Parametric arc length and surfaces of revolution: VCE Specialist Mathematics Units 3 & 4

Loav · VCE Specialist Mathematics · Units 3 & 4

S34-I06 · Recommended sequence: Unit 4

1. Length accumulates small distances

For a parametric plane curve x = x(t), y = y(t), a small change in parameter produces small coordinate changes Δx and Δy. The straight-line distance between nearby points is approximately √((Δx)² + (Δy)²). Dividing by the parameter change, then taking the limiting relationship, gives the speed along the curve. Accumulating that speed produces arc length.

ds = √[(dx/dt)² + (dy/dt)²] dt.
L = ∫ab √[(x′(t))² + (y′(t))²] dt, a ≤ b.

Use a smooth or piecewise smooth parametrisation with finite length, and split at any needed piece boundaries. The parameter need not represent physical time. If it does, the square root is speed and its integral is distance travelled. If it is a geometric parameter, the same expression measures length. Squaring the coordinate derivatives prevents negative velocity components from cancelling the distance.

The straight-line distance between the endpoints is only a lower bound for the length, with equality for a straight path traversed without reversal. A curve returning to its starting point has zero endpoint displacement but generally positive length. This distinction is the geometric version of the difference between displacement and distance in motion.

The integral counts the traversal described by the parameter interval. If the curve retraces itself, repeated sections are counted repeatedly. Decide whether the question asks for distance travelled or the length of the geometric curve. A restricted interval may be necessary to describe each part of a geometric curve once. Orientation can reverse without changing the length, but multiplicity does matter.

2. Cartesian formulas are special parametrisations

For y = f(x), use x itself as the parameter: the curve is (x, f(x)). Then dx/dx = 1 and the length formula becomes ∫√(1 + [f′(x)]²) dx. For x = g(y), use y as the parameter instead. These are not separate principles to memorise; they are the same Pythagorean distance formula with a convenient parameter.

L = ∫pq √[1 + (dy/dx)²] dx.
L = ∫rs √[1 + (dx/dy)²] dy.

Choose a variable that describes the required curve without missing branches. A complete circle is not a single y-function of x, but a circular parametrisation covers it naturally. A vertical tangent may make dy/dx inconvenient even while a parametric derivative remains well behaved. Avoid dividing by x′(t) merely to create a Cartesian formula when the original parameter already works.

Keep the square root around the whole sum of squares. In general √(a² + b²) is not |a| + |b|. Also √(t²) = |t|, so simplification must use the parameter interval. A sign error here can turn a non-negative speed into a signed expression and cause an impossible negative length.

Worked example 1: Parametric length with a sign check

A curve is given by x = t² and y = (2/3)t³ for 0 ≤ t ≤ 2. Its derivatives are x′ = 2t and y′ = 2t². Therefore speed is √(4t² + 4t⁴) = 2|t|√(1 + t²). On the stated interval t is non-negative, so this simplifies to 2t√(1 + t²).

Hence L = ∫02 2t√(1 + t²) dt. With u = 1 + t² and bounds 1 and 5, L = [2u3/2/3]15 = (2/3)(5√5 − 1), approximately 6.787 length units. The endpoint distance is √(4² + (16/3)²) = 20/3, slightly smaller, as required.

3. Surface area uses circumference times slant length

Rotating a short curve segment around an axis produces a narrow band. Its area is approximately its circumference 2πr multiplied by its slant length ds. Accumulating these bands gives the curved surface area S = ∫2πr ds. The radius is the perpendicular distance from the point on the curve to the axis of rotation.

About the x-axis: S = 2π∫ab |y(t)|√[(x′)² + (y′)²] dt.
About the y-axis: S = 2π∫ab |x(t)|√[(x′)² + (y′)²] dt.

The absolute values express distances. If the chosen curve lies above the x-axis or to the right of the y-axis, they can be removed after that sign has been stated. Do not square the radius as in a volume formula. Surface area uses circumference and slant length; volume uses cross-sectional disk or washer area and perpendicular thickness.

These formulas measure the surface swept out by the curve. They do not automatically include flat end caps. A cone generated by a line segment has its curved surface counted; a circular base must be added separately if the question asks for a closed total surface. Read whether the object is open, closed, hollow or only a curved sheet.

The generating description should cover the intended surface once. Even a curve that does not retrace itself can generate overlapping surface patches under rotation. For example, rotating both upper and lower semicircles about their common horizontal diameter generates the same sphere twice. One semicircle is enough. Count distinct surface area, rather than repeated parametrised coverage, unless multiplicity is explicitly intended.

Surface area from a thin bandA small sloping curve element has slant length ds and radius r from the axis. On rotation its narrow band area is approximately circumference two pi r times ds.A surface band follows the sloperdsaxis of rotationcircumference 2πrdsBand area ≈ 2πr ds
The band’s width follows the curve. Replacing ds by dx generally underestimates the curved area.

Worked example 2: The same segment, two axes

The segment x = 3t, y = 4t, 0 ≤ t ≤ 2, is rotated first about the x-axis and then, separately, about the y-axis. Its speed is √(3² + 4²) = 5. Both coordinates are non-negative. About the x-axis, the radius is y = 4t, so Sₓ = 2π∫02 (4t)(5) dt = 80π square units.

About the y-axis, the radius is x = 3t, so Sᵧ = 2π∫02 (3t)(5) dt = 60π square units. The segment has slant length 10. The generated cones have base radii 8 and 6, so their familiar curved areas πrl give the same answers. Neither calculation includes a circular base.

4. Match the radius to the integration variable

For a Cartesian graph y = f(x) rotated about the y-axis, the radius is x but the arc-length element can still be written √(1 + [f′(x)]²) dx. There is no requirement to integrate with respect to y merely because the rotation axis is vertical. That requirement belonged to perpendicular washers for volume, not to surface bands.

Likewise, a parametric surface-area integral can stay entirely in t. Insert x(t), y(t), their derivatives and the original parameter bounds. Switching to a new variable is optional and should simplify the calculation. If you switch, transform all bounds and ensure the selected branch describes the same generating arc exactly once.

A radius crossing zero is not itself a reason for negative area. Use its absolute value, splitting at sign changes if necessary. More importantly, check whether the rotation of different curve sections overlaps. A larger integral caused by duplicate surface coverage is not corrected by merely taking an absolute value at the end.

Worked example 3: Cartesian surface area about the y-axis

Rotate the arc y = x²/2, 0 ≤ x ≤ 2, about the y-axis. The radius is x ≥ 0 and dy/dx = x. Thus S = 2π∫02 x√(1 + x²) dx. This x-variable setup is simple and covers the generated curved surface once.

Let u = 1 + x², so du = 2x dx and the bounds become 1 and 5. Then S = π∫15 √u du = (2π/3)(5√5 − 1), approximately 21.322 square units. If a flat circular lid at x-radius 2 were explicitly required, its area would be 4π in addition. It is not part of the curved arc’s generated area.

5. Numerical integration is often the natural finish

Many valid arc-length and surface-area integrals do not simplify to elementary antiderivatives. An ellipse is a familiar example: parametrising x = a cos t, y = b sin t produces speed √(a²sin²t + b²cos²t). Unless the geometry simplifies, state the exact definite integral and evaluate it numerically with suitable technology.

Use radians, retain full parameter precision and show the integrand before giving a decimal. A numerical answer should include a rounding instruction and units. Check against an endpoint chord for an open arc, or suitable enclosing geometry for a closed curve. If the result is shorter than its endpoint chord, the setup or evaluation is wrong.

A useful computational check samples the parameter in order and adds the straight-line distances between consecutive points. Increasing the number of segments should approach the arc length for a well-behaved curve. This polygonal approximation tests the geometry in a different way from symbolic integration. It also makes retracing visible: the sum follows the full traversal rather than automatically recognising duplicate pieces.

For surface area, inspect the parameter interval and rotation geometry before trusting numerical output. An integrator can accurately evaluate an incorrectly doubled integral. The strongest solution combines a clear geometric description, the correct non-negative integrand, an exact setup and a numerical or symbolic check.

6. Parameter speed and geometric invariance

A curve’s length should not depend on how quickly you describe it. If a new parameter s gives t = h(s), the chain rule multiplies both coordinate derivatives by h′(s). The new speed is therefore the old speed multiplied by |h′(s)|. That factor compensates for the change in parameter width when the integral is transformed. A one-to-one reparametrisation preserves length; a repeated traversal changes what is being counted.

For an increasing change of parameter, the absolute value is simply h′. For a decreasing change, its sign is negative and the order of endpoint values reverses. Tracking the non-negative speed and the bounds together preserves a positive length. This is the same care with orientation used in substitution, now given a geometric interpretation.

Units also clarify the formula. If x and y are in centimetres and t is in seconds, both derivatives are in centimetres per second, speed has those same units, and multiplying by dt gives centimetres. Multiplying ds by a radius in centimetres gives square centimetres for surface area. If t is dimensionless instead, the coordinate derivatives have length units directly; the final geometric units are unchanged.

Report numerical precision honestly. The exact integral describes the geometric quantity independently of a calculator, while a rounded decimal reports only an approximation to it. Retain several guard digits during evaluation and round once at the end. For a measured physical object, mathematical precision does not remove uncertainty in the measured dimensions or the suitability of the chosen curve model. The surface-area formula is exact for the stated mathematical surface, subject to the coverage conditions already checked.

Practise, then check

Distinguish geometric length, distance travelled and generated curved area. Do not add end caps unless asked.

  1. Find the length of x = 3t − 1, y = 4t + 2, 0 ≤ t ≤ 3.
  2. Find the arc length of x = 2cos t, y = 2sin t, 0 ≤ t ≤ π.
  3. Find the length of x = t, y = (2/3)t3/2, 0 ≤ t ≤ 3.
  4. Rotate y = x, 1 ≤ x ≤ 3, about the x-axis. Find the curved surface area.
  5. Rotate x = 2 + cos t, y = sin t, 0 ≤ t ≤ π, about the y-axis. Find the generated curved area.
  6. A point follows x = cos t, y = sin t for 0 ≤ t ≤ 4π. Find its distance travelled and the length of the distinct geometric curve.
  7. Give an exact integral for the first-quadrant ellipse arc x = 3cos t, y = 2sin t, 0 ≤ t ≤ π/2, and evaluate it to three decimal places.
Reasoned answers
  1. The speed is √(3² + 4²) = 5, so L = ∫03 5 dt = 15 length units. This is a straight segment with no reversal, so the endpoint distance agrees.
  2. The speed is √(4sin²t + 4cos²t) = 2. Thus L = 2π length units, matching a semicircle of radius 2. The parameter covers it once.
  3. Derivatives are 1 and √t, so L = ∫03 √(1 + t) dt = (2/3)[(1 + t)3/2]03 = 14/3 length units.
  4. Radius x is positive and ds = √2 dx. Hence S = 2π√2∫13 x dx = 8π√2 square units. This is the curved area of a conical frustum, excluding its two circular ends.
  5. The speed is 1 and the radius 2 + cos t is positive. S = 2π∫0π (2 + cos t) dt = 4π² square units. This generating upper semicircle sweeps the upper half of a torus once.
  6. The speed is 1, so distance travelled is 4π. The point traverses the unit circle twice; the distinct geometric curve has circumference 2π. The interval determines multiplicity.
  7. L = ∫0π/2 √(9sin²t + 4cos²t) dt ≈ 3.966 length units. Its endpoint chord is √13 ≈ 3.606, smaller than the arc. The integral is exact; the decimal requires numerical evaluation.

Curriculum and sources

VCE Specialist Mathematics Units 3 and 4, Areas of Study 4;5 as applicable, with Outcomes 1–3. Scope follows the VCAA Mathematics Study Design 2023–2027 and the current Specialist Mathematics FAQ. Curriculum checked 15 October 2026. Teaching reference: Cambridge Senior Mathematics VCE Specialist Mathematics 3&4, second edition (2023), sections 10E–10F (pp458–466). All teaching prose, examples, practice questions and diagrams are original. Independent teaching material; not a VCAA publication. The VCAA planning support provides additional teaching guidance.