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Arc length and surface area questions

Chop a curve into tiny straight pieces and Pythagoras measures each one. Add them with an integral for length, or spin them for the area of a surface of revolution.

7 original questions · 32 marks · the arc length and surface area notes · Further Pure 2

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• 4 marks · find · warm-up

  Find the length of the curve y = (2/3)x^3/2 from x = 0 to x = 3.

  Worked answer

  dy/dx = x^1/2, so 1 + (dy/dx)² = 1 + x and the integrand is √(1 + x). Integrating gives (2/3)(1 + x)^3/2, which runs from 2/3 to 16/3. The length is 14/3, about 4.667. B1 for dy/dx, M1 for the arc length integrand, M1 for the integration, A1 for 14/3.

• 3 marks · find · warm-up

  Find the length of y = cosh x from x = 0 to x = 1.

  Worked answer

  dy/dx = sinh x and 1 + sinh²x = cosh²x, so the integrand is cosh x itself. The catenary is the one curve on this specification where the square root disappears without a substitution. The length is [sinh x] from 0 to 1 = sinh 1, about 1.175. M1 for using the identity, M1 for integrating cosh x, A1 for sinh 1.

• 5 marks

  A curve has x = t², y = t³ for 0 ≤ t ≤ 1. Find its length.

  Worked answer

  dx/dt = 2t and dy/dt = 3t², so the integrand is √(4t² + 9t⁴) = t√(4 + 9t²) for t ≥ 0. Substituting u = 4 + 9t² gives (1/27)(4 + 9t²)^3/2, which runs from 8/27 to 13^3/2/27. The length is (13√13 − 8)/27, about 1.440. M1 for the parametric integrand, A1 for t√(4 + 9t²), M1 for the substitution, A1 for the antiderivative, A1 for the length.

• 4 marks · find

  Find the length of the polar curve r = e^θ from θ = 0 to θ = π.

  Worked answer

  In polar form the integrand is √(r² + (dr/dθ)²). Here dr/dθ = e^θ as well, so the bracket is 2e^2θ and the integrand is √2 e^θ. Integrating gives √2(e^π − 1), about 31.31. M1 for the polar integrand, A1 for √2 e^θ, M1 for integrating, A1 for the length. Every polar arc length needs r and its derivative, never r alone.

• 5 marks

  The arc of y = x³ from x = 0 to x = 1 is rotated once about the x-axis. Find the area of the surface generated.

  Worked answer

  The surface integral is 2π∫y ds = 2π∫x³√(1 + 9x⁴) dx from 0 to 1. Substituting u = 1 + 9x⁴ gives (π/27)(1 + 9x⁴)^3/2, which runs from π/27 to 10√10 π/27. The area is (π/27)(10√10 − 1), about 3.563 square units. M1 for the surface integral, A1 for the integrand, M1 for the substitution, A1 for the antiderivative, A1 for the area.

• 5 marks · stretch

  The arc of y = cosh x from x = 0 to x = 1 is rotated once about the x-axis. Find the area of the surface generated, in terms of sinh 2.

  Worked answer

  Since ds = cosh x dx, the integral is 2π∫cosh²x dx from 0 to 1. Writing cosh²x = (cosh 2x + 1)/2 gives 2π[x/2 + sinh(2x)/4], so the area is 2π(1/2 + (sinh 2)/4) = π + π(sinh 2)/2, about 8.839 square units. M1 for the surface integral, M1 for the double angle identity, A1 for the integration, M1 for the limits, A1 for the exact area. Without the double angle identity there is nothing to integrate.

• 6 marks · find · stretch

  Find the total perimeter of the cardioid r = 2(1 + cos θ). You may use the identity 1 + cos θ = 2 cos²(θ/2).

  Worked answer

  dr/dθ = −2 sin θ, so r² + (dr/dθ)² = 4(1 + cos θ)² + 4 sin²θ = 8 + 8 cos θ = 16 cos²(θ/2). The square root is 4|cos(θ/2)|, and cos(θ/2) ≥ 0 for 0 ≤ θ ≤ π, so the upper half has length ∫4 cos(θ/2) dθ = [8 sin(θ/2)] = 8. By symmetry in the initial line the perimeter is 16. M1 for r² + (dr/dθ)², A1 for 16 cos²(θ/2), B1 for the modulus and its range of validity, M1 for integrating, A1 for the half length, A1 for the perimeter. Dropping the modulus and running the integral straight from 0 to 2π returns 0, since cos(θ/2) is negative on the second half.

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