Integration · Arc Length

Arc Length — Measuring the Length of a Curve

Every smooth curve has a definite length. Using integration, we can compute it exactly: from the humble parabola to the elegant catenary. This page derives the arc length formula from scratch, extends it to parametric curves and surface area, then walks through 8 fully worked examples.

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§ 01Why Arc Length Needs Integration

Measuring a straight line is trivial: you use a ruler. But curves bend. The arc length of a curve is the total distance you would travel if you walked along it, and that distance can only be captured exactly by an integral.

The core insight is this: zoom in on any smooth curve far enough and it looks like a straight line. That tiny straight-line segment has length given by the Pythagorean theorem. Arc length is found by summing infinitely many such segments: which is exactly what an integral does.

Arc length appears across science and engineering: the length of a cable suspended between two pylons (catenary), the perimeter of an ellipse, the circumference of a circle computed by calculus, the path length of a planetary orbit, and the length of a roller-coaster track. Wherever a curve lives in the real world, arc length quantifies how long it actually is.

Prerequisites Before tackling arc length, you should be comfortable with: the Pythagorean theorem, differentiation (chain rule, trig derivatives), and basic definite integration. Parametric differentiation is needed for § 03.

§ 02Deriving the Arc Length Formula

Let y = f(x) be a smooth curve on the interval [a, b]. We want to find the total length L of the curve between these two endpoints.

The Pythagoras Argument

Divide [a, b] into n sub-intervals of width Δx. On each sub-interval, approximate the curve by a chord: a tiny straight-line segment. The horizontal change across the segment is Δx and the vertical change is Δy, so by Pythagoras, the length of the chord is:

Chord Length
Δs = √( (Δx)² + (Δy)² )

Now factor out (Δx)² inside the square root:

Factoring
Δs = √( 1 + (Δy/Δx)² ) · Δx

As the partition gets finer, Δy/Δx → dy/dx = f′(x). The sum of all chord lengths approaches the definite integral:

The Arc Length Formula
L = ∫ab √( 1 + [f′(x)]² ) dx

Valid when f is continuously differentiable on [a, b].

x y Δx Δy Δs a b Curve y = f(x) Chord approximations
Each chord has length Δs = √(Δx² + Δy²). As the partition is refined, the sum of chord lengths converges to the definite integral for arc length.
A Common Trap The arc length integral often looks simple but is hard to evaluate analytically. For many curves, the integrand √(1 + [f′(x)]²) does not have a nice antiderivative — you'll need numerical integration. Always simplify [f′(x)]² fully before declaring the integral unsolvable.

§ 03All Four Arc Length Formulas

Depending on how a curve is described, the arc length formula takes different forms. Here are all four versions you need to know.

Cartesian — x form
L = ∫ab √(1 + [f′(x)]²) dx

Use when y is given as a function of x on [a, b].

Cartesian — y form
L = ∫cd √(1 + [g′(y)]²) dy

Use when x is given as a function of y on [c, d]. Often easier for sideways parabolas.

Parametric
L = ∫αβ √((dx/dt)² + (dy/dt)²) dt

Use when x = x(t), y = y(t) are given as functions of a parameter t.

Polar
L = ∫αβ √(r² + (dr/dθ)²) dθ

Use when r = f(θ) is given in polar coordinates.

The Arc Length Element ds

All four formulas are really the same idea dressed differently. In each case we are integrating the infinitesimal arc length element ds. In Cartesian coordinates:

Arc Length Element
ds = √(dx² + dy²) = √(1 + (dy/dx)²) dx

For parametric curves: ds = √((dx/dt)² + (dy/dt)²) dt

Think of ds as the length of an infinitesimal piece of the curve. The full arc length is simply L = ∫ ds across the relevant interval.

§ 04Arc Length for Parametric Curves

When a curve is defined parametrically as x = x(t), y = y(t) for t ∈ [α, β], the derivation is analogous. The key step is to write the chord length in terms of dt:

  • Write the chord length: Δs = √(Δx² + Δy²).

  • Divide and multiply inside by (Δt)²:

    Δs = √( (Δx/Δt)² + (Δy/Δt)² ) · Δt

  • Take the limit as Δt → 0: Δx/Δt → dx/dt and Δy/Δt → dy/dt.

  • Integrate over [α, β]:

    L = ∫αβ √( (dx/dt)² + (dy/dt)² ) dt
Circle Verification A circle of radius r has parametric equations x = r cos t, y = r sin t for t ∈ [0, 2π]. Then dx/dt = −r sin t and dy/dt = r cos t. The integrand is √(r²sin²t + r²cos²t) = r. So L = ∫₀²π r dt = 2πr. ✓ The arc length formula recovers the circumference formula.

§ 05Surface Area of Revolution

When a curve is rotated about an axis, it sweeps out a surface. The area of that surface can be computed using arc length as the key ingredient.

Imagine rotating a thin strip of arc length ds around the x-axis. The strip sits at height y above the axis, so it traces a thin ring of radius y and circumference 2πy. The area of this ring is 2πy ds. Integrating over the whole curve gives:

Surface Area — Rotation About the x-axis
S = 2π ∫ab f(x) · √(1 + [f′(x)]²) dx

where f(x) ≥ 0 on [a, b].

Surface Area — Rotation About the y-axis
S = 2π ∫ab x · √(1 + [f′(x)]²) dx

Notice: same integrand, but we multiply by x (the distance from the y-axis) instead of y.

x y = f(x) ds ring: area = 2πy ds a b
Rotating the curve y = f(x) about the x-axis. Each infinitesimal arc segment ds sweeps a ring of area 2πy ds. Integrating gives the total surface area.
Quick Memory Aid Surface area = 2π × (distance from axis) × (arc length element). For x-axis rotation multiply by y; for y-axis rotation multiply by x. The arc length element √(1 + [f′(x)]²) dx is always the same.

§ 06Worked Examples

Example 1 Straight line — a sanity check

Find the arc length of y = 3x + 2 from x = 0 to x = 4.

We know the answer should be the length of a straight line segment: √(4² + 12²) = √(16 + 144) = √160 = 4√10.

  1. Differentiate: f′(x) = 3.
  2. Compute 1 + [f′(x)]² = 1 + 9 = 10.
  3. Integrate: L = ∫₀⁴ √10 dx = 4√10
L = 4√10 ≈ 12.649 ✓
Example 2 Parabola arc length

Find the arc length of y = x²/2 from x = 0 to x = 1.

  1. f′(x) = x, so 1 + [f′(x)]² = 1 + x².
  2. L = ∫₀¹ √(1 + x²) dx
  3. Use the standard formula: ∫√(1+x²) dx = ½[x√(1+x²) + ln(x + √(1+x²))] + C.
  4. Evaluate from 0 to 1: L = ½[1·√2 + ln(1 + √2)] − ½[0 + ln(1)] L = ½[√2 + ln(1 + √2)]
L = ½(√2 + ln(1 + √2)) ≈ ½(1.4142 + 0.8814) ≈ 1.148
Example 3 Catenary — the hanging cable curve

Find the arc length of y = cosh(x) from x = −1 to x = 1. Recall cosh(x) = (eˣ + e⁻ˣ)/2 and d/dx[cosh x] = sinh x.

  1. f′(x) = sinh x, so [f′(x)]² = sinh²x.
  2. Use the identity 1 + sinh²x = cosh²x.
  3. √(1 + sinh²x) = cosh x (since cosh x > 0 always).
  4. L = ∫₋₁¹ cosh x dx = [sinh x]₋₁¹ = sinh(1) − sinh(−1) = 2 sinh(1).
L = 2 sinh(1) ≈ 2 × 1.1752 ≈ 2.350
Example 4 Parametric — circle circumference

Verify the circumference of a circle of radius r using the parametric arc length formula.

Set x = r cos t, y = r sin t for t ∈ [0, 2π].

  1. dx/dt = −r sin t, dy/dt = r cos t.
  2. (dx/dt)² + (dy/dt)² = r²sin²t + r²cos²t = r².
  3. √(r²) = r (since r > 0).
  4. L = ∫₀²π r dt = 2πr
L = 2πr ✓
Example 5 Parametric — cycloid one arch

A cycloid has equations x = r(t − sin t), y = r(1 − cos t). Find the arc length of one full arch (t from 0 to 2π).

  1. dx/dt = r(1 − cos t), dy/dt = r sin t.
  2. (dx/dt)² + (dy/dt)² = r²(1 − cos t)² + r²sin²t
  3. Expand: r²(1 − 2cos t + cos²t + sin²t) = r²(2 − 2cos t) = 2r²(1 − cos t).
  4. Use the identity 1 − cos t = 2sin²(t/2): √(2r² · 2sin²(t/2)) = 2r |sin(t/2)| = 2r sin(t/2) for t ∈ [0, 2π].
  5. L = ∫₀²π 2r sin(t/2) dt = 2r [−2cos(t/2)]₀²π = 2r(−2cos π + 2cos 0) = 2r(2 + 2) = 8r.
L = 8r — the famous result that one arch of a cycloid has length exactly eight times the generating radius.
Example 6 Surface area — rotating y = √x

Find the surface area generated when y = √x, 1 ≤ x ≤ 4, is rotated about the x-axis.

  1. f′(x) = 1/(2√x), so [f′(x)]² = 1/(4x).
  2. 1 + [f′(x)]² = 1 + 1/(4x) = (4x + 1)/(4x).
  3. √((4x+1)/(4x)) = √(4x+1) / (2√x).
  4. S = 2π ∫₁⁴ √x · √(4x+1)/(2√x) dx = π ∫₁⁴ √(4x+1) dx.
  5. Let u = 4x + 1, du = 4 dx. Limits: x=1 → u=5, x=4 → u=17. S = π/4 ∫₅¹⁷ √u du = π/4 · [⅔u^(3/2)]₅¹⁷ = π/6 [17^(3/2) − 5^(3/2)]
S = π/6 (17√17 − 5√5) ≈ π/6 (70.09 − 11.18) ≈ π/6 × 58.91 ≈ 30.85
Example 7 Using the y-form for x = g(y)

Find the arc length of x = y³/³ from y = 0 to y = 2. (This is easier to integrate in the y-direction.)

Here x = g(y) = y³/3, so g′(y) = y².

  1. 1 + [g′(y)]² = 1 + y⁴.
  2. L = ∫₀² √(1 + y⁴) dy
  3. This integral has no closed form: use numerical integration (Simpson's Rule with n = 4):
    h = 0.5; nodes at y = 0, 0.5, 1, 1.5, 2.
    Values of √(1+y⁴): 1, 1.0308, 1.4142, 2.2776, 4.1231.
    S₄ ≈ (0.5/3)[1 + 4(1.0308) + 2(1.4142) + 4(2.2776) + 4.1231] ≈ (1/6)[1 + 4.123 + 2.828 + 9.110 + 4.123] ≈ (1/6)(21.184) ≈ 3.53.
L ≈ 3.53 (numerical result; the true value is approximately 3.5391)
Example 8 Surface area of a sphere from calculus

Verify that the surface area of a sphere of radius r is 4πr² by rotating the upper semicircle y = √(r² − x²) about the x-axis.

  1. f′(x) = −x/√(r²−x²), so [f′(x)]² = x²/(r²−x²).
  2. 1 + [f′(x)]² = (r²−x²+x²)/(r²−x²) = r²/(r²−x²).
  3. √(r²/(r²−x²)) = r/√(r²−x²).
  4. S = 2π ∫₋ᵣʳ √(r²−x²) · r/√(r²−x²) dx = 2πr ∫₋ᵣʳ 1 dx = 2πr · 2r = 4πr².
S = 4πr² ✓ — the surface area formula derived rigorously from integration.

§ 07Common Mistakes

Mistake 1 — Forgetting the square root Arc length is L = ∫√(1 + [f′]²) dx, not ∫(1 + [f′]²) dx. The square root is the Pythagorean theorem at work and cannot be dropped.
Mistake 2 — Squaring incorrectly If f′(x) = 2x + 1, then [f′(x)]² = (2x+1)² = 4x² + 4x + 1, NOT 4x² + 1. Always expand the square fully before simplifying.
Mistake 3 — Wrong limits in parametric problems When using the parametric formula, the limits must be values of the parameter t, not values of x or y. Read the problem carefully.
Mistake 4 — Confusing arc length with area Arc length uses √(1 + [f′]²); area under a curve uses just f(x). Surface area of revolution uses 2πy · √(1 + [f′]²). Do not mix these up.
Mistake 5 — Dropping the absolute value in |sin(t/2)| When simplifying √(sin²θ), you get |sin θ|, not sin θ. Check whether the function is non-negative on your interval before dropping the absolute value.

§ 08Quick Reference Table

Curve Type Formula Key Step
Cartesian y = f(x) L = ∫ab √(1+[f′]²) dx Compute f′(x) and simplify 1 + [f′]²
Cartesian x = g(y) L = ∫cd √(1+[g′]²) dy Differentiate w.r.t. y
Parametric x(t), y(t) L = ∫αβ √(ẋ²+ẏ²) dt Limits must be t-values
Polar r = f(θ) L = ∫αβ √(r²+(r′)²) dθ Don't forget the r² term
Surface area (x-axis) S = 2π∫ f(x)√(1+[f′]²) dx Multiply arc length element by 2πy
Surface area (y-axis) S = 2π∫ x√(1+[f′]²) dx Multiply arc length element by 2πx

§ 09Test Yourself — 10-Question Quiz

Questions alternate between fill-in-the-blank computations and multiple-choice conceptual questions.

Score: 0 / 0
Question 1 of 10

The arc length of y = 4x + 1 from x = 0 to x = 3 equals k√17. What is k?

  1. f′(x) = 4, so [f′(x)]² = 16.
  2. 1 + 16 = 17, √17 is constant.
  3. L = ∫₀³ √17 dx = 3√17. So k = 3.
Question 2 of 10

Which integral correctly gives the arc length of y = x³ from x = 0 to x = 2?

  1. f(x) = x³, so f′(x) = 3x².
  2. [f′(x)]² = 9x⁴.
  3. Arc length = ∫₀² √(1 + 9x⁴) dx.
Question 3 of 10

For the curve y = (2/3)x^(3/2), find the arc length from x = 0 to x = 3. Enter your answer as a decimal rounded to 2 decimal places.

  1. f′(x) = x^(1/2) = √x, so [f′]² = x.
  2. L = ∫₀³ √(1+x) dx. Let u = 1+x, du = dx. Limits: 1 to 4.
  3. L = ∫₁⁴ √u du = [⅔u^(3/2)]₁⁴ = ⅔(8 − 1) = ⅔ × 7 = 14/3 ≈ 4.67.
Question 4 of 10

The catenary y = cosh(x) has arc length 2 sinh(1) on [−1, 1]. Why does this integral simplify so cleanly?

The identity 1 + sinh²x = cosh²x (analogous to 1 + tan²x = sec²x) means that √(1 + sinh²x) = cosh x. The integrand becomes simply cosh x, whose antiderivative is sinh x.

Question 5 of 10

A circle of radius 5 has arc length formula L = ∫₀²π 5 dt using the parametric form. What is L? (Give an exact numerical answer in terms of π, enter the coefficient of π.)

  1. L = ∫₀²π 5 dt = 5 × 2π = 10π.
  2. So the coefficient of π is 10.
Question 6 of 10

To find the surface area when y = f(x) is rotated about the y-axis, which formula applies?

When rotating about the y-axis, each arc element ds sweeps a ring at distance x from the axis (not y). So the area element is 2πx ds, giving S = 2π ∫ x √(1 + [f′]²) dx. Option (b) is the x-axis rotation formula.

Question 7 of 10

The cycloid arc length for one arch equals kr. What is the integer k?

As derived in Example 5, the arc length of one arch of a cycloid is 8r. So k = 8.

Question 8 of 10

For the parametric curve x = t², y = t³, the arc length element ds equals:

  1. dx/dt = 2t, dy/dt = 3t².
  2. (dx/dt)² + (dy/dt)² = 4t² + 9t⁴.
  3. ds = √(4t² + 9t⁴) dt.
Question 9 of 10

The surface area of a sphere of radius r, obtained by rotating y = √(r²−x²) about the x-axis, equals k·πr². What is k?

As shown in Example 8, the surface area of a full sphere is 4πr². So k = 4.

Question 10 of 10

Which statement about arc length integrals is TRUE?

The integrand √(1 + [f′(x)]²) is only integrable in closed form for special functions, the catenary, straight lines, and a handful of others. For most curves (e.g., the ellipse), numerical integration is required. This is why arc length is a classic motivation for numerical methods.

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