Parametric Integration

This lesson covers how to find areas under curves defined by parametric equations. This extends your knowledge of integration to situations where the curve is not given as y = f(x) but in parametric form. The key idea is to replace dx with (dx/dt) dt and change the limits from x-values to t-values.

The Parametric Integration Formula

If a curve is defined parametrically by x = f(t) and y = g(t), then the area under the curve between x = a and x = b is

Area = ∫ y dx = ∫ y (dx/dt) dt

The limits of integration must be converted from x-values to t-values.

Important Steps:

Write y and dx/dt in terms of t.
Find the t-values corresponding to the x-limits.
Integrate y × (dx/dt) with respect to t.
Ensure the area is positive — take the modulus if necessary, or reverse limits.

Basic Examples

Example 1: Find the area under the curve x = t², y = 2t between x = 0 and x = 4.

dx/dt = 2t

When x = 0: t² = 0 ⟹ t = 0
When x = 4: t² = 4 ⟹ t = 2

Area = ∫₀² y × (dx/dt) dt
     = ∫₀² 2t × 2t dt
     = ∫₀² 4t² dt
     = [4t³/3]₀²
     = 4(8)/3 − 0
     = 32/3

Verification: The Cartesian equation is y² = 4x, so y = 2√x (upper half).

Area = ∫₀⁴ 2√x dx = 2 × [2x^(3/2)/3]₀⁴ = (4/3)(8) = 32/3  ✓

Example 2: Find the area enclosed by the ellipse x = 3cos θ, y = 2sin θ.

By symmetry, the total area is 4 times the area in the first quadrant.

In the first quadrant: x goes from 0 to 3.
When x = 0: cos θ = 0 ⟹ θ = π/2
When x = 3: cos θ = 1 ⟹ θ = 0

dx/dθ = −3sin θ

Area of first quadrant = ∫ y (dx/dθ) dθ

Note: as x goes from 0 to 3, θ goes from π/2 to 0 (decreasing).

Area = ∫_{π/2}^{0} (2sin θ)(−3sin θ) dθ
     = ∫_{π/2}^{0} (−6sin²θ) dθ
     = 6∫₀^{π/2} sin²θ dθ    (reversing limits changes sign)

Using sin²θ = (1 − cos 2θ)/2:

     = 6∫₀^{π/2} (1 − cos 2θ)/2 dθ
     = 3∫₀^{π/2} (1 − cos 2θ) dθ
     = 3[θ − sin 2θ/2]₀^{π/2}
     = 3[(π/2 − 0) − (0 − 0)]
     = 3π/2

Total area = 4 × 3π/2 = 6π

This matches the known formula for the area of an ellipse: πab = π × 3 × 2 = 6π. ✓

Changing Limits Carefully

When converting limits, you must ensure:

Each x-limit corresponds to a unique t-value (within the range considered).
The direction of integration is consistent — if t decreases as x increases, the t-limits will be in reverse order. You can either keep them reversed and account for the sign, or swap them and adjust.

Example 3: Find the area between the curve x = 1 + sin t, y = cos t (for 0 ≤ t ≤ π) and the x-axis.

dx/dt = cos t

When does y = cos t = 0 (crossing the x-axis)?
cos t = 0  ⟹  t = π/2  (within 0 ≤ t ≤ π)

For 0 ≤ t < π/2: y = cos t > 0 (above x-axis)
For π/2 < t ≤ π: y = cos t < 0 (below x-axis)

Area above x-axis (t from 0 to π/2):
∫₀^{π/2} cos t × cos t dt = ∫₀^{π/2} cos²t dt

Using cos²t = (1 + cos 2t)/2:
= ∫₀^{π/2} (1 + cos 2t)/2 dt
= [t/2 + sin 2t/4]₀^{π/2}
= (π/4 + 0) − 0
= π/4

Area Enclosed by a Closed Parametric Curve

For a closed curve traced as t goes from α to β, the enclosed area can be found using:

Area = |∫_α^β y (dx/dt) dt|

or equivalently

Area = ½|∫_α^β (x dy/dt − y dx/dt) dt|

The second formula (the shoelace-type formula) works for any closed curve and does not depend on the curve being above or below the x-axis.

Example 4: Find the area enclosed by the curve x = cos t, y = sin t for 0 ≤ t ≤ 2π.

dx/dt = −sin t

Area = |∫₀^{2π} sin t × (−sin t) dt|
     = |−∫₀^{2π} sin²t dt|
     = ∫₀^{2π} sin²t dt

Using sin²t = (1 − cos 2t)/2:
= ∫₀^{2π} (1 − cos 2t)/2 dt
= [t/2 − sin 2t/4]₀^{2π}
= (π − 0) − (0 − 0)
= π

This is the area of a unit circle: π × 1² = π. ✓

Volumes of Revolution (Extension)

For a curve defined parametrically, the volume of revolution about the x-axis is:

V = π∫ y² (dx/dt) dt

with limits converted to t-values.

Example 5: Find the volume generated when the curve x = t², y = t, from t = 0 to t = 2, is rotated about the x-axis.

dx/dt = 2t

V = π∫₀² t² × 2t dt
  = 2π∫₀² t³ dt
  = 2π[t⁴/4]₀²
  = 2π × 4
  = 8π

Worked Examination-Style Problem

Problem: A curve is defined by x = 3t², y = 6t for t ≥ 0.

(a) Show that the Cartesian equation is y² = 12x.

(b) Find the area enclosed between the curve, the x-axis, and the line x = 12.

Solution:

(a)

From y = 6t: t = y/6
x = 3(y/6)² = 3y²/36 = y²/12
So y² = 12x  ✓

(b)

When x = 12: 3t² = 12 ⟹ t² = 4 ⟹ t = 2 (since t ≥ 0)
When x = 0: t = 0

dx/dt = 6t

Area = ∫₀² y × (dx/dt) dt
     = ∫₀² 6t × 6t dt
     = ∫₀² 36t² dt
     = [12t³]₀²
     = 12 × 8
     = 96

But this gives the area between the curve (upper branch) and the x-axis.
Alternatively, since the parabola is symmetric about the x-axis, the area between the upper branch and the x-axis from x = 0 to x = 12 is:

Area = ∫₀¹² 2√(12x)/2 ...

Actually let us proceed correctly:
Area = ∫₀² 6t × 6t dt = 36∫₀² t² dt = 36 × [t³/3]₀² = 36 × 8/3 = 96

Verification: ∫₀¹² √(12x) dx = √12 × [2x^(3/2)/3]₀¹² = 2√3 × (2 × 12√12/3)
= 2√3 × (24√12/3) — let's compute directly:

∫₀¹² (12x)^(1/2) dx = ∫₀¹² √12 × x^(1/2) dx = 2√3 [2x^(3/2)/3]₀¹² = (4√3/3)(12)^(3/2)
= (4√3/3)(12√12) = (4√3/3)(24√3) = (4 × 3 × 24)/3 = 96  ✓

Summary

Parametric integration: ∫ y dx = ∫ y (dx/dt) dt with limits converted to t-values.
Always convert the limits of integration from x-values to the corresponding t-values.
Take care with the direction — if t decreases as x increases, account for the sign.
For closed curves, use |∫ y (dx/dt) dt| or the shoelace formula.
For volumes of revolution: V = π∫ y² (dx/dt) dt.

Exam Tip: The most common mistake in parametric integration is forgetting to change the limits. Always write down the conversion from x-limits to t-limits explicitly. If your answer comes out negative, consider whether you need to reverse the limits or take the modulus. Show the substitution dx = (dx/dt) dt clearly in your working.

A-Level Deep Dive: Parametric Integration

Spec mapping

AQA 7357 specification, Pure Mathematics Section H — Integration (Year 2) covers using parametric equations (refer to the official specification document for exact wording). This is an extension topic within the AQA Pure content; not all AQA candidates may meet this exact framing in classroom delivery, but the underlying technique — converting $\int y\,dx$ into $\int y(t)\dfrac{dx}{dt}\,dt$ — is examinable on Paper 2 (Pure). It draws on Section G (parametric differentiation), Section F (chain rule and substitution in integration) and connects forward to volumes of revolution $V = \pi\int y^2\,dx$ when the curve is given parametrically. The AQA formula booklet does not list the parametric area formula explicitly — it must be derived from $dx = \dfrac{dx}{dt}\,dt$ on the spot.

Worked example with full mark scheme

Question (8 marks):

A curve $C$ has parametric equations $x = 2t^2,\ y = 4t$ for $0 \le t \le 2$ . The region $R$ is bounded by $C$ , the $x$ -axis and the line $x = 8$ .

(a) State $\dfrac{dx}{dt}$ and identify the parameter limits corresponding to $0 \le x \le 8$ . (2)

(b) Using $\displaystyle A = \int_{t_0}^{t_1} y\,\dfrac{dx}{dt}\,dt$ , find the exact area of $R$ . (6)

Solution with mark scheme:

(a) $\dfrac{dx}{dt} = 4t$ . B1.

When $x = 0$ : $2t^2 = 0 \implies t = 0$ . When $x = 8$ : $2t^2 = 8 \implies t = 2$ (positive root, since $t \in [0, 2]$ ). B1.

(b) Step 1 — set up the parametric integral.

$A = \int_{0}^{2} y \cdot \dfrac{dx}{dt}\,dt = \int_{0}^{2} (4t)(4t)\,dt = \int_{0}^{2} 16t^2\,dt$

M1 — correct substitution $y\,dx \to y(t)(dx/dt)\,dt$ . Examiners typically demand the conversion be written down; silently changing limits or omitting the $\dfrac{dx}{dt}$ factor loses this mark.

A1 — integrand $16t^2$ correct.

Step 2 — integrate.

$\int_{0}^{2} 16t^2\,dt = \dfrac{16t^3}{3}\Big|_{0}^{2}$

M1 — correct antiderivative $\tfrac{16t^3}{3}$ .

Step 3 — evaluate.

$= \dfrac{16 \cdot 8}{3} - 0 = \dfrac{128}{3}$

A1 — final exact answer $\dfrac{128}{3}$ .

Step 4 — sanity check. Eliminating the parameter: $y = 2\sqrt{2}\sqrt{x}$ , so $\int_{0}^{8} 2\sqrt{2}\sqrt{x}\,dx = \tfrac{4\sqrt{2}}{3} \cdot 8^{3/2} = \tfrac{4\sqrt{2}}{3} \cdot 16\sqrt{2} = \tfrac{128}{3}$ .

A1 (AO2.1) — for either the sanity check or an explicit comment that $\dfrac{dx}{dt} \ge 0$ on the parameter range so the integral equals the geometric area directly.

Total: 8 marks (M2 A3 B2 plus 1 reasoning mark).

Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): A curve $C$ has parametric equations $x = \sin t,\ y = \sin 2t$ for $0 \le t \le \dfrac{\pi}{2}$ .

(a) Find $\dfrac{dx}{dt}$ . (1)

(b) Show that the area of the region between $C$ and the $x$ -axis is given by $\displaystyle \int_{0}^{\pi/2} \sin 2t \cos t\,dt$ . (2)

Mark scheme decomposition by AO:

(a) $\dfrac{dx}{dt} = \cos t$ . B1 (AO1.1b).

(b)

M1 (AO1.1a) — substituting $y = \sin 2t$ and $\dfrac{dx}{dt} = \cos t$ into $\int y\,\dfrac{dx}{dt}\,dt$ to produce $\int \sin 2t \cos t\,dt$ .
A1 (AO2.1) — limits $t = 0$ to $t = \pi/2$ identified explicitly via $x = 0 \to t = 0$ , $x = 1 \to t = \pi/2$ .

(c)

M1 (AO1.1b) — using $\sin 2t = 2\sin t \cos t$ to rewrite the integrand as $2\sin t \cos^2 t$ .
M1 (AO1.1b) — substitution $u = \cos t$ , $du = -\sin t\,dt$ , giving $\int -2u^2\,du = -\tfrac{2u^3}{3} = -\tfrac{2\cos^3 t}{3}$ .
A1 (AO1.1b) — evaluating limits: $\left[-\dfrac{2\cos^3 t}{3}\right]_{0}^{\pi/2} = 0 - \left(-\dfrac{2}{3}\right) = \dfrac{2}{3}$ .

Total: 6 marks split AO1 = 5, AO2 = 1. Parametric area questions on AQA 7357 are AO1-heavy: the conceptual move (multiply by $dx/dt$ ) is rewarded once, and the bulk of the marks come from executing the resulting standard integral cleanly.

Synoptic links

Connects to:

Section G — Parametric differentiation: the chain-rule identity $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ and the area formula $\int y\dfrac{dx}{dt}\,dt$ are dual statements of the same principle. Both express how $x$ and $y$ relate via $t$ when neither is explicitly a function of the other.
Section F — Integration by substitution: parametric integration is a substitution $x \to t$ with $dx = \dfrac{dx}{dt}\,dt$ . The mechanical rule for changing limits — replace the $x$ -limits with the corresponding $t$ -values — is identical to the substitution rule taught for $\int f(g(x))g'(x)\,dx$ .
Volumes of revolution: rotating a parametric curve about the $x$ -axis gives $V = \pi\int y^2\,dx = \pi\int [y(t)]^2 \dfrac{dx}{dt}\,dt$ . Same conversion, with $y$ squared. About the $y$ -axis it becomes $V = \pi\int x^2\,dy = \pi\int [x(t)]^2\dfrac{dy}{dt}\,dt$ .
Mechanics — work integral: in 7357 Paper 3, work done by a variable force along a path can be set up as $W = \int F\,dx$ and converted to a parameter $t$ (often time) via $W = \int F(t)\dfrac{dx}{dt}\,dt$ . The integrand is force times velocity — this is power integrated over time, the exact same calculus pattern.
Section D — Trigonometry: parametric curves are frequently trigonometric ( $x = a\cos t,\ y = b\sin t$ for an ellipse), so executing the integral often reduces to standard trigonometric integration via double-angle formulae $\sin 2t = 2\sin t\cos t$ or the identity $\cos^2 t = \tfrac{1}{2}(1 + \cos 2t)$ .

Mark-scheme literacy

Parametric integration questions on 7357 split AO marks as follows:

Parametric Integration

Parametric Integration

The Parametric Integration Formula

Important Steps:

Basic Examples

Changing Limits Carefully

Area Enclosed by a Closed Parametric Curve

Volumes of Revolution (Extension)

Worked Examination-Style Problem

Summary

A-Level Deep Dive: Parametric Integration

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 2 format

Synoptic links

Mark-scheme literacy

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