Pappus's Centroid Theorems

Pappus's centroid theorems are results from geometry about the surface area and volume of solids of revolution. These quantities can be computed using the distance traveled by the centroids of the curve and region being revolved.

Let \( C\) be a curve in the plane. The area of the surface obtained when \( C\) is revolved around an external axis is equal to the product of the arc length of \( C\) and the distance traveled by the centroid of \( C.\)

Let \( R\) be a region in the plane. The volume of the solid obtained when \( R\) is revolved around an external axis is equal to the product of the area of \( R\) and the distance traveled by the centroid of \( R.\)

An illustration of Pappus's first theorem. Source: wikipedia

Consider the cylinder obtained by revolving a rectangle with horizontal side \( r\) and vertical side \( h\) around one of its vertical sides (say its left side). The surface area of the cylinder, not including the top and bottom, can be computed from Pappus's theorem since the surface is obtained by revolving its right side around its left side. The arc length of its right side is \( h \) and the distance traveled by its centroid is simply \( 2\pi r,\) so its area is \( 2 \pi r h.\)

The volume of the cylinder is the area \( rh\) of the rectangle multiplied by the distance traveled by its centroid. The centroid of the rectangle is its center, which is a distance of \( \frac r2\) from the axis of revolution. So it travels a distance of \( 2\pi\big(\frac r2\big) = \pi r\) as it revolves. The volume of the cylinder is \( (rh)(\pi r) = \pi r^2 h.\)

To compute the volume of a solid formed by rotating a region \( R\) around an external axis (a similar argument applies for surface area), one can break the region up into small regions of area \( \Delta A\) that are located a distance \( r\) from the axis. Since these regions travel a distance \( 2\pi r \) when revolved around the axis, their contribution to the volume of the solid is roughly \( 2\pi r \Delta A.\) Adding up all the contributions gives the volume (as usual, passing to the limit where \( \Delta A \) becomes \( dA,\) and the sum becomes an integral, makes this an exact computation).

The difficulty is that all the different small regions are different distances away from the axis. To make this method computationally feasible, one would need to know the average distance \( r_{\text{avg}} \) of all the small pieces, so that the volume of the region is \( 2\pi r_{\text{avg}} \) times the area \( A.\) This is precisely what Pappus' centroid theorem gives: it identifies \( r_{\text{avg}}\) as the distance from the centroid of the region to the axis of revolution.

Of course, this does not make the computation trivial in general, since computing the centroid of a region (or curve) is not easy, even for relatively simple shapes. However, there are sometimes symmetry considerations or other computational aids that make Pappus' theorem an effective shortcut for solving problems involving volumes and areas of revolution.

The surface area and volume of a torus are quite easy to compute using Pappus' theorem. A torus is a circle of radius \( r< R,\) centered at \( (R,0)\) and rotated around the \(y\)-axis. The centroid of both the surface of the circle and the region enclosed by the circle is just the center of the circle. This travels a distance of \( 2\pi R \) when it revolves, so the surface area is \( 2\pi R \) times the arc length \(2\pi r\) of the circle, and the volume is \( 2\pi R \) times the area \( \pi r^2\) of the circle. That is,

\[ \begin{align} \text{surface area of the torus} &= 4\pi^2 Rr \\ \text{volume of the torus} &= 2\pi^2 Rr^2. \end{align} \]

Revolving a right triangle with legs of length \( r\) and \( h\) around the leg of length \( h\) produces a cone. The surface of the cone (not including the circular base) is obtained by revolving the hypotenuse around that leg. The centroid of the hypotenuse is just the midpoint, located halfway up the side of the cone, which travels a distance \( \frac{2\pi r}2\) as it rotates. So the surface area is \( \pi r \sqrt{r^2+h^2} \) by Pappus' theorem.

The centroid of the triangle is the center of mass of the three vertices (see the Triangle Centroid wiki), which is located at a distance of \( \frac r3\) from the axis of revolution \(\big(\)and \(\frac h3\) above the base\(\big).\) So the volume is \( 2\pi \big(\frac r3\big)\) times the area of the triangle, which is \( \frac{rh}2.\) The product is \( \frac13 \pi r^2 h,\) the familiar formula for the volume of a cone.

The volume and surface area of a sphere are computable using Pappus's theorems, but the computations involve nontrivial integrals; Pappus's theorems do not provide a "shortcut" in this case.

Let \( C \) be the curve \( y = \sqrt{r^2-x^2}.\) The sphere of radius \(r\) is obtained by revolving \( C\) around the \( x\)-axis. The arc length \(L\) of \( C\) is just \( \pi r\) since it is half a circle.

The centroid of \( C\) is on the \(y\)-axis by symmetry. Its \(y\)-coordinate is given by the formula

\[y = \frac1{L} \int_{-r}^r y\sqrt{1+\left( \frac{dy}{dx}\right)^2} \, dx.\]

The reasoning here is that the centroid is a center of mass and the mass of a small piece of the arc over a small length \( dx\) on the \(x\)-axis is proportional to the arc length, which is \( \sqrt{1+\left( \frac{dy}{dx}\right)^2} \, dx.\) (See the Arc Length wiki for a thorough explanation.) So the location of the centroid is

\[\begin{align} \frac1{\pi r} \int_{-r}^r \sqrt{r^2-x^2} \sqrt{1+\frac{x^2}{r^2-x^2}} dx &= \frac1{\pi r} \int_{-r}^r \sqrt{r^2-x^2} \sqrt{\frac{r^2}{r^2-x^2}} dx \\ &= \frac1{\pi r} \int_{-r}^r r \, dx\\t &= \frac{2r^2}{\pi r} = \frac{2r}{\pi}. \end{align}\]

So the surface area is \( 2\pi\) times the \(y\)-coordinate of the centroid, times the arc length, which is

\[2\pi \left( \frac{2r}{\pi} \right) (\pi r) = 4\pi r^2.\]

The \(y\)-coordinate of the centroid of the disk is the (double) integral of \(y\) over the region, divided by the area of the region:

\[\begin{align} \frac1{\frac12 \pi r^2} \iint_R y \, dy \, dx &= \frac2{\pi r^2} \left. \int_{-r}^r \frac{y^2}2\right|_0^{\sqrt{r^2-x^2}} dx \\ &= \frac1{\pi r^2} \int_{-r}^r \big(r^2-x^2\big) dx \\ &= \frac1{\pi r^2}\left. \left(r^2 x - \frac{x^3}3\right)\right|_{-r}^r \\ &= \frac{4r}{3\pi}. \end{align}\]

So the volume is \(2\pi\) times the \(y\)-coordinate of the centroid, times the area, which is

\[2\pi \left( \frac{4r}{3\pi} \right) \frac{\pi r^2}2 = \frac43 \pi r^3,\]

as expected.