The Attraction of Surfaces of Revolution

Adam Coffman

                    CoffmanA@ipfw.edu

Department of Mathematical Sciences Indiana University Purdue University Fort Wayne Fort Wayne, IN 46805-1499

July 24, 2000

In my lectures for the first-year calculus sequence, I state and solve physics problems. After the section on surface area, the following problem generated some interest:

Assuming an inverse square law of attraction, what is the force exerted by a massive surface of revolution on a point mass m located on the axis of symmetry?

An important special case is the attractive force of gravity exerted by a spherical shell on a point mass m. Since any line through the center is an axis of symmetry, m can be anywhere in space.

For the general case, here are some preliminary assumptions:

1. The surface of revolution is defined by a nonnegative function f(x) on a closed interval [a, b], such that f′ exists on (a, b). The graph of f is revolved around the x-axis as in the Figure.

2. The surface’s mass is distributed evenly, in the sense that it has a constant “planar density,” d ≥ 0. The units on d might be kilograms per square meter, for example, to distinguish it from linear or spatial density.

3. The “inverse square law” refers to a force exerted on a point mass m by another point mass M separated by distance r > 0. Then the magnitude of the force is GmMr^{−2}, for a positive constant G. M, m will be assumed nonnegative, and the direction of the force on m is toward M.

4. To simplify calculation, the point mass m can be assumed to be at the origin, by translating f left or right if necessary.

Figure 1

To start with the solution to the physics problem, we slice the surface with planes parallel to the yz-plane, and review a (sketchy) derivation of the integral formula for surface area.

The Riemann sum procedure is to partition [a, b] into n subintervals [x_{i−1}, x_{i}], with length Δx_{i}, for i = 1 to n, and then select a sample point x_{i}^{∗} — the midpoint will be convenient. The graph of f can be approximated by n line segments L_{i} connecting (x_{i−1}, f(x_{i−1})) to (x_{i}, f(x_{i})). Revolving each segment L_{i} gives a truncated cone C_{i}, which approximates a slice S_{i} of the surface of revolution. Each C_{i} has surface area

π(f(x_{i−1}) + f(x_{i}))\sqrt{(Δx_{i})^{2} + (Δf_{i})^{2}},

where Δf_{i} abbreviates f(x_{i}) − f(x_{i−1}). (This well-known formula for the area of a truncated cone can be derived without calculus.) The average \frac{1}{2}(f(x_{i−1}) + f(x_{i})) is the distance from (x_{i}^{∗}, 0) to the midpoint of L_{i}, which can be approximated by f(x_{i}^{∗}). Then, the approximate area of C_{i}, and the

slice S_{i}, is 2πf(x_{i}^{∗})\sqrt{1 + (\frac{Δf_{i}}{Δx_{i}} )^{2}}Δx_{i}, and the mass of S_{i}, denoted M_{i}, is approximately the density times this area:

M_{i} ≈ d · 2πf(x_{i}^{∗})\sqrt{1 + (\frac{Δf_{i}}{Δx_{i}} )^{2}}Δx_{i}.

The total area of the surface is the n → ∞ (and max Δx_{i} → 0) limit of the sum of the approximate areas, and its total mass, denoted M(f) = ∑ M_{i}, is equal to d times this area:

M(f) = d · ∫_{a}^{b} 2πf(x)\sqrt{1 + (\frac{df}{dx})^{2}}dx.

The force F_{i}, exerted by each slice S_{i} on the mass m at the origin, will be directed along the x-axis. This is obvious by the rotational symmetry, and also follows from the following approximation of F_{i} as a vector sum. The slice S_{i} can itself be subdivided “radially” into 2N pieces by N planes through the x-axis. When n and N are large, each of these pieces can be treated as a point with mass \frac{M_{i}}{2N}, for the purposes of approximating F_{i} using the inverse square law. Every piece of S_{i} will be represented by one of its points, with horizontal coordinate x_{i}^{∗} and at distance f(x_{i}^{∗}) from the x-axis, so that the line from m to this point is at an angle θ_{i} with the x-axis. The force exerted on m by the piece has approximate magnitude Gm\frac{M_{i}}{2N} (\sqrt{(x_{i}^{∗})^{2} + (f(x_{i}^{∗}))^{2}})^{−2}. Its horizontal component (along the x-axis) is cos(θ_{i}) times the magnitude, and its radial component is sin(θ_{i}) times the magnitude. The force exerted by the opposite piece (rotating the piece and its representative point by 180^{∘}) has the same horizontal component, but an oppositely directed radial component. In the sum over 2N pieces, the radial components all cancel, and the approximate horizontal components total to

F_{i} ≈ Gm\frac{M_{i}}{(x_{i}^{∗})^{2} + (f(x_{i}^{∗}))^{2}} cos(θ_{i})

≈ Gm\frac{d2πf(x_{i}^{∗})\sqrt{1 + (\frac{Δf_{i}}{Δx_{i}} )^{2}}Δx_{i}}{(x_{i}^{∗})^{2} + (f(x_{i}^{∗}))^{2}} \frac{x_{i}^{∗}}{\sqrt{(x_{i}^{∗})^{2} + (f(x_{i}^{∗}))^{2}}}

= 2πGmd\frac{x_{i}^{∗}f(x_{i}^{∗})\sqrt{1 + (\frac{Δf_{i}}{Δx_{i}} )^{2}}}{((x_{i}^{∗})^{2} + (f(x_{i}^{∗}))^{2})^{(3/2)}} Δx_{i}.

The second step uses the earlier approximation for M_{i}, and the ratio for the cosine: cos(θ_{i}) = \frac{x_{i}^{∗}}{\sqrt{(x_{i}^{∗})^{2}+(f(x_{i}^{∗}))^{2}}} . This formula for F_{i} is actually a signed

quantity, with the formula for the cosine taking into account the direction of the force acting on m: to the right for x_{i}^{∗} > 0, and to the left for x_{i}^{∗} < 0.

So, in the n → ∞ limit, the answer to the physics question is

∫_{a}^{b} 2πGmd\frac{xf(x)\sqrt{1 + (f′(x))^{2}}}{(x^{2} + (f(x))^{2})^{(3/2)}} dx,

assuming that this definite integral exists, which (mathematically) is a nontrivial condition required of f.

As an application of this formula, consider a sphere with center (c, 0) (on the positive x-axis, c > 0) and radius R > 0. Using the above formula, with f(x) = \sqrt{R^{2} − (x − c)^{2}}, and [a, b] = [c − R, c + R] gives f′(x) = \frac{c−x}{\sqrt{R^{2}−(x−c)^{2}}} , and total force

F = ∫_{c−R}^{c+R} 2πGmd\frac{x\sqrt{R^{2} − (x − c)^{2}}\sqrt{1 + (\frac{c−x}{\sqrt{R^{2}−(x−c)^{2}}} )^{2}}}{(x^{2} + R^{2} − (x − c)^{2})^{(3/2)}} dx
= 2πGmdR ∫_{c−R}^{c+R} \frac{x}{(R^{2} + 2xc − c^{2})^{(3/2)}} dx
= \frac{2πGmdR}{c^{2}} \frac{R^{2} + xc − c^{2}}{\sqrt{R^{2} + 2xc − c^{2}}} |_{c−R}^{c+R}

= \{ \begin{matrix}\frac{2πGmdR^{2}}{c^{2}} (\frac{c + R}{\sqrt{(c + R)^{2}}} + \frac{c − R}{\sqrt{(c − R)^{2}}} ) & if c ≠ R \\ 2πGmd & if c = R. \end{matrix}

= \{ \begin{matrix}4πGmdR^{2}c^{−2} & if c > R \\ 0 & if c < R \\ 2πGmd & if c = R. \end{matrix}

The total mass of the sphere is M(f) = 4πR^{2}d, and if this mass were concentrated at the center (c, 0) with c > R, the force on the mass m at (0, 0) would be GmM(f)c^{−2}. This is the same as the above integral, so we have a single-variable derivation of a result of Newton, that the external gravitational attraction of a sphere is equal to the attractive force of a point with the same mass at the sphere’s center. This was part of Newton’s argument that a solid ball has the same property.

The same integral also demonstrates the fact that if the particle of mass m is inside the sphere, so c < R, then it feels no force acting in any direction. (This fact was interesting and surprising to many students.) At c = R, the particle is on the sphere, and the force is \frac{1}{2}GmM(f)c^{−2}; plotting F as a function of c, there is a discontinuity at c = R. The c = 0 and c < 0 cases follow from similar calculations.

Other surfaces of revolution for which the above integral formula might be tractable are cylinders, f(x) = K, truncated cones, f(x) = kx + K, or funnel shapes, f(x) = k/x, over intervals where f(x) ≥ 0. The construction also could be applied to a repelling force.