Mass, Centroids & Moments of Inertia

What a multiple integral was really for

For four guides you have been learning the mechanics of the double integral and its three-dimensional sibling — how to slice a region into iterated slabs, how to flip the order, how to bend the coordinate grid with a Jacobian, how to reach for polar or spherical when a problem is round. All of that machinery exists to do one humble thing: take a quantity spread out over a region and add it up. In Volume I the definite integral added a quantity spread along a line; the multiple integral simply does the same over an area or a volume. This guide is where the bookkeeping finally pays a physical dividend.

Here is the one idea everything in this guide flows from. Imagine a thin metal plate occupying a region R in the plane, not necessarily uniform — its density rho(x, y) (mass per unit area) can vary from place to place, thick here, thin there. Chop the plate into tiny tiles. A tile sitting near the point (x, y) with area dA holds a tiny mass dm = rho(x, y) dA — density times the little bit of area, exactly the way mass equals density times area for a uniform piece, applied locally. Every physical quantity in this guide is built by attaching some weight to each tile's mass dm and summing over all tiles. Summing over all tiles is precisely what the multiple integral does, so each quantity is just an integral of (something) times rho dA.

Total mass, and the balance point

Start with the simplest weighting of all: weight each tile by 1. Adding up every dm gives the total mass M = double integral over R of rho(x, y) dA. If the plate is uniform, rho is a constant you can pull outside, and M = rho times (area of R) — mass really is density times area, just as you would guess; the integral only earns its keep when rho varies. This is the warm-up, and it is also a sanity check you should run first: if you cannot set up the mass integral over your region R, you are not ready to set up the harder ones, because they all integrate over the very same R with the very same dA.

Now weight each tile's mass by its position. The [[calc-center-of-mass|center of mass]] is the single point where the plate would balance on a pin — the mass-weighted average location. Its x-coordinate is x-bar = (1/M) times the double integral over R of x rho dA, and likewise y-bar = (1/M) times the integral of y rho dA. Read the formula as what it is: each tile votes for its own x-position, but a heavy tile's vote counts more (it carries more dm), and dividing by M normalizes the tally into an honest average. The quantity in the numerator, the integral of x rho dA, even has its own name — the first moment of mass about the y-axis — because position appears to the first power.

Moment of inertia: where the square changes everything

Mass measures how hard it is to push an object in a straight line. The [[moment-of-inertia|moment of inertia]] measures how hard it is to spin it about an axis — it is rotation's version of mass, the I in the rotational law torque equals I times angular acceleration, the exact twin of force equals mass times acceleration. The crucial new ingredient is that it depends not just on how much mass there is, but on where that mass sits relative to the axis. For a single point mass m at distance d from the axis, the moment of inertia is m d^2: mass times the square of its distance. To get a whole body, weight each tile's mass by the square of its distance from the axis and integrate.

Concretely, for our plate the distance from the x-axis is just y, so the moment of inertia about the x-axis is I_x = double integral over R of y^2 rho dA. About the y-axis the distance is x, giving I_y = integral of x^2 rho dA. And the polar moment about the origin — resistance to spinning about the vertical axis through the origin — uses the squared distance x^2 + y^2 from that axis, so I_0 = integral of (x^2 + y^2) rho dA = I_x + I_y. Because distance enters squared, this is also called the second moment of mass, in deliberate contrast to the first moments that built the center of mass. That square is not a detail; it is the entire personality of the quantity. Mass twice as far from the axis contributes four times as much to the moment of inertia, so faraway mass dominates utterly.

This single fact — faraway mass weighted by the square — explains a wardrobe of everyday physics. A figure skater spins faster the instant she pulls her arms in, because moving mass closer to the axis shrinks I, and with angular momentum conserved a smaller I means a larger spin. A flywheel meant to store rotational energy is built heavy at the rim, where each kilogram buys the most I. And the polar moment I_0 = integral of (x^2 + y^2) rho dA is exactly the integrand that begs for polar coordinates: x^2 + y^2 collapses to r^2, and you are about to see that same collapse perform real magic.

The Gaussian integral, cracked open by polar

Now the reward. The [[gaussian-integral|Gaussian integral]] is the total area under the bell curve, integral over the whole real line of e^{-x^2} dx, and it equals the strange, clean number sqrt(pi). Strange because e^{-x^2} has no elementary antiderivative — there is no formula in elementary functions for the area out to a finite point, which is exactly why the error function erf had to be invented to name that partial area. (To be precise about the word: 'non-elementary' means no elementary closed form, NOT that the integral is uncomputable — it is a perfectly definite, finite number you can compute to any precision.) And yet the area all the way out to infinity is this immaculate sqrt(pi). How can the total be pristine when no partial sum is? The resolution is one of the most beautiful tricks in mathematics, and it is a multiple integral in disguise.

Call the answer I, so I = integral over the line of e^{-x^2} dx. The one-dimensional integral fights back, so square it — and squaring is what opens the door, because a product of two one-dimensional integrals is a double integral. Using a dummy variable y in the second copy, I^2 = (integral of e^{-x^2} dx) times (integral of e^{-y^2} dy) = double integral over the whole plane of e^{-x^2} e^{-y^2} dA = double integral over the plane of e^{-(x^2 + y^2)} dA. We have promoted an intractable one-dimensional integral into a two-dimensional one over the entire plane — which sounds worse, until you notice the integrand depends only on x^2 + y^2, the squared distance from the origin. That is the signature of a problem that is secretly round, and round problems are what polar coordinates were built for. The change of variables to polar then supplies, for free, an extra factor of r from the area element — and that very factor is the derivative-shaped piece you needed to integrate e^{-r^2} by an ordinary u = r^2 substitution. The geometry hands you the algebra: the radial integral collapses to 1/2, the angular sweep contributes a full 2 pi, so I^2 = pi and I = sqrt(pi). Halving (the bell curve is symmetric) gives the form you will quote forever, integral from 0 to infinity of e^{-x^2} dx = sqrt(pi)/2.

I  =  integral_{-inf}^{inf}  e^{-x^2} dx           ( the goal )

Square it and pair the two copies into one double integral:

  I^2 = ( int e^{-x^2} dx )( int e^{-y^2} dy )
      = int int_{plane}  e^{-(x^2 + y^2)} dA

Switch to polar:   x^2 + y^2 = r^2,   dA = r dr dtheta
                                            ^^^^^^^^^ the Jacobian factor

  I^2 = int_{0}^{2pi} int_{0}^{inf}  e^{-r^2} . r dr dtheta

The stray r is exactly what makes the radial part elementary.
Let u = r^2,  du = 2r dr :

  int_{0}^{inf} e^{-r^2} r dr = (1/2) int_{0}^{inf} e^{-u} du = 1/2

  I^2 = int_{0}^{2pi} (1/2) dtheta = (1/2)(2pi) = pi

  ==>  I = sqrt(pi)        and       int_{0}^{inf} e^{-x^2} dx = sqrt(pi)/2

One recipe for all of them

Notice that mass, center of mass, and moment of inertia are not three separate skills — they are one skill applied with three different weights. Every one is the integral over R of (weight) times rho dA: weight 1 gives mass; weight x or y gives a first moment, and dividing by M gives a coordinate of the center of mass; weight (distance to axis)^2 gives a moment of inertia. The hard part is never the physics formula; it is always describing the region R and choosing coordinates that make the integral pleasant — exactly the muscles you built in the earlier guides of this rung. So the procedure is short, and the same every time.

1. Draw the region R and pick coordinates that fit its shape — Cartesian for rectangles and graphs, polar for disks and wedges, spherical or cylindrical for round solids. This single choice decides whether the integral is friendly or fierce.
2. Write the correct area or volume element for those coordinates — dx dy, or r dr dtheta in polar, or rho^2 sin(phi) drho dphi dtheta in spherical — and never forget this Jacobian factor; it is the most common silent error.
3. Compute the total mass M = integral over R of rho dA first. You will reuse it as the denominator for the center of mass, and computing it is a low-stakes rehearsal of the harder integrals to come.
4. Attach the weight you need — x or y for a first moment (then divide by M), or (distance to the axis)^2 for a moment of inertia — and integrate over the very same R with the very same element. Same region, same dA, different weight: that is the entire game.

And that closes the rung. You came in able to evaluate an iterated integral over a rectangle; you leave able to describe a curved region, change its coordinates with a Jacobian, and read off the mass, balance point, and rotational stubbornness of a physical body — and, as a bonus, to evaluate by hand the integral that anchors all of probability and statistics. The same r dr dtheta that rescaled area for the change of variables turned out to be the secret hero of the Gaussian integral. Next, in vector calculus, these same regions grow a third dimension and a sense of direction: we will integrate not just densities over regions but flows across surfaces, and the great theorems of Green, Stokes, and Gauss will tie boundary to interior the way the fundamental theorem of calculus once tied endpoints to a line.