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Arc length and surface area: √(1 + (y′)²) under the integral, and 2πr for the surface

The length of a curve is found like an area: cut it into tiny pieces, treat each piece as a straight segment, and add. Pythagoras gives each segment's length, and turning a segment around an axis gives a thin band whose area is its circumference times its width. The formulas are short. The mistakes come from simplifying a square root that doesn't simplify, leaving the slant out of a surface, and measuring the radius from the wrong axis.

Short answer

For y = f(x) on [a, b], the arc length is L = ∫ab √(1 + (y′)2) dx. Square the derivative, add 1, and look for a perfect square under the root; never split √(1 + (y′)2) into 1 + y′. Turning the curve about an axis gives a surface of area S = ∫ 2πr ds, where ds = √(1 + (y′)2) dx is the slant width of each band and r is the distance to the axis of turning: r = y about the x-axis, r = x about the y-axis.

Why it works

Over a short step dx the curve rises about dy = y′ dx, so the piece of curve is close to the hypotenuse of a right triangle with legs dx and dy. Its length is ds = √(dx2 + dy2) = √(1 + (y′)2) dx, and adding the pieces is integrating. That's also the quickest check: an arc is never shorter than the straight chord between its ends, and never longer than the path along the axes.

When the piece of length ds is turned about an axis, it sweeps out a band shaped like a slice of a cone. Cut open and laid flat, the band is nearly a strip of length 2πr and width ds, not dx: the band runs along the slanted curve. Where the curve is flat the two are close, and where it is steep ds is much bigger.

Example 1: a substitution under the root

Find the length of y = 23x3/2 from x = 0 to x = 8.

  1. y′ = x1/2, 1 + (y′)2 = 1 + xSquare the derivative before adding 1.
  2. L = ∫08 (1 + x)1/2 dx = 23[(1 + x)3/2]08The power rule, with u = 1 + x.
  3. L = 23(27 − 1) = 523 ≈ 17.33Check: the curve ends at (8, 15.08), and the chord from (0, 0) has length 17.07, a little shorter.

Example 2: a perfect square under the root

Find the length of y = x48 + 14x2 from x = 1 to x = 2.

  1. y′ = x32 − 12x3Differentiate term by term.
  2. 1 + (y′)2 = x64 + 12 + 14x6 = (x32 + 12x3)2The middle term of (y′)² is −1/2, and adding 1 flips its sign. Curves like this are built so that it happens.
  3. L = ∫12 (x32 + 12x3) dx = [x48 − 14x2]12The square root is gone.
  4. L = (2 − 116) − (18 − 14) = 3316 = 2.0625Check: the chord from (1, 3/8) to (2, 33/16) has length about 1.96.

Example 3: a trig identity under the root

Find the length of y = ln(cos x) from x = 0 to x = π4.

  1. y′ = −sin xcos x = −tan xThe chain rule on ln.
  2. 1 + tan2 x = sec2 x, so √(1 + (y′)2) = sec xsec x > 0 on this interval, so no absolute value is needed.
  3. L = ∫0π/4 sec x dx = ln(sec x + tan x)|0π/4 = ln(√2 + 1) ≈ 0.8814At 0, ln(1 + 0) = 0. Check: the chord to (π/4, ln(√2/2)) has length about 0.858.

Example 4: a surface about the x-axis

Find the area of the surface made by turning y = x3, 0 ≤ x ≤ 1, about the x-axis.

  1. r = y = x3, ds = √(1 + 9x4) dxAbout the x-axis the radius is the height of the curve; y′ = 3x².
  2. S = ∫01 2πx3√(1 + 9x4) dxCircumference times slant width.
  3. u = 1 + 9x4, du = 36x3 dx: S = 2π36∫110 u1/2 duThe x³ from the radius is exactly what the substitution needs.
  4. S = π18 · 23(103/2 − 1) = π27(10√10 − 1) ≈ 3.563Units of length squared, as an area should have.

Example 5: a cone about the y-axis, checked by geometry

Turn the segment y = 2x, 0 ≤ x ≤ 1, about the y-axis, and find the area of the cone it makes.

  1. r = x, ds = √(1 + 22) dx = √5 dxAbout the y-axis the radius is horizontal, the distance x.
  2. S = ∫01 2πx√5 dx = π√5The integral of 2x from 0 to 1 is 1.
  3. S = π√5 ≈ 7.025Check with the cone formula πrℓ: the base radius is 1 and the slant height is √(1² + 2²) = √5.

Common mistakes, and the exact line they happen on

1. Taking the square root term by term

The curve from Example 1.

  1. 1 + (y′)2 = 1 + xRight so far.
  2. √(1 + x) = 1 + √x, so L = ∫08 (1 + √x) dx ≈ 23.08This is the mistake. √(a + b) is not √a + √b: squaring 1 + √x gives 1 + 2√x + x. The 23.08 is 8 + 15.08, exactly the path along the axes, which an arc can never exceed. The length is 52/3 ≈ 17.33.

2. Leaving out the slant factor

The surface from Example 4.

  1. r = x3Right so far.
  2. S = ∫01 2πx3 dx = π2 ≈ 1.571This is the mistake. The band's width was taken as dx, as in a volume by shells. Along the surface the band is ds = √(1 + 9x⁴) dx wide, which near x = 1 is more than three times dx. With it, S ≈ 3.563.

3. Measuring the radius from the wrong axis

The cone from Example 5.

  1. ds = √5 dxRight so far.
  2. S = ∫01 2πy ds = ∫01 2π(2x)√5 dx = 2π√5 ≈ 14.05This is the mistake. y is the distance to the x-axis, so this is the area of a different cone, the one made by turning the segment about the x-axis, with base radius 2. About the y-axis, each point travels on a circle of radius x, and S = π√5 ≈ 7.025.

Practice

  1. Find the length of y = 3x − 2 from x = 1 to x = 4, and check it with the distance formula.

    Show answer

    √(1 + 9) = √10, so L = 3√10 ≈ 9.487. The ends are (1, 1) and (4, 10), at distance √(9 + 81) = 3√10.

  2. Find the length of y = 13(x2 + 2)3/2 from x = 0 to x = 1.

    Show answer

    y′ = x√(x2 + 2), so 1 + (y′)2 = x4 + 2x2 + 1 = (x2 + 1)2, and L = ∫01 (x2 + 1) dx = 43.

  3. Find the length of y = x22 − ln x4 from x = 1 to x = e.

    Show answer

    y′ = x − 14x, and 1 + (y′)2 = (x + 14x)2. So L = [x22 + ln x4]1e = 2e2 − 14 ≈ 3.445.

  4. Turn y = √(4 − x2), −1 ≤ x ≤ 1, about the x-axis. Find the area of this band of a sphere.

    Show answer

    ds = 2√(4 − x2) dx, so 2πy ds = 4π dx and S = 4π · 2 = 8π ≈ 25.13.

  5. Turn y = 2√x, 0 ≤ x ≤ 3, about the x-axis, and find the surface area.

    Show answer

    y′ = 1√x, so 2πy√(1 + (y′)2) = 4π√x · √(x + 1x) = 4π√(x + 1), and S = 4π · 23(8 − 1) = 56π3 ≈ 58.64.

Practice this

Printable PDF: arc-length-and-surface-area.pdf

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