Mass, Moments, and Centers of Mass, calculus lesson

A seesaw balances when the heavier child sits closer to the middle. Weight times distance is what matters, and that product is called a moment. For a rod whose material gets denser toward one end, or for a flat metal plate cut into a curved shape, the weight is spread out continuously. Integrals add up the moments, and the balance point follows from one division.

Key terms

Variable density

Definition A thin rod lying along $[a, b]$ on the $x$ -axis has linear density $\rho(x)$ , a nonnegative continuous function measured in mass per unit length, if the mass of every piece $[c, d] \subseteq [a, b]$ equals

\int_c^d \rho(x) \, dx.

For a flat plate (lamina) the density is measured in mass per unit area. A uniform object has constant density.

In plain words Density tells you how heavy the material is at each point. A small piece of rod of length $dx$ near $x$ has mass about $\rho(x)\,dx$ .

Example A rod on $[0, 2]$ meters with $\rho(x) = 1 + x$ kilograms per meter has density $1$ kilogram per meter at its left end and $3$ kilograms per meter at its right end.

Mass

Definition The mass of a rod on $[a, b]$ with continuous linear density $\rho$ is

m = \int_a^b \rho(x) \, dx.

For a lamina of constant density $\rho$ occupying the region $g(x) \le y \le f(x)$ , $a \le x \le b$ , with $f$ and $g$ continuous, the mass is $m = \rho A$ , where

A = \int_a^b \left(f(x) - g(x)\right) dx

is its area.

In plain words Chop the object into small pieces, multiply each length or area by its density, and add.

Example For the rod with $\rho(x) = 1 + x$ on $[0, 2]$ ,

m = \int_0^2 (1 + x) \, dx = 2 + 2 = 4 \text{ kilograms}.

Moments of mass

Definition For point masses $m_1, \dots, m_n$ at positions $x_1, \dots, x_n$ on a line, the moment about the origin is

M_0 = \sum_{i=1}^n m_i x_i.

For a rod on $[a, b]$ with density $\rho$ ,

M_0 = \int_a^b x \, \rho(x) \, dx.

For a lamina of constant density $\rho$ on the region $g(x) \le y \le f(x)$ , $a \le x \le b$ , with $f$ and $g$ continuous, the moments about the $y$ -axis and the $x$ -axis are

M_y = \rho\int_a^b x \left(f(x) - g(x)\right) dx, \qquad M_x = \rho\int_a^b \frac{1}{2}\left(f(x)^2 - g(x)^2\right) dx.

The second formula uses the fact that a thin vertical strip has its center at height $(f(x) + g(x))/2$ , and

\frac{f + g}{2}\,(f - g) = \frac{1}{2}\left(f^2 - g^2\right).

In plain words A moment is mass times signed distance, added over all pieces. It measures how strongly the mass tends to turn the object about a pivot. Note the naming. The moment $M_y$ uses $x$ distances because it is measured from the $y$ -axis.

Example Masses of $2$ kg at $x = 1$ and $3$ kg at $x = 4$ have $M_0 = 2 \cdot 1 + 3 \cdot 4 = 14$ . For the rod with $\rho(x) = 1 + x$ on $[0, 2]$ ,

M_0 = \int_0^2 x(1 + x) \, dx = 2 + \frac{8}{3} = \frac{14}{3}.

Center of mass

Definition The center of mass of a rod or system on a line with total mass $m > 0$ is at

\bar x = \frac{M_0}{m}.

For a lamina with total mass $m > 0$ the center of mass is the point

(\bar x, \bar y) = \left(\frac{M_y}{m}, \frac{M_x}{m}\right).

It is the point at which the object would balance on a pin.

In plain words The center of mass is the weighted average position, with density as the weight. Heavy regions pull it toward themselves.

Example For the two point masses, $\bar x = \frac{14}{5} = 2.8$ . For the rod with $\rho(x) = 1 + x$ ,

\bar x = \frac{14/3}{4} = \frac{7}{6} \approx 1.167,

to the right of the midpoint $1$ , as expected since the right end is heavier.

Centroid

Definition The centroid of a plane region $R$ with area $A > 0$ is the center of mass of a lamina of constant density occupying $R$ . Since the density cancels, for $R$ given by $g(x) \le y \le f(x)$ , $a \le x \le b$ ,

\bar x = \frac{1}{A}\int_a^b x\left(f(x) - g(x)\right) dx, \qquad \bar y = \frac{1}{A}\int_a^b \frac{1}{2}\left(f(x)^2 - g(x)^2\right) dx.

The centroid of a smooth curve of length $L > 0$ is defined the same way with arc length in place of area,

\bar x = \frac{1}{L}\int x \, ds, \qquad \bar y = \frac{1}{L}\int y \, ds.

In plain words The centroid is the balance point of the shape itself, if it were cut from a uniform sheet. It depends only on geometry.

Example For the region under $y = x^2$ on $[0, 1]$ , $A = \frac{1}{3}$ and

\bar x = 3\int_0^1 x^3 \, dx = \frac{3}{4}, \qquad \bar y = 3\int_0^1 \frac{x^4}{2} \, dx = \frac{3}{10}.

The centroid $\left(\frac{3}{4}, \frac{3}{10}\right)$ sits toward the right and low, where most of the area is.

Symmetry

Definition If a region is symmetric about a line (its mirror image in the line is the region itself), then its centroid lies on that line. The same holds for the center of mass of a lamina whose density is also symmetric about the line.

In plain words A symmetric shape cannot prefer one side, so its balance point sits on the line of symmetry. This often gives one coordinate for free.

Example The half disk $0 \le y \le \sqrt{r^2 - x^2}$ is symmetric about the $y$ -axis, so $\bar x = 0$ with no computation. For $\bar y$ , with $A = \frac{\pi r^2}{2}$ ,

\bar y = \frac{2}{\pi r^2}\int_{-r}^r \frac{1}{2}(r^2 - x^2) \, dx = \frac{2}{\pi r^2} \cdot \frac{2r^3}{3} = \frac{4r}{3\pi}.

For $r = 3$ that is $\frac{4}{\pi} \approx 1.273$ .

Pappus’s theorems

Definition Let $R$ be a plane region of area $A$ and let $\ell$ be a line in its plane such that $R$ lies entirely on one side of $\ell$ (it may touch $\ell$ along its boundary). If $d$ is the distance from the centroid of $R$ to $\ell$ , the solid obtained by rotating $R$ about $\ell$ has volume

V = 2\pi d A.

Likewise, if a smooth curve of length $L$ lies on one side of $\ell$ and does not cross it, and its centroid is at distance $d$ from $\ell$ , the surface obtained by rotating the curve has area $S = 2\pi d L$ . The one-side hypothesis is essential.

In plain words Volume is area times the distance the centroid travels in one turn. Surface area is length times the distance the curve’s centroid travels. If the axis passes through the region, the formula fails.

Example Rotate the half disk of radius $r$ about its straight edge. Then $d = \frac{4r}{3\pi}$ and

V = 2\pi \cdot \frac{4r}{3\pi} \cdot \frac{\pi r^2}{2} = \frac{4}{3}\pi r^3,

the volume of a ball, as it should be. For a torus, a disk of radius $r$ with center at distance $R \ge r$ from the axis gives

V = 2\pi R \cdot \pi r^2 = 2\pi^2 R r^2.

But rotating a whole disk about one of its diameters would give $V = 0$ by the formula, because the centroid lies on the axis. The disk crosses the axis, so the theorem does not apply.

Check yourself

A rod on $[0, 3]$ has density $\rho(x) = x^2$ . Find its mass and center of mass.

Show the answer

The mass and moment are

m = \int_0^3 x^2 \, dx = 9, \qquad M_0 = \int_0^3 x^3 \, dx = \frac{81}{4},

so the center of mass is at $\bar x = \frac{81}{4} \cdot \frac{1}{9} = \frac{9}{4}$ .

Find the centroid of the triangle under $y = x$ on $[0, 2]$ .

Show the answer

The area is $A = 2$ , so

\bar x = \frac{1}{2}\int_0^2 x^2 \, dx = \frac{4}{3}, \qquad \bar y = \frac{1}{2}\int_0^2 \frac{x^2}{2} \, dx = \frac{2}{3}.

The centroid is $\left(\frac{4}{3}, \frac{2}{3}\right)$ .

Use Pappus’s theorem to find the volume when that triangle is rotated about the line $x = -1$ .

Show the answer

The triangle lies to the right of the line, and its centroid is at distance $\frac{4}{3} + 1 = \frac{7}{3}$ from it. Pappus gives

V = 2\pi \cdot \frac{7}{3} \cdot 2 = \frac{28\pi}{3}.

Shells agree, since

2\pi\int_0^2 (x + 1)x \, dx = \frac{28\pi}{3}.

Mass, Moments, and Centers of Mass

Key terms

Variable density

Mass

Moments of mass

Center of mass

Centroid

Symmetry

Pappus’s theorems

Check yourself

How this lesson connects

Builds on