[CALCULUS] Mass, centroids and probability

Interpret double integrals as mass, moments, centres of mass, population and expected value.

This page integrates a density over a region to find total mass, centres of mass and expected values. The area formula uses unit density, so every point has the same weight. A varying density weights some parts of the region more than others.

The same weighted integrals describe a thin plate, a population spread over a region and a continuous probability model. They are evaluated by slicing, as in Fubini's theorem, often after a change of variables.

Density on a lamina

A lamina is a flat sheet of material with negligible thickness. If $\rho(x,y)$ is its mass density, then a small area $\Delta A$ near $(x,y)$ has approximate mass $\rho(x,y)\Delta A$ .

[DEFINITION]

Mass density of a lamina

A mass density $\rho(x,y)$ assigns mass per unit area to a lamina occupying a region $\Omega$ . For a physical lamina, $\rho\ge0$ and $\rho$ is integrable. The total mass is

M=\iint_\Omega\rho(x,y)\,dA.

When $\rho=1$ , the mass is the area of $\Omega$ .

[1]:Centre of mass and moments
[2]:Normalising a density

[THEOREM]

Centre of mass and moments

For a lamina with mass density $\rho$ and total mass $M>0$ , the centre of mass is

{ \bar x=\frac1M\iint_\Omega x\rho(x,y)\,dA, \qquad \bar y=\frac1M\iint_\Omega y\rho(x,y)\,dA }.

The density-weighted first moments are

M_y=\iint_\Omega x\rho(x,y)\,dA, \qquad M_x=\iint_\Omega y\rho(x,y)\,dA.

The factor $x$ weights mass by horizontal position and the factor $y$ weights it by vertical position. For uniform density, the centre of mass is the centroid, the geometric centre of the region.

Units give a check: if $x$ and $y$ are measured in units of length and $\rho$ in mass per unit area, then $M$ has units of mass, each first moment has units of mass, $\times$ ,length and $(\bar x,\bar y)$ has units of length.

[EXAMPLE]

A symmetric region with nonconstant density

Let

\Omega=\{(x,y):|x|\le y\le1\}, \qquad \rho(x,y)=x^2+y^2.

Horizontal slices give

0\le y\le1,\qquad-y\le x\le y.

The mass is

\begin{aligned} M &=\int_0^1\int_{-y}^{y}(x^2+y^2)\,dx\,dy\\ &=\int_0^1\left[\frac{x^3}{3}+xy^2\right]_{-y}^{y}dy\\ &=\int_0^1\frac{8y^3}{3}\,dy =\frac23. \end{aligned}

The region and density are symmetric under $x\mapsto-x$ , so

\bar x=0.

The vertical moment is

\begin{aligned} \iint_\Omega y\rho\,dA &=\int_0^1\int_{-y}^{y}y(x^2+y^2)\,dx\,dy\\ &=\int_0^1\frac{8y^4}{3}\,dy =\frac{8}{15}. \end{aligned}

Thus

\bar y=\frac{8/15}{2/3}=\frac45,

and the centre of mass is

{\left(0,\frac45\right)}.

For uniform density, this triangle's centroid has $y$ -coordinate $2/3$ . Here the density increases with $y$ , so the centre of mass lies higher.

[exam]

[1]:Centre of mass and moments

A semicircular lamina in three densities

Let $\Omega$ be the upper half of the unit disk:

0\le r\le1,\qquad0\le\theta\le\pi.

The example compares three densities on this region. Each is symmetric under $x\mapsto-x$ , so $\bar x=0$ and only $\bar y$ needs computing. Coordinates are treated as pure numbers; in a physical model, the coefficients multiplying $y$ or $r$ would carry the units needed to make $\rho$ a mass per unit area.

[EXAMPLE]

Uniform, vertical and radial density

For uniform density $\rho=1$ , the mass is the area:

M=\frac{\pi}{2}.

The vertical moment is

\begin{aligned} \iint_\Omega y\,dA &=\int_0^\pi\int_0^1 (r\sin\theta)r\,dr\,d\theta\\ &=\left(\int_0^\pi\sin\theta\,d\theta\right) \left(\int_0^1r^2\,dr\right) =\frac23. \end{aligned}

Therefore

\bar y=\frac{4}{3\pi}.

For $\rho=y$ , the mass is

\begin{aligned} M &=\iint_\Omega y\,dA\\ &=\int_0^\pi\int_0^1r^2\sin\theta\,dr\,d\theta =\frac23. \end{aligned}

The vertical moment becomes

\begin{aligned} \iint_\Omega y\rho\,dA &=\iint_\Omega y^2\,dA\\ &=\int_0^\pi\int_0^1 r^3\sin^2\theta\,dr\,d\theta =\frac{\pi}{8}. \end{aligned}

Hence

\bar y=\frac{\pi/8}{2/3}=\frac{3\pi}{16}.

For radial density

\rho=\sqrt{x^2+y^2}=r,

the mass is

\begin{aligned} M &=\int_0^\pi\int_0^1r\cdot r\,dr\,d\theta =\frac{\pi}{3}, \end{aligned}

and the vertical moment is

\begin{aligned} \iint_\Omega y\rho\,dA &=\int_0^\pi\int_0^1 (r\sin\theta)\,r\,r\,dr\,d\theta =\frac12. \end{aligned}

Thus

\bar y=\frac{1/2}{\pi/3}=\frac{3}{2\pi}.

The three heights are approximately $0.424$ , $0.589$ and $0.477$ . Radial density moves weight away from the origin and density proportional to $y$ moves it more strongly towards the top.

[exam]

[1]:Centre of mass and moments
[2]:Polar area element

Population and probability

[PROPOSITION]

Normalising a density

If $\rho\ge0$ on $\Omega$ and

0<M=\iint_\Omega\rho(x,y)\,dA<\infty,

then

p(x,y)=\frac{\rho(x,y)}{M}

is a two-dimensional probability density on $\Omega$ , because

\iint_\Omega p(x,y)\,dA=1.

If $\rho$ is a population density, $\iint_R\rho\,dA$ is the population in a subregion $R\subseteq\Omega$ . The normalised density $p$ has units of inverse area, so $p\,dA$ is dimensionless.

After normalisation, the centre-of-mass coordinates become expected values:

\mathbb E[X]=\iint_\Omega xp(x,y)\,dA, \qquad \mathbb E[Y]=\iint_\Omega yp(x,y)\,dA.

For an event represented by a subregion $E\subseteq\Omega$ , the probability is

\mathbb P((X,Y)\in E)=\iint_E p(x,y)\,dA.

This is the population formula with the normalisation making the whole region have probability $1$ . Mass, a population centre and an expected value are all the same weighted-average construction.

[1]:Mass density of a lamina

Mass, centroids and probability

On this page