x-3)^2 + y^2 \leq 1$ about the $y$-axis.
The disk has radius $r = 1$ and its center is at $(3,0)$, so the centroid is at distance $\bar{d} = 3$ from the $y$-axis.
Area of disk: $A = \pi r^2 = \pi$.
Length of boundary circle (used for surface area): $L = 2\pi r = 2\pi$.
Volume: $V = 2\pi\bar{d}\cdot A = 2\pi(3)(\pi) = 6\pi^2$.
Surface area: $S = 2\pi\bar{d}\cdot L = 2\pi(3)(2\pi) = 12\pi^2$.
Boxed answers: $V = 6\pi^2 \approx 59.2$ cubic units; $S = 12\pi^2 \approx 118.4$ square units.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Using the center of the bounding box rather than the centroid | Taking $\bar{d}$ as the midpoint of the region’s width | $\bar{d}$ is the centroid’s distance, which is a weighted average of distances across the region |
| Forgetting the axis must not intersect the region | Applying Pappus to a region that crosses the axis | If the axis intersects the region, parts of the region generate different volumes and the formula does not apply directly |
| Confusing the two theorems | Using area $A$ in the surface area theorem | Surface area theorem uses arc length $L$ of the boundary curve, not area of the enclosed region |
Common Misconceptions
the distance $\bar{d}$ in Pappus’s theorem is the midpoint of the region’s width, that is, the average of its nearest and farthest distances from the axis.
This is the average-rate-as-arithmetic-mean error. The correct $\bar{d}$ is the $x$-coordinate of the centroid, a weighted average of distance from the axis where each point is weighted by the area it contributes. For a semicircle of radius $r$ above the $x$-axis rotated about the $x$-axis, the midpoint of the height range would be $r/2$, but the centroid is at $\bar{y} = 4r/(3\pi) \approx 0.424r$, which is less than $r/2$ because the semicircle concentrates more area near the flat base. Using the midpoint in place of the centroid gives the wrong volume.
Pappus’s volume theorem and Pappus’s surface area theorem both use the area $A$ of the region.
This is the concept-image-conflicts-definition error. The volume theorem uses the area $A$ of the rotating region: $V = 2\pi\bar{d}A$. The surface area theorem uses the arc length $L$ of the curve generating the surface, not the area enclosed: $S = 2\pi\bar{d}L$. For a disk of radius $r$ centered at distance $R$ from the axis, the volume of the resulting torus is $2\pi R\cdot\pi r^2$ and the surface area is $2\pi R\cdot 2\pi r$; substituting the area $\pi r^2$ into the surface formula gives $2\pi R\pi r^2$, which has wrong units and is numerically incorrect.
Leveled Practice
Level 1 -- Torus
Problem 1. Find the volume of the solid of revolution formed by rotating the rectangle $1 \leq x \leq 3$, $0 \leq y \leq 2$ about the $y$-axis.
Show answer
The centroid of the rectangle is at $\bar{x} = 2$, $\bar{y} = 1$.
Area $A = (3-1)(2-0) = 4$.
$V = 2\pi\bar{x}\cdot A = 2\pi(2)(4) = 16\pi$.
(Check via shells: $V = 2\pi\int_1^3 x\cdot 2\,dx = 4\pi[x^2/2]_1^3 = 4\pi(4.5) = 18\pi$. Wait -- let me recheck the centroid.)
Actually $\bar{x} = \dfrac{1+3}{2} = 2$. $V = 2\pi(2)(4) = 16\pi$. Check with shells: $V = 2\pi\int_1^3 2x\,dx = 4\pi[x^2/2]_1^3 = 4\pi(9/2-1/2) = 4\pi\cdot 4 = 16\pi$. Confirmed.
Boxed answer: $V = 16\pi$.
Level 2 -- Finding a Centroid
Problem 2. The centroid of a right triangle with legs $a$ and $b$ is at $(\bar{x},\bar{y}) = (a/3, b/3)$ from the right-angle vertex. Verify Pappus’s volume theorem by rotating the triangle with vertices $(0,0)$, $(a,0)$, $(0,b)$ about the $y$-axis and comparing to the known formula for a cone.
Show answer
Area $A = \dfrac{ab}{2}$. Centroid at $\bar{x} = \dfrac{a}{3}$.
$V_{\text{Pappus}} = 2\pi\cdot\dfrac{a}{3}\cdot\dfrac{ab}{2} = \dfrac{\pi a^2 b}{3}$.
Known formula for a cone with base radius $a$ and height $b$: $V = \dfrac{1}{3}\pi a^2 b$. These agree. Pappus is verified.
Mastery Checklist
Mental Model
Pappus’s theorem says: volume = (cross-section area) times (distance traveled by centroid). The centroid travels in a circle of circumference $2\pi\bar{d}$. So the volume is just area times circumference of centroid’s path. Think of sweeping a cross-section along a circular track: the swept volume equals the cross-section times the track length. The centroid is the right “representative point” because it accounts for how the cross-section is distributed in the direction perpendicular to the axis.
Connections
Looking back
- Center of mass / centroid (Section 8.3): Pappus’s theorems use the centroid as a key input.
- Volumes of revolution (Section 6.2-6.3): Pappus is an alternative to the disk, washer, and shell methods.
- Surface area of revolution (Section 8.2): The surface area theorem uses the arc length $L$ and the centroid of the arc.
Looking ahead
This topic completes the rotation-and-area cluster of MATH162 Section 8.