---
title: Work, Average Value, Arc Length, and Surface Area
module: Applications of Integration
moduleNumber: 5
lessonNumber: 2
order: 502
summary: >
  The work done by a force that varies with position, the average value of a
  function and the Mean Value Theorem it satisfies, the length of a curve, and
  the area of a surface swept out by revolving that curve. Each is a limit of
  Riemann sums, hence a definite integral.
topics: [Applications of Integration]
sources:
  - book: Stewart
    ref: "Ch. 5 — Applications of Integration; §5.4 Work, §5.5 Average Value of a Function"
  - book: Stewart
    ref: "Ch. 8 — Further Applications of Integration; §8.1 Arc Length, §8.2 Area of a Surface of Revolution"
---

Work, average value, arc length, and surface area are each built the same way:
cut the quantity into thin pieces, approximate each piece, sum, and take the
limit. The integrand is whatever a single small piece contributes.

## Work done by a variable force

In physics, **work** is force times distance — but only when the force is
constant. Lifting a weight $F$ through a height $d$ does work $W = Fd$, measured
in joules ($1\,\text{J} = 1\,\text{N}\cdot\text{m}$) or foot-pounds. When the
force varies with position, no single value of $F$ applies over the whole path,
and the product breaks into pieces.

Let a force $f(x)$ act along the $x$-axis as an object moves from $x = a$ to
$x = b$. Over a short subinterval $[x_{i-1}, x_i]$ the continuous force is nearly
constant at $f(x_i^\ast)$, so the work on that piece is about
$f(x_i^\ast)\, \Delta x$. Summing and taking the limit gives an integral.

> **Definition (Work).** The work done by a continuous force $f(x)$ in moving an
> object along the $x$-axis from $a$ to $b$ is
>
> $$
> W = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^\ast)\, \Delta x = \int_a^b f(x)\, \d x.
> $$

By **Hooke's Law** the force to hold a spring stretched $x$ units beyond its
natural length is $f(x) = kx$, with $k$ the spring constant.

> **Worked example.** A $40\,\text{N}$ force holds a spring stretched
> $5\,\text{cm} = 0.05\,\text{m}$ beyond its natural length. Find the work to
> stretch it from $0.05\,\text{m}$ to $0.08\,\text{m}$.
>
> The force law fixes $k$: from $k(0.05) = 40$, $k = 800\,\text{N/m}$. Then
>
> $$
> W = \int_{0.05}^{0.08} 800x\, \d x
>   = 400 \bigl[(0.08)^2 - (0.05)^2\bigr] = 1.56\ \text{J}.
> $$

$$
% caption: A spring force f(x) = kx rises linearly with stretch, so the work to
% stretch from a to b is the shaded area under the line.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (-0.3,0) -- (6.2,0) node[right] {$x$};
\draw[->, black] (0,-0.3) -- (0,4.0) node[above] {force};
% force line f = kx
\draw[black, thick] (0,0) -- (5.6,3.6) node[right, black] {$f(x) = kx$};
% work = area under the line from a to b
\filldraw[acc!14, draw=acc!60] (2.0,0) -- (2.0,1.286) -- (4.4,2.829) -- (4.4,0) -- cycle;
\node[font=\scriptsize, acc] at (3.25,0.65) {work};
\draw[black, dashed] (2.0,0) -- (2.0,1.286);
\draw[black, dashed] (4.4,0) -- (4.4,2.829);
\node[font=\scriptsize, black, below] at (2.0,0) {$a$};
\node[font=\scriptsize, black, below] at (4.4,0) {$b$};
\end{tikzpicture}
$$

### Lifting and pumping

Two families of problems supply the force implicitly: lifting a heavy cable, and
pumping fluid out of a tank. In both, the object is cut into horizontal layers,
and each layer contributes weight times the distance _it_ must be lifted — a
distance that changes from layer to layer.

- **Cable.** A $200\,\text{lb}$ cable $100\,\text{ft}$ long hangs from a
  building; it weighs $2\,\text{lb/ft}$. Measuring $x$ downward from the top, the
  layer at depth $x$ weighs $2\,\Delta x$ and rises a distance $x$, so
  $W = \int_0^{100} 2x\, \d x = \bigl[x^2\bigr]_0^{100} = 10{,}000\ \text{ft-lb}$.
- **Fluid.** For fluid the layer's weight is its density times its volume
  (cross-sectional area times $\Delta x$), and the lift distance is measured to
  the spout.

A load carried by the cable adds a second contribution.

> **Worked example.** A cable weighing $2\,\text{lb/ft}$ lifts $800\,\text{lb}$
> of coal up a $500\,\text{ft}$ mine shaft. Find the total work.
>
> The coal is a constant $800\,\text{lb}$ raised $500\,\text{ft}$, doing
> $800 \cdot 500 = 400{,}000\,\text{ft-lb}$. For the cable, the layer at depth
> $x$ weighs $2\,\Delta x$ and rises $x$, contributing
> $\int_0^{500} 2x\, \d x = \bigl[x^2\bigr]_0^{500} = 250{,}000\,\text{ft-lb}$.
> The two add:
>
> $$
> W = 400{,}000 + 250{,}000 = 650{,}000\ \text{ft-lb}.
> $$

> **Worked example.** An inverted cone of height $10\,\text{m}$ and base radius
> $4\,\text{m}$ holds water to a depth of $8\,\text{m}$. Find the work to pump
> all the water to the rim.
>
> Measure $x$ downward from the top. By similar triangles the layer at depth
> $x$ is a disk of radius $r = \tfrac{2}{5}(10 - x)$, so its volume is
> $\pi r^2\, \Delta x$. At water density $1000\,\text{kg/m}^3$ with
> $g = 9.8\,\text{m/s}^2$, the layer weighs
> $1000 \cdot 9.8 \cdot \tfrac{4\pi}{25}(10 - x)^2\, \Delta x$ and rises a
> distance $x$. The water occupies depths $2 \le x \le 10$, so
>
> $$
> W = \int_2^{10} 1568\pi\, x\, (10 - x)^2\, \d x \approx 3.4 \times 10^6\ \text{J}.
> $$

$$
% caption: Pumping empties the tank layer by layer; the layer at depth x rises a
% distance x, and its radius shrinks with depth by similar triangles.
\begin{tikzpicture}[scale=0.62, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% cone outline (apex at bottom)
\draw[black, thick] (-4,6) -- (0,0) -- (4,6);
\draw[black, thick] (-4,6) arc (180:360:4 and 0.7);
\draw[black, thick, dashed] (-4,6) arc (180:0:4 and 0.7);
% water surface at depth 2 (x measured down from top rim at y=6)
\draw[black] (-3.2,4.6) -- (0,0) -- (3.2,4.6) -- cycle;
\draw[black] (-3.2,4.6) arc (180:360:3.2 and 0.56);
\draw[black, dashed] (-3.2,4.6) arc (180:0:3.2 and 0.56);
% representative layer
\filldraw[acc!22, draw=acc] (-1.9,2.05) -- (1.9,2.05) -- (1.75,1.75) -- (-1.75,1.75) -- cycle;
\draw[<->, black!70] (0,1.9) -- (0,7.6) node[midway, right, font=\scriptsize] {lift $= x$};
% radius label
\draw[black!70, ->] (0,1.9) -- (1.9,1.9);
\node[black!70, font=\scriptsize, above] at (1.0,1.9) {$r$};
% depth axis
\draw[->, black] (5.2,6) -- (5.2,-0.4) node[right, font=\scriptsize] {$x$ (depth)};
\node[font=\scriptsize, black] at (5.2,6.35) {$0$};
\end{tikzpicture}
$$

In the pumping integral, two things vary with the slice at once: the layer's
size (through $r$) and the distance it travels (through $x$). The integral
multiplies them before summing.

## Average value of a function

The average of finitely many numbers is their sum over their count. A continuous
function has infinitely many values, so the sum becomes an integral. Sampling
$f$ at $n$ equally spaced points and averaging gives
$\tfrac{1}{n}\sum f(x_i^\ast)$; writing $n = (b - a)/\Delta x$ turns this into
$\tfrac{1}{b-a}\sum f(x_i^\ast)\, \Delta x$, and the limit is an integral.

> **Definition (Average value).** The average value of a continuous function $f$
> on $[a, b]$ is
>
> $$
> f_{\text{ave}} = \frac{1}{b - a} \int_a^b f(x)\, \d x.
> $$

For a positive function this is the height of the rectangle over $[a, b]$ whose
area equals the area under the curve: $\text{area} = f_{\text{ave}} \cdot (b-a)$.

> **Worked example.** Find the average value of $f(x) = 1 + x^2$ on $[-1, 2]$.
>
> $$
> f_{\text{ave}} = \frac{1}{3} \int_{-1}^{2} (1 + x^2)\, \d x
>   = \frac{1}{3} \left[ x + \frac{x^3}{3} \right]_{-1}^{2} = 2.
> $$

### The Mean Value Theorem for Integrals

A continuous function actually attains its average value somewhere on the
interval. This is the integral analogue of the Mean Value Theorem for
[derivatives](/calculus/applications-of-derivatives/extrema-and-the-mean-value-theorem).

> **Theorem (Mean Value Theorem for Integrals).** If $f$ is continuous on
> $[a, b]$, there exists a number $c \in [a, b]$ with
>
> $$
> f(c) = f_{\text{ave}} = \frac{1}{b - a} \int_a^b f(x)\, \d x,
> \qquad\text{equivalently}\qquad
> \int_a^b f(x)\, \d x = f(c)\,(b - a).
> $$

Geometrically, the rectangle of height $f(c)$ over $[a, b]$ has exactly the area
under the graph — the top of the graph can be shaved off at height $f(c)$ to
fill in its own valleys. For $f(x) = 1 + x^2$ on $[-1, 2]$, the value $c$
satisfies $1 + c^2 = 2$, so $c = \pm 1$, and both lie in the interval.

$$
% caption: The Mean Value Theorem for Integrals: the rectangle of height f(c)
% over [a,b] has the same area as the region under the curve.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (-0.3,0) -- (6.4,0) node[right] {$x$};
\draw[->, black] (0,-0.3) -- (0,3.9) node[above] {$y$};
% curve
\draw[black, thick] (0.6,1.0) .. controls (2.2,3.4) and (3.8,0.6) .. (5.7,2.9)
  node[right, black] {$y = f(x)$};
% equal-area rectangle at height f(c)=fave
\draw[black, dashed] (0.6,1.95) -- (5.7,1.95);
\filldraw[acc!10, draw=acc!60] (0.6,0) rectangle (5.7,1.95);
\node[font=\scriptsize, acc] at (3.15,1.55) {area $= f(c)\,(b-a)$};
\draw[black, dashed] (0.6,0) -- (0.6,1.0);
\draw[black, dashed] (5.7,0) -- (5.7,2.9);
\node[font=\scriptsize, black, below] at (0.6,0) {$a$};
\node[font=\scriptsize, black, below] at (5.7,0) {$b$};
\node[font=\scriptsize, black, left] at (0,1.95) {$f(c)$};
% mark c
\draw[black!70] (2.55,0) -- (2.55,1.95);
\node[font=\scriptsize, black!70, below] at (2.55,0) {$c$};
\end{tikzpicture}
$$

Average velocity is the cleanest instance. If $s(t)$ is displacement, the
average value of the velocity $v = s'$ over $[t_1, t_2]$ is

$$
v_{\text{ave}} = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} s'(t)\, \d t
  = \frac{s(t_2) - s(t_1)}{t_2 - t_1}
$$

by the [Net Change Theorem](/calculus/integrals/the-fundamental-theorem-of-calculus)
— exactly the elementary average-velocity formula, recovered from the integral
definition.

## Arc length

The length of a curve $y = f(x)$ is defined the way the circumference of a
circle is: inscribe a polygon, add its segment lengths, and refine. Partition
$[a, b]$ and join the points $P_i = (x_i, f(x_i))$; the polygon's length
approaches the curve's length as the mesh shrinks.

$$
% caption: A curve's length is the limit of the lengths of inscribed polygons;
% one segment has horizontal run dx and rise dy.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (-0.3,0) -- (6.6,0) node[right] {$x$};
% smooth curve
\draw[black, thick] (0.4,0.7) .. controls (2.0,3.3) and (4.0,0.4) .. (6.1,2.6)
  node[right, black] {$y = f(x)$};
% inscribed polygon vertices
\foreach \p in {(0.4,0.7),(1.55,2.35),(2.9,1.75),(4.2,0.95),(5.2,1.35),(6.1,2.6)}
  \fill[black!70] \p circle (1.6pt);
\draw[black!70, thick]
  (0.4,0.7) -- (1.55,2.35) -- (2.9,1.75) -- (4.2,0.95) -- (5.2,1.35) -- (6.1,2.6);
% one segment's legs
\draw[black, dashed] (2.9,1.75) -- (4.2,1.75) -- (4.2,0.95);
\node[font=\scriptsize, black, above] at (3.55,1.75) {$dx$};
\node[font=\scriptsize, black, right] at (4.2,1.35) {$dy$};
\node[font=\scriptsize, acc] at (3.35,1.05) {$\sqrt{dx^2 + dy^2}$};
\end{tikzpicture}
$$

Each segment has length
$\sqrt{(\Delta x)^2 + (\Delta y)^2}$. When $f$ is smooth, the Mean Value Theorem
gives $\Delta y_i = f'(x_i^\ast)\, \Delta x$ for some $x_i^\ast$, so the segment
length is $\sqrt{1 + [f'(x_i^\ast)]^2}\, \Delta x$. Summing and taking the limit
produces the arc-length integral.

> **Theorem (Arc Length Formula).** If $f'$ is continuous on $[a, b]$, the length
> of the curve $y = f(x)$, $a \le x \le b$, is
>
> $$
> L = \int_a^b \sqrt{1 + [f'(x)]^2}\, \d x
>   = \int_a^b \sqrt{1 + \left(\frac{\d y}{\d x}\right)^2}\, \d x.
> $$
>
> For a curve $x = g(y)$, $c \le y \le d$, the roles of $x$ and $y$ swap:
> $L = \int_c^d \sqrt{1 + (\d x/\d y)^2}\, \d y$.

> **Worked example.** Find the length of the semicubical parabola $y = x^{3/2}$
> (the top of $y^2 = x^3$) from $(1, 1)$ to $(4, 8)$.
>
> Here $\d y/\d x = \tfrac{3}{2}x^{1/2}$, so
>
> $$
> L = \int_1^4 \sqrt{1 + \tfrac{9}{4}x}\, \d x.
> $$
>
> The [substitution](/calculus/integrals/the-substitution-rule)
> $u = 1 + \tfrac94 x$, $\d u = \tfrac94\, \d x$, carries the limits
> $x = 1, 4$ to $u = \tfrac{13}{4}, 10$:
>
> $$
> L = \frac{4}{9} \int_{13/4}^{10} \sqrt{u}\, \d u
>   = \frac{4}{9} \cdot \frac{2}{3} \left[ u^{3/2} \right]_{13/4}^{10}
>   = \frac{1}{27}\bigl(80\sqrt{10} - 13\sqrt{13}\bigr) \approx 7.63.
> $$
>
> The straight-line distance from $(1,1)$ to $(4,8)$ is
> $\sqrt{58} \approx 7.62$, so the arc is slightly longer, as it must be.

For most curves $\sqrt{1 + [f'(x)]^2}$ has no elementary antiderivative, and the
length must be estimated numerically.

> **Worked example.** Find the length of $xy = 1$ from $(1, 1)$ to
> $(2, \tfrac12)$.
>
> With $y = 1/x$, $\d y/\d x = -1/x^2$, so
>
> $$
> L = \int_1^2 \sqrt{1 + \frac{1}{x^4}}\, \d x.
> $$
>
> The integrand has no elementary antiderivative;
> [Simpson's Rule](/calculus/techniques-of-integration/approximate-and-improper-integrals)
> gives $L \approx 1.1321$.

### The differential of arc length

Differentiating the arc-length function $s(x) = \int_a^x \sqrt{1 + [f'(t)]^2}\, \d t$
gives the differential

$$
\d s = \sqrt{1 + \left(\frac{\d y}{\d x}\right)^2}\, \d x,
\qquad\text{symmetrically}\qquad
(\d s)^2 = (\d x)^2 + (\d y)^2.
$$

The differential $\d s$ recurs: writing $L = \int \d s$ and choosing which variable
to solve for recovers either arc-length formula, and it reappears directly in
the surface-area integral below.

## Area of a surface of revolution

Revolve a curve about a line and it sweeps out a surface. To find its area,
approximate the curve by the same inscribed polygon; revolving one segment
produces a **frustum** of a cone — a conical band with slant height
$\ell = |P_{i-1}P_i|$ and average radius $r = \tfrac12(y_{i-1} + y_i)$. The
lateral area of such a band is $2\pi r \ell$.

$$
% caption: Revolving one polygon segment sweeps a conical frustum of slant
% height l and radii r1, r2; its lateral area is 2 pi times the average radius
% times l.
\begin{tikzpicture}[scale=0.9, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% axis
\draw[->, black] (-0.4,0) -- (6.6,0) node[right] {axis};
% frustum: two ellipses joined by slanted sides
\draw[black, thick] (1.2,0) ++(0,0.9) arc (90:270:0.32 and 0.9);
\draw[black, thick, dashed] (1.2,0) ++(0,0.9) arc (90:-90:0.32 and 0.9);
\draw[black, thick] (4.8,0) ++(0,1.7) arc (90:270:0.42 and 1.7);
\draw[black, thick, dashed] (4.8,0) ++(0,1.7) arc (90:-90:0.42 and 1.7);
\draw[black, thick] (1.2,0.9) -- (4.8,1.7);
\draw[black, thick] (1.2,-0.9) -- (4.8,-1.7);
\filldraw[acc!10, draw=none] (1.2,0.9) -- (4.8,1.7) -- (4.8,-1.7) -- (1.2,-0.9) -- cycle;
% radii
\draw[black!70, ->] (1.2,0) -- (1.2,0.9); \node[font=\scriptsize, black!70, left] at (1.2,0.5) {$r_1$};
\draw[black!70, ->] (4.8,0) -- (4.8,1.7); \node[font=\scriptsize, black!70, left] at (4.8,1.0) {$r_2$};
% slant height
\node[font=\scriptsize, acc, above, sloped] at (3.0,1.42) {$\ell$};
\end{tikzpicture}
$$

Because $y_{i-1}, y_i \approx f(x_i^\ast)$ and
$\ell = \sqrt{1 + [f'(x_i^\ast)]^2}\, \Delta x$, the band's area is about
$2\pi f(x_i^\ast) \sqrt{1 + [f'(x_i^\ast)]^2}\, \Delta x$. Summing gives the
surface-area integral, compactly $S = \int 2\pi r\, \d s$ with $r$ the distance
from the curve to the axis.

> **Definition (Surface area of revolution).** For a curve $y = f(x) \ge 0$ with
> $f'$ continuous on $[a, b]$, the surface generated by revolving it has area
>
> $$
> \text{about the } x\text{-axis:}\quad S = \int_a^b 2\pi\, f(x)\sqrt{1 + [f'(x)]^2}\, \d x = \int 2\pi y\, \d s,
> $$
> $$
> \text{about the } y\text{-axis:}\quad S = \int 2\pi x\, \d s.
> $$

> **Worked example.** Revolve $y = x^2$, $1 \le x \le 2$, about the $y$-axis and
> find the area of the surface generated.
>
> The radius is $x$ and $\d y/\d x = 2x$, so $\d s = \sqrt{1 + 4x^2}\, \d x$ and
>
> $$
> S = \int_1^2 2\pi x \sqrt{1 + 4x^2}\, \d x.
> $$
>
> With $u = 1 + 4x^2$, $\d u = 8x\, \d x$, the limits $x = 1, 2$ become
> $u = 5, 17$:
>
> $$
> S = \frac{\pi}{4} \int_5^{17} \sqrt{u}\, \d u
>   = \frac{\pi}{6} \left[ u^{3/2} \right]_5^{17}
>   = \frac{\pi}{6}\bigl(17\sqrt{17} - 5\sqrt{5}\bigr) \approx 30.85.
> $$

> **Worked example.** Revolve the circular arc $y = \sqrt{4 - x^2}$,
> $-1 \le x \le 1$, about the $x$-axis (a zone of a sphere of radius $2$).
>
> Here $\d y/\d x = -x/\sqrt{4 - x^2}$, and the integrand collapses:
>
> $$
> S = \int_{-1}^1 2\pi \sqrt{4 - x^2}\,\sqrt{1 + \frac{x^2}{4 - x^2}}\, \d x
>   = \int_{-1}^1 2\pi \sqrt{4 - x^2}\cdot\frac{2}{\sqrt{4 - x^2}}\, \d x
>   = \int_{-1}^1 4\pi\, \d x = 8\pi.
> $$
>
> The radius cancels the square root exactly: a spherical zone's area depends
> only on the width of the band, not on where along the sphere it sits.

## The shared method

Each quantity isolates the contribution of a single small piece and integrates
it over the interval.

| Quantity | Piece contributed | Integral |
| --- | --- | --- |
| Work | force over a small displacement | $\int_a^b f(x)\, \d x$ |
| Average value | $\tfrac{1}{b-a}$ of the strip area | $\tfrac{1}{b-a}\int_a^b f(x)\, \d x$ |
| Arc length | segment length $\sqrt{1 + (y')^2}\,\d x$ | $\int_a^b \sqrt{1 + (y')^2}\, \d x$ |
| Surface area | frustum band $2\pi r\, \d s$ | $\int 2\pi r\, \d s$ |