Arc Length and Surface Area

Why a derivative enters length

A small change along $y=f(x)$ has horizontal component $dx$ and vertical component $dy\approx f'(x)dx$ . Pythagoras gives $ds=\sqrt{1+[f'(x)]^2}\,dx$ . For a continuously differentiable graph,

L=\int_a^b\sqrt{1+[f'(x)]^2}\,dx.

If the graph has corners, split it into smooth pieces. If a derivative becomes unbounded at an endpoint, the resulting integral requires an improper-limit check. A valid length integral often has no elementary antiderivative; numerical evaluation is a legitimate conclusion.

Visual guide

VISUAL GUIDEA short chord approximates a curve segment

y = x²
Four chord segments

Arc length sums √(dx² + dy²), giving the factor √(1 + (y′)²) dx. The polygonal approximation follows y = x² on [0, 2]; finer segments approach the curve’s length. Surface area of revolution adds the circumference factor 2πr.

Worked example: a straight segment

For $y=3x+2$ on $0\le x\le2$ , $f'=3$ , so $L=\int_0^2\sqrt{10}dx=2\sqrt{10}$ . The endpoint displacement is $(2,6)$ , whose length is also $\sqrt{4+36}=2\sqrt{10}$ . This checks the formula against ordinary geometry.

Worked example: a surface of revolution

A narrow curve segment rotated about the $x$ -axis produces a band with area approximately $2\pi r\,ds$ . Therefore, when the surface is traced once,

S=2\pi\int_a^b|f(x)|\sqrt{1+[f'(x)]^2}\,dx.

For $y=x$ , $0\le x\le1$ , the result is $2\pi\sqrt2\int_0^1x dx=\pi\sqrt2$ . This is the lateral surface area of a cone with radius $1$ and slant height $\sqrt2$ . It does not include the base disk.

For rotation about a different horizontal line $y=c$ , the radius is $|f(x)-c|$ . The absolute value expresses a distance. A general parametrization must also be checked for repeated coverage, since integrating a curve twice counts its length or swept area twice.

Practice

Find the length of $y=4$ from $x=-2$ to $x=3$ .
Rotate that segment about the $x$ -axis. Find the swept surface area.
Set up, without claiming an elementary answer, the length of $y=x^2$ on $[0,1]$ .

Show worked solutions

$L=\int_{-2}^3 1dx=5$ .
$S=2\pi\cdot4\cdot5=40\pi$ , the lateral area of a cylinder.
$L=\int_0^1\sqrt{1+4x^2}dx$ . Numerical quadrature can approximate it; a hyperbolic substitution also evaluates it exactly.