---
title: Applications to Physics, Economics, and Probability
module: Applications of Integration
moduleNumber: 5
lessonNumber: 3
order: 503
summary: >
  Definite integrals in physics, economics, and statistics: the force a fluid
  exerts on a submerged plate, the balance point of a plane region, the money
  consumers save at a market price, and the probability that a continuous random
  variable lands in an interval, together with its mean.
topics: [Applications of Integration]
sources:
  - book: Stewart
    ref: "Ch. 8 — Further Applications of Integration; §8.3 Applications to Physics and Engineering"
  - book: Stewart
    ref: "§8.4 Applications to Economics and Biology; §8.5 Probability"
---

Hydrostatic force, centers of mass, consumer surplus, and probability all reduce
to one construction: cut a quantity into small pieces, approximate each, sum,
integrate. Only the setting changes. The pieces are strips of a submerged plate,
point masses on a lamina, dollars saved at a price, and slices of probability.

## Hydrostatic pressure and force

Water pressure grows with depth, because the weight of the fluid above a point
increases. A thin horizontal plate of area $A$ at depth $d$ in a fluid of
density $\rho$ supports a column of fluid of weight $\rho g A d$, so the
**pressure** (force per unit area) is

$$
P = \rho g d.
$$

A key experimental fact makes the definite integral usable: at a fixed depth,
fluid pressure acts equally in _all_ directions. So the pressure on a vertical
surface at depth $d$ is also $\rho g d$. This lets the force on a vertical wall
or dam be computed even though the pressure is not constant over the wall.

The wall is cut into thin horizontal strips. A strip at depth $x$ has nearly
constant pressure $\rho g x$ and area $w(x)\, \Delta x$, where $w(x)$ is the
strip's width, so its force is about $\rho g x\, w(x)\, \Delta x$.

> **Definition (Hydrostatic force).** A vertical plate submerged in fluid of
> density $\rho$, with width $w(x)$ at depth $x$ and extending from depth $a$ to
> depth $b$, experiences hydrostatic force
>
> $$
> F = \int_a^b \rho g\, x\, w(x)\, \d x.
> $$

$$
% caption: Pressure grows linearly with depth, so a submerged plate is cut into
% horizontal strips; the strip at depth x contributes force (density)(g)(x)(width).
\begin{tikzpicture}[scale=0.95, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% water surface
\draw[black] (-0.4,0) -- (6.6,0);
\node[font=\scriptsize, black, above left] at (6.6,0) {surface};
\foreach \x in {0.2,1.0,1.8,2.6} \draw[black] (\x-0.6,0.18) .. controls (\x-0.35,0.32) and (\x-0.1,0.05) .. (\x+0.15,0.18);
% depth axis (downward)
\draw[->, black] (0,0.2) -- (0,-4.4) node[below] {$x$ (depth)};
% the plate (a trapezoid dam)
\draw[black, thick] (1.4,-0.7) -- (5.4,-0.7) -- (4.7,-4.0) -- (2.1,-4.0) -- cycle;
% representative strip
\filldraw[acc!22, draw=acc] (1.75,-2.05) rectangle (5.02,-2.32);
\draw[<->, black!70] (1.75,-2.55) -- (5.02,-2.55) node[midway, below, font=\scriptsize] {$w(x)$};
\draw[black, dashed] (0,-2.18) -- (1.75,-2.18);
\node[font=\scriptsize, black, left] at (0,-2.18) {$x$};
% pressure arrows increasing with depth
\foreach \y/\len in {-1.1/0.4,-1.8/0.7,-2.5/1.0,-3.2/1.3,-3.8/1.55}
  \draw[black!70, ->] (\len+5.55,\y) -- (5.55,\y);
\node[font=\scriptsize, black, right] at (7.0,-3.5) {pressure};
\node[font=\scriptsize, black, right] at (7.05,-3.85) {grows with $x$};
\end{tikzpicture}
$$

> **Worked example.** A dam is a trapezoid $20\,\text{m}$ tall, $50\,\text{m}$
> wide at the top and $30\,\text{m}$ at the bottom; the water reaches
> $4\,\text{m}$ below the top. Find the hydrostatic force on the dam.
>
> Put $x = 0$ at the water surface, pointing down, so the water runs from
> $x = 0$ to $x = 16$. Similar triangles give the half-width increase
> $a = 8 - \tfrac{x}{2}$, so the strip width is $w(x) = 2(15 + a) = 46 - x$.
> With $\rho = 1000$ and $g = 9.8$,
>
> $$
> F = \int_0^{16} 1000(9.8)\, x\,(46 - x)\, \d x
>   = 9800 \left[ 23x^2 - \frac{x^3}{3} \right]_0^{16}
>   \approx 4.43 \times 10^7\ \text{N}.
> $$

## Moments and centers of mass

A plate balances horizontally on a single point: its **center of mass**. The
one-dimensional law is Archimedes' Law of the Lever: two masses on a rod balance
about the fulcrum when $m_1 d_1 = m_2 d_2$.

$$
% caption: Two masses balance when m1 d1 = m2 d2; the balance point is the
% mass-weighted average of the two positions.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% the rod
\draw[black, thick] (0,0) -- (7,0);
% masses
\filldraw[acc!14, draw=acc, thick] (0.9,0) rectangle (1.55,0.78);
\node[font=\scriptsize, black] at (1.22,0.4) {$m_1$};
\filldraw[acc!14, draw=acc, thick] (5.6,0) rectangle (6.05,0.5);
\node[font=\scriptsize, black] at (5.82,0.27) {$m_2$};
% fulcrum at the balance point
\draw[black, thick] (2.6,-0.05) -- (2.35,-0.68) -- (2.85,-0.68) -- cycle;
\node[font=\scriptsize, black, below] at (2.6,-0.7) {fulcrum};
% distances measured above the rod
\draw[black, dashed] (1.22,0.78) -- (1.22,1.4);
\draw[black, dashed] (2.6,0) -- (2.6,1.4);
\draw[black, dashed] (5.82,0.5) -- (5.82,1.4);
\draw[<->, black!70] (1.22,1.25) -- (2.6,1.25) node[midway, above, font=\scriptsize] {$d_1$};
\draw[<->, black!70] (2.6,1.25) -- (5.82,1.25) node[midway, above, font=\scriptsize] {$d_2$};
\end{tikzpicture}
$$

Placing masses $m_i$ at coordinates $x_i$ on the $x$-axis, the balance point is
the mass-weighted average

$$
\bar{x} = \frac{\sum_i m_i x_i}{\sum_i m_i} = \frac{M}{m},
\qquad M = \sum_i m_i x_i,
$$

where $M$ is the **moment** about the origin and $m$ the total mass. In the
plane, masses $m_i$ at $(x_i, y_i)$ have two moments, one about each axis:

- **Moment about the $y$-axis:** $M_y = \sum_i m_i x_i$ (measures the tendency to
  rotate about the $y$-axis).
- **Moment about the $x$-axis:** $M_x = \sum_i m_i y_i$.

The center of mass is $(\bar{x}, \bar{y}) = (M_y/m,\, M_x/m)$ — the point where a
single particle of mass $m$ would have the same moments as the whole system.

### The centroid of a lamina

For a flat plate of uniform density occupying a region under $y = f(x)$ over
$[a, b]$, replace the discrete masses by strips. A strip at $x$ has mass
proportional to its area $f(x)\, \Delta x$, and its own center of mass sits at
its midpoint height $\tfrac12 f(x)$. Summing the strip moments and passing to the
limit, the density cancels, and the balance point — the **centroid** — depends
only on the shape.

> **Definition (Centroid of a region).** The centroid of the region under
> $y = f(x)$ on $[a, b]$, with area $A = \int_a^b f(x)\, \d x$, is
> $(\bar{x}, \bar{y})$ where
>
> $$
> \bar{x} = \frac{1}{A} \int_a^b x\, f(x)\, \d x,
> \qquad
> \bar{y} = \frac{1}{A} \int_a^b \tfrac{1}{2}\,[f(x)]^2\, \d x.
> $$
>
> For a region between $f(x) \ge g(x)$, the strip height is $f - g$ and its
> midpoint is $\tfrac12(f + g)$, giving
> $\bar{x} = \tfrac1A\int x(f - g)\, \d x$ and
> $\bar{y} = \tfrac1A\int \tfrac12\bigl([f]^2 - [g]^2\bigr)\, \d x$.

$$
% caption: The centroid is the balance point of the region; each vertical strip
% contributes a moment through its position x and its midpoint height f(x)/2.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (-0.3,0) -- (6.4,0) node[right] {$x$};
\draw[->, black] (0,-0.3) -- (0,3.9) node[above] {$y$};
% region under a curve
\draw[black, thick] (0.5,0.5) .. controls (2.4,3.6) and (4.2,2.2) .. (5.8,3.2)
  node[right, black] {$y = f(x)$};
\filldraw[acc!10, draw=none] (0.5,0) -- (0.5,0.5)
  .. controls (2.4,3.6) and (4.2,2.2) .. (5.8,3.2) -- (5.8,0) -- cycle;
% representative strip
\filldraw[acc!22, draw=acc] (2.7,0) rectangle (2.96,2.95);
\fill[black!70] (2.83,1.48) circle (1.3pt);
\node[font=\scriptsize, black, right] at (2.96,1.48) {$\tfrac12 f(x)$};
\draw[black!70] (2.83,0) -- (2.83,-0.05) node[below, font=\scriptsize] {$x$};
% centroid
\fill[acc] (3.05,1.62) circle (2.6pt);
\node[acc, font=\scriptsize, above right] at (3.05,1.62) {$(\bar{x}, \bar{y})$};
% balance triangle
\draw[black!70] (2.75,-0.32) -- (3.35,-0.32) -- (3.05,-0.02) -- cycle;
\end{tikzpicture}
$$

> **Worked example.** Find the centroid of the half-disk under
> $y = \sqrt{r^2 - x^2}$ on $[-r, r]$.
>
> Symmetry forces $\bar{x} = 0$. With $A = \tfrac12\pi r^2$,
>
> $$
> \bar{y} = \frac{1}{\tfrac12\pi r^2} \int_{-r}^{r} \tfrac12 (r^2 - x^2)\, \d x
>   = \frac{1}{\pi r^2} \cdot 2 \int_0^r (r^2 - x^2)\, \d x
>   = \frac{4r}{3\pi}.
> $$
>
> The centroid sits at $\bigl(0, \tfrac{4r}{3\pi}\bigr)$, below the geometric
> midline, since more of the half-disk lies near the diameter.

### The Theorem of Pappus

Once the centroid is known, a volume of revolution follows without an integral.

> **Theorem (Pappus).** If a plane region of area $A$ lies entirely on one side
> of a line and is revolved about that line, the volume of the resulting solid is
>
> $$
> V = A\, d,
> $$
>
> where $d = 2\pi \bar{x}$ is the distance traveled by the centroid in one
> revolution.

$$
% caption: Pappus theorem: revolving a region of area A about an axis it does
% not meet gives volume A times 2 pi R, with R the centroid distance.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% axis of revolution
\draw[black, thick] (0,-1.7) -- (0,2.1) node[above, font=\scriptsize] {axis};
% torus outline (schematic top view)
\draw[black!45] (0,0) ellipse (3.4 and 0.95);
\draw[black!45] (0,0) ellipse (1.7 and 0.46);
% generating region
\filldraw[acc!14, draw=acc, thick] (2.55,0) circle (0.62);
\fill[acc] (2.55,0) circle (1.6pt);
\node[font=\scriptsize, acc, above right] at (2.6,0.05) {$C$};
\draw[black!70, ->] (2.55,0) -- (2.13,0.44);
\node[font=\scriptsize, black!70] at (2.2,0.2) {$r$};
\node[font=\scriptsize, black] at (2.55,-0.95) {area $A$};
% centroid distance R
\draw[<->, black!70] (0,-1.35) -- (2.55,-1.35) node[midway, below, font=\scriptsize] {$R$};
\draw[black, dashed] (2.55,-0.62) -- (2.55,-1.35);
\end{tikzpicture}
$$

> **Worked example.** Find the volume of a torus of tube radius $r$ formed by
> revolving a circle about a line a distance $R > r$ from its center.
>
> The disk has area $A = \pi r^2$, and its centroid is the circle's center,
> which travels $d = 2\pi R$. By Pappus,
>
> $$
> V = (\pi r^2)(2\pi R) = 2\pi^2 r^2 R,
> $$
>
> obtained without evaluating a single integral.

## Economics: consumer and producer surplus

The **demand function** $p(x)$ gives the price at which $x$ units sell; it
decreases, since selling more requires a lower price. At the market price
$P = p(X)$, consumers who would have paid more than $P$ keep the difference. Cut
$[0, X]$ into intervals: the buyers of the units near $x_i$ valued them at
$p(x_i)$ but paid only $P$, saving $[p(x_i) - P]\, \Delta x$. Summing gives the
total saving.

> **Definition (Consumer surplus).** At sales level $X$ and price $P = p(X)$, the
> consumer surplus is
>
> $$
> \int_0^X \bigl[p(x) - P\bigr]\, \d x,
> $$
>
> the area between the demand curve and the horizontal line $p = P$.

$$
% caption: Consumer surplus is the area under the demand curve and above the
% market price line p = P, out to the sales level X.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (-0.3,0) -- (6.4,0) node[right] {$x$};
\draw[->, black] (0,-0.3) -- (0,4.2) node[above] {$p$};
% demand curve (decreasing)
\draw[black, thick] (0.2,3.9) .. controls (2.2,2.5) and (3.4,1.7) .. (5.9,1.0)
  node[right, black] {$p = p(x)$};
% price level P and sales X
\coordinate (X) at (4.3,0);
\coordinate (P) at (0,1.55);
\draw[black, dashed] (0,1.55) -- (4.3,1.55) -- (4.3,0);
% shaded consumer surplus
\filldraw[acc!14, draw=acc!60]
  (0.2,3.9) .. controls (2.2,2.5) and (3.4,1.7) .. (4.3,1.55) -- (0,1.55) -- cycle;
\node[font=\scriptsize, acc] at (1.55,2.35) {consumer};
\node[font=\scriptsize, acc] at (1.6,1.95) {surplus};
% point (X,P)
\fill[black!70] (4.3,1.55) circle (1.6pt);
\node[font=\scriptsize, black!70, above right] at (4.3,1.55) {$(X, P)$};
\node[font=\scriptsize, black, below] at (4.3,0) {$X$};
\node[font=\scriptsize, black, left] at (0,1.55) {$P$};
\end{tikzpicture}
$$

> **Worked example.** For the demand function
> $p(x) = 1200 - 0.2x - 0.0001x^2$ and sales level $X = 500$, find the consumer
> surplus.
>
> The market price is $P = p(500) = 1200 - 0.2(500) - 0.0001(500)^2 = 1075$, so
>
> $$
> \int_0^{500} \bigl[p(x) - 1075\bigr]\, \d x
>   = \int_0^{500} (125 - 0.2x - 0.0001x^2)\, \d x
>   = \text{\textdollar}33{,}333.33.
> $$

The mirror image on the supply side is **producer surplus**,
$\int_0^X [P - p_S(x)]\, \d x$, the extra earned by producers willing to sell below
$P$. Their sum, the total surplus, is maximized when supply meets demand at the
equilibrium price.

The same slicing measures biological flow. Blood in a vessel of radius $R$ moves
fastest at the center: $v(r) = \tfrac{P}{4\eta l}(R^2 - r^2)$ at distance $r$
from the axis.

> **Worked example.** Find the volume of blood flowing past a cross-section of a
> vessel of radius $R$ per unit time.
>
> The flow through a thin ring of radius $r$ and area $2\pi r\, \Delta r$ is
> $v(r)$ times that area. Summing over rings gives the flux
>
> $$
> F = \int_0^R 2\pi r\, v(r)\, \d r = \frac{\pi P R^4}{8\eta l}.
> $$
>
> This is **Poiseuille's Law**: flux scales with the _fourth_ power of the
> radius, so a modest narrowing of a vessel sharply cuts flow.

## Probability

A **continuous random variable** $X$ — a height, a lifetime, a waiting time —
takes values across an interval, and probabilities attach to intervals, not
points. They are areas under a **probability density function**.

> **Definition (Probability density function).** A function $f$ is a probability
> density function for $X$ if $f(x) \ge 0$ for all $x$,
>
> $$
> \int_{-\infty}^{\infty} f(x)\, \d x = 1,
> \qquad\text{and}\qquad
> P(a \le X \le b) = \int_a^b f(x)\, \d x.
> $$

The first condition says total probability is $1$; the second reads the
probability of any interval as the area over it.

$$
% caption: A probability density function; the shaded area over [a,b] is the
% probability P(a <= X <= b), and the mean mu is the balance point.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (-0.3,0) -- (7.0,0) node[right] {$x$};
\draw[->, black] (0,-0.3) -- (0,3.2) node[above] {$y$};
% bell-shaped density
\draw[black, thick]
  (0.2,0.12) .. controls (2.0,0.2) and (2.4,2.8) .. (3.5,2.8)
  .. controls (4.6,2.8) and (5.0,0.2) .. (6.8,0.12)
  node[right, black] {$y = f(x)$};
% shaded interval [a,b]
\begin{scope}
  \clip (2.6,0) rectangle (4.4,3.2);
  \filldraw[acc!16, draw=none]
    (0.2,0) -- (0.2,0.12) .. controls (2.0,0.2) and (2.4,2.8) .. (3.5,2.8)
    .. controls (4.6,2.8) and (5.0,0.2) .. (6.8,0.12) -- (6.8,0) -- cycle;
\end{scope}
\draw[black, dashed] (2.6,0) -- (2.6,1.95);
\draw[black, dashed] (4.4,0) -- (4.4,1.95);
\node[font=\scriptsize, black, below] at (2.6,0) {$a$};
\node[font=\scriptsize, black, below] at (4.4,0) {$b$};
\node[font=\scriptsize, acc] at (3.5,1.05) {$P(a \le X \le b)$};
% mean marker
\draw[black!70] (3.5,0) -- (3.5,2.8);
\node[font=\scriptsize, black!70, below] at (3.5,0) {mean};
\end{tikzpicture}
$$

> **Worked example.** Verify that $f(x) = 0.006\,x(10 - x)$ for
> $0 \le x \le 10$ (and $0$ elsewhere) is a probability density, then find
> $P(4 \le X \le 8)$.
>
> The function is nonnegative on $[0, 10]$, and
>
> $$
> \int_0^{10} 0.006\,x(10 - x)\, \d x
>   = 0.006 \left[ 5x^2 - \tfrac13 x^3 \right]_0^{10} = 1,
> $$
>
> so it integrates to $1$ and is a valid density. The probability of the
> interval is the same integrand over $[4, 8]$:
>
> $$
> P(4 \le X \le 8) = \int_4^8 0.006\,x(10 - x)\, \d x = 0.544.
> $$

### The mean

Balancing the density like a lamina locates its center. The **mean** is the
$x$-coordinate of the centroid of the region under $f$ — the value about which
$f$ balances.

> **Definition (Mean).** The mean (or expected value) of a random variable $X$
> with density $f$ is
>
> $$
> \mu = \int_{-\infty}^{\infty} x\, f(x)\, \d x.
> $$

Waiting times and failure times are commonly modeled by the **exponential
density** $f(t) = c e^{-ct}$ for $t \ge 0$ (and $0$ for $t < 0$).

> **Worked example.** A call center answers calls in a mean time of $5$ minutes,
> exponentially distributed. Find the probability a call is answered within the
> first minute.
>
> The mean of $f(t) = ce^{-ct}$, via
> [integration by parts](/calculus/techniques-of-integration/integration-by-parts),
> is
>
> $$
> \mu = \int_0^{\infty} t\, c e^{-ct}\, \d t = \frac{1}{c},
> $$
>
> the boundary term vanishing by
> [l'Hospital's Rule](/calculus/exponential-logarithmic-and-inverse-functions/lhospitals-rule).
> So $c = 1/\mu$ and $f(t) = \tfrac1\mu e^{-t/\mu}$. With $\mu = 5$,
>
> $$
> P(0 \le T \le 1) = \int_0^1 0.2\, e^{-t/5}\, \d t = 1 - e^{-1/5} \approx 0.181.
> $$
>
> About $18\%$ of calls are answered in the first minute, while
> $P(T > 5) = e^{-1} \approx 0.37$; the long right tail pulls the mean above the
> typical wait.

**The normal distribution.** Heights, test scores, and measurement errors follow
the bell-shaped normal density

$$
f(x) = \frac{1}{\sigma\sqrt{2\pi}}\, e^{-(x - \mu)^2 / (2\sigma^2)},
$$

with mean $\mu$ and standard deviation $\sigma$ controlling the spread. The
integrand has no elementary antiderivative, so interval probabilities are found
numerically — the reason the normal table exists.

## The common construction

The same integral recurs across every quantity above, differing only in the
contribution named for a single piece.

| Quantity | Strip / piece contributes | Integral |
| --- | --- | --- |
| Hydrostatic force | $\rho g\, x\, w(x)\, \d x$ | $\int_a^b \rho g\, x\, w(x)\, \d x$ |
| Centroid $\bar x$ | strip moment $x\, f(x)\, \d x$ | $\tfrac1A \int_a^b x\, f(x)\, \d x$ |
| Consumer surplus | saving $[p(x) - P]\, \d x$ | $\int_0^X [p(x) - P]\, \d x$ |
| Probability | $f(x)\, \d x$ over the interval | $\int_a^b f(x)\, \d x$ |
| Mean | $x\, f(x)\, \d x$ | $\int_{-\infty}^{\infty} x\, f(x)\, \d x$ |