Work, Average Value, Arc Length, and Surface Area

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.

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.

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 $W={\int }_0^{100}2x\mathbf{d}x=[x^2{]}_0^{100}=10,000\text{ft-lb}$.
• 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.

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 $\frac{1}{n}\sum f(x_i^{\ast })$; writing $n=(b-a)/\Delta x$ turns this into $\frac{1}{b-a}\sum f(x_i^{\ast })\Delta x$, and the limit is an integral.

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

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.

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.

Average velocity is the cleanest instance. If $s(t)$ is displacement, the average value of the velocity $v=s^{\prime }$ over $[t_1,t_2]$ is

by the Net Change Theorem — 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.

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

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

The differential of arc length

Differentiating the arc-length function $s(x)={\int }_a^x\sqrt{1+[f^{\prime }(t){]}^2}\mathbf{d}t$ gives the differential

The differential $\mathbf{d}s$ recurs: writing $L=\int \mathbf{d}s$ 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 $r=\frac{1}{2}(y_{i-1}+y_i)$. The lateral area of such a band is $2\pi r\ell$.

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

The shared method

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