Questions on arc length and surface area

Question 1

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

Accepted 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.

Question 2

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

Accepted Answer

dy/dx = sinh x and 1 + sinh²x = cosh²x, so the integrand is cosh x itself. For the catenary, 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.

Question 3

Write down the formula for the length of the arc of y = f(x) from x = a to x = b, and the formula for the area of the surface generated when that arc is rotated once about the x-axis.

Accepted Answer

The arc length is ∫√(1 + (dy/dx)²) dx taken between the two limits, and the surface area is 2π∫y√(1 + (dy/dx)²) dx taken between the same limits. B1 for the arc length integral, B1 for the surface integral. The second is the first with an extra factor of 2πy, the circumference of the circle each point of the arc sweeps out as it turns. That factor is a distance from the axis, so an arc that dips below the x-axis needs |y| in its place.

Question 4

Find the length of the line y = 2x from x = 0 to x = 3 using the arc length formula, and check the answer by Pythagoras' theorem.

Accepted Answer

dy/dx = 2, so 1 + (dy/dx)² = 5 and the integrand is the constant √5. The length is √5 × 3 = 3√5, about 6.708. The line runs 3 across and 6 up, so Pythagoras gives √(3² + 6²) = √45 = 3√5, the same number. M1 for the integrand √5, A1 for 3√5, B1 for the check against Pythagoras.

Question 5

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

Accepted 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.

Question 6

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

Accepted 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.

Question 7

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.

Accepted 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.

Question 8

A curve has parametric equations x = t², y = 2t for 0 ≤ t ≤ 1. The curve is rotated once about the x-axis. Find the exact area of the surface generated.

Accepted Answer

For a parametric curve the element of arc is ds = √((dx/dt)² + (dy/dt)²) dt, and the surface integral is still 2π∫y ds, with everything in terms of t. Here dx/dt = 2t and dy/dt = 2, so ds = √(4t² + 4) dt = 2√(t² + 1) dt. M1 for the parametric ds inside 2π∫y ds.
With y = 2t the area is 2π∫2t × 2√(t² + 1) dt = 8π∫t√(t² + 1) dt from 0 to 1. A1 for the integrand.
Substituting u = t² + 1, with du = 2t dt, gives the antiderivative (1/3)(t² + 1)^3/2, which runs from 1/3 to 2√2/3. M1 for the substitution.
The area is 8π × (2√2 − 1)/3 = (8π/3)(2√2 − 1), about 15.32 square units. A1 for the exact area.
The limits are values of t, not of x, so they stay 0 and 1 and no conversion is needed; what does need care is that y is written as 2t and not left as y. Eliminating t gives y² = 4x, and the cartesian form 2π∫2√x √(1 + 1/x) dx from 0 to 1 gives the same number, but that route is harder to integrate, which is the usual reason a curve is handed over in parametric form.

Question 9

A curve has parametric equations x = 3 cos t and y = 3 sin t for 0 ≤ t ≤ π/2. Find its length, and check the answer against the circumference of a circle.

Accepted Answer

dx/dt = −3 sin t and dy/dt = 3 cos t, so (dx/dt)² + (dy/dt)² = 9(sin²t + cos²t) = 9 and the integrand is the constant 3. The length is 3 × π/2 = 3π/2, about 4.712. The equations describe the circle of radius 3 centred at the origin, and t running from 0 to π/2 traces a quarter of it, so the length should be (1/4)(2π × 3) = 3π/2, which it is. M1 for the parametric integrand, A1 for the bracket reducing to 9, A1 for 3π/2, B1 for the check against a quarter of the circumference.

Question 10

The arc of y = x² from x = 0 to x = 2 is rotated once about the y-axis. Find the exact area of the surface generated.

Accepted Answer

Turning about the y-axis makes the radius of the circle swept by each point its distance from that axis, which is x, so the integral is 2π∫x ds and not 2π∫y ds. Here ds = √(1 + (dy/dx)²) dx = √(1 + 4x²) dx, so the area is 2π∫x√(1 + 4x²) dx from 0 to 2. Substituting u = 1 + 4x², with du = 8x dx, gives the antiderivative (1/12)(1 + 4x²)^3/2, which runs from 1/12 to 17√17/12. The area is (2π/12)(17√17 − 1) = (π/6)(17√17 − 1), about 36.18 square units. B1 for the radius being x, M1 for the surface integral, M1 for the substitution, A1 for the antiderivative, A1 for the exact area. Using y as the radius here would integrate x²√(1 + 4x²) instead and answer a question about rotation round the other axis.

Question 11

State the radius to use in the surface area formula 2π∫r ds when the axis of rotation is the x-axis, and when it is the y-axis, and explain why that radius has to be a distance rather than a coordinate.

Accepted Answer

About the x-axis the radius is the distance from that axis, which is |y|; about the y-axis it is |x|. B1 for y about the x-axis, B1 for x about the y-axis. Each point of the arc travels once round a circle whose radius is its distance from the axis, and the thin band it sweeps has area 2πr ds. A distance is never negative, so where a coordinate goes negative the modulus is required, or the integral subtracts area that the surface actually has. B1 for the distance argument. An arc lying wholly on one side of the axis is the case in which the modulus can be dropped without comment, which is why most questions are set that way.

Question 12

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.

Accepted 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.

Question 13

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

Accepted 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.

Question 14

The arc of the circle x² + y² = a² with y ≥ 0 is rotated once about the x-axis, generating a sphere of radius a. Show that the area of the surface is 4πa².

Accepted Answer

Differentiating implicitly, 2x + 2y dy/dx = 0, so dy/dx = −x/y. Then 1 + (dy/dx)² = 1 + x²/y² = (y² + x²)/y² = a²/y², and for y > 0 the square root is a/y. The integrand of 2π∫y ds is therefore 2πy × a/y = 2πa, a constant, and the y has cancelled. Integrating from x = −a to x = a gives 2πa × 2a = 4πa². M1 for implicit differentiation, A1 for dy/dx = −x/y, M1 for the bracket collapsing to a²/y², A1 for the constant integrand 2πa, A1 for 4πa². The cancellation is what makes the classical result true: every band of the sphere cut between two planes a fixed distance apart carries the same area, whether it is taken at the equator or near a pole.

Question 15

The line y = (r/h)x from x = 0 to x = h is rotated once about the x-axis, generating a cone of base radius r and height h. Show that the curved surface area is πrl, where l = √(r² + h²) is the slant height.

Accepted Answer

dy/dx = r/h, a constant, so 1 + (dy/dx)² = 1 + r²/h² = (h² + r²)/h² and the square root is √(h² + r²)/h, that is l/h. The surface integral is 2π∫y ds = 2π∫(r/h)x × (l/h) dx from 0 to h = 2π(rl/h²)[x²/2] = 2π(rl/h²)(h²/2) = πrl. M1 for dy/dx, A1 for the square root written as l/h, M1 for the surface integral, A1 for the integration, A1 for πrl. Checking with r = 3 and h = 4 gives l = 5 and an area of 15π, about 47.12 square units. B1 for a numerical check. Quoting πr(r + l) answers a different question: that is the total surface area, and the flat base is not part of the surface of revolution.