Parametric Integration

Builds on: aqa-alevel-maths-pure-2 / parametric-equations — this lesson is the A2-level deep dive.

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.

Spec Mapping — AQA 7357 Sections C (Coordinate geometry) and H (Integration). This lesson is synoptic — it covers the parametric-integration content, linking parametric curve work in Section C with definite-integral techniques in Section H. Refer to the official AQA specification document for exact wording.

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 = int y dx = int 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.

\begin{aligned} dx/dt = 2t \\ When x = 0: t^{2} = 0 ⟹ t = 0 \\ When x = 4: t^{2} = 4 ⟹ t = 2 \\ Area = \int _{0}^{2} y \times (dx/dt) dt \\ = \int _{0}^{2} 2t \times 2t dt \\ = \int _{0}^{2} 4t^{2} dt \\ = [4t^{3}/3]_{0}^{2} \\ = 4(8)/3 - 0 \\ = 32/3 \end{aligned}

Verification: The Cartesian equation is $y^2 = 4x$ , so $y = 2\sqrt{x}$ (upper half).

Area = \int _{0}^{4} 2 \sqrt{x} dx = 2 \times [2x^(3/2)/3]_{0}^{4} = (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.

\begin{aligned} &\text{In the first quadrant: } x \text{ goes from } 0 \text{ to } 3. \\ &\text{When } x = 0: \cos\theta = 0 \implies \theta = \tfrac{\pi}{2} \\ &\text{When } x = 3: \cos\theta = 1 \implies \theta = 0 \\[4pt] &\frac{dx}{d\theta} = -3\sin\theta \\[6pt] &\text{Area of first quadrant} = \int y\, \frac{dx}{d\theta}\, d\theta \\[4pt] &\text{As } x \text{ goes from } 0 \text{ to } 3,\ \theta \text{ goes from } \tfrac{\pi}{2} \text{ to } 0 \text{ (decreasing).} \\[4pt] \text{Area} &= \int_{\pi/2}^{0} (2\sin\theta)(-3\sin\theta)\, d\theta \\ &= \int_{\pi/2}^{0} (-6\sin^{2}\theta)\, d\theta \\ &= 6\int_{0}^{\pi/2} \sin^{2}\theta\, d\theta \quad \text{(reversing limits changes sign)} \\[4pt] &\text{Using } \sin^{2}\theta = \frac{1 - \cos 2\theta}{2}: \\[4pt] &= 6\int_{0}^{\pi/2} \frac{1 - \cos 2\theta}{2}\, d\theta \\ &= 3\int_{0}^{\pi/2} (1 - \cos 2\theta)\, d\theta \\ &= 3\left[ \theta - \frac{\sin 2\theta}{2} \right]_{0}^{\pi/2} \\ &= 3\left[ \left(\tfrac{\pi}{2} - 0\right) - (0 - 0) \right] \\ &= \frac{3\pi}{2} \\[6pt] \text{Total area} &= 4 \times \frac{3\pi}{2} = 6\pi \end{aligned}

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.

\begin{aligned} &\frac{dx}{dt} = \cos t \\[4pt] &\text{When does } y = \cos t = 0 \text{ (crossing the } x\text{-axis)?} \\ &\cos t = 0 \implies t = \tfrac{\pi}{2} \quad (\text{within } 0 \leq t \leq \pi) \\[4pt] &\text{For } 0 \leq t < \tfrac{\pi}{2}: y = \cos t > 0 \text{ (above } x\text{-axis)} \\ &\text{For } \tfrac{\pi}{2} < t \leq \pi: y = \cos t < 0 \text{ (below } x\text{-axis)} \\[6pt] &\text{Area above } x\text{-axis } (t \text{ from } 0 \text{ to } \tfrac{\pi}{2}): \\ &\int_{0}^{\pi/2} \cos t \times \cos t\, dt = \int_{0}^{\pi/2} \cos^{2} t\, dt \\[4pt] &\text{Using } \cos^{2} t = \frac{1 + \cos 2t}{2}: \\ &= \int_{0}^{\pi/2} \frac{1 + \cos 2t}{2}\, dt \\ &= \left[ \frac{t}{2} + \frac{\sin 2t}{4} \right]_{0}^{\pi/2} \\ &= \left(\tfrac{\pi}{4} + 0\right) - 0 \\ &= \frac{\pi}{4} \end{aligned}

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:

\text{Area} = \left| \int_{\alpha}^{\beta} y\, \frac{dx}{dt}\, dt \right|

or equivalently

\text{Area} = \frac{1}{2} \left| \int_{\alpha}^{\beta} \left( x\, \frac{dy}{dt} - y\, \frac{dx}{dt} \right) dt \right|

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.

\begin{aligned} &\frac{dx}{dt} = -\sin t \\[4pt] \text{Area} &= \left| \int_{0}^{2\pi} \sin t \times (-\sin t)\, dt \right| \\ &= \left| -\int_{0}^{2\pi} \sin^{2} t\, dt \right| \\ &= \int_{0}^{2\pi} \sin^{2} t\, dt \\[4pt] &\text{Using } \sin^{2} t = \frac{1 - \cos 2t}{2}: \\ &= \int_{0}^{2\pi} \frac{1 - \cos 2t}{2}\, dt \\ &= \left[ \frac{t}{2} - \frac{\sin 2t}{4} \right]_{0}^{2\pi} \\ &= (\pi - 0) - (0 - 0) \\ &= \pi \end{aligned}

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 = \pi \int y^{2} (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.

\begin{aligned} dx/dt = 2t \\ V = \pi \int _{0}^{2} t^{2} \times 2t dt \\ = 2 \pi \int _{0}^{2} t^{3} dt \\ = 2 \pi [t^{4}/4]_{0}^{2} \\ = 2 \pi \times 4 \\ = 8 \pi \end{aligned}

Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

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)

\begin{aligned} \text{From } y = 6t &: t = y/6 \\ x &= 3(y/6)^{2} = 3y^{2}/36 = y^{2}/12 \\ &\therefore y^{2} = 12x \;\checkmark \end{aligned}

(b)

\begin{aligned} &\text{When } x = 12: 3t^{2} = 12 \implies t^{2} = 4 \implies t = 2 \ (\text{since } t \geq 0) \\ &\text{When } x = 0: t = 0 \\[4pt] &\frac{dx}{dt} = 6t \\[4pt] \text{Area} &= \int_{0}^{2} y \times \frac{dx}{dt}\, dt \\ &= \int_{0}^{2} 6t \times 6t\, dt \\ &= \int_{0}^{2} 36t^{2}\, dt \\ &= \left[ 12t^{3} \right]_{0}^{2} \\ &= 12 \times 8 = 96 \\[6pt] &\textbf{Verification: } \\ \int_{0}^{12} \sqrt{12x}\, dx &= \sqrt{12}\left[ \frac{2x^{3/2}}{3} \right]_{0}^{12} = \frac{4\sqrt{3}}{3}(12)^{3/2} \\ &= \frac{4\sqrt{3}}{3}(24\sqrt{3}) = \frac{4 \times 3 \times 24}{3} = 96 \quad \checkmark \end{aligned}

Synoptic Links

This topic connects to:

Parametric equations (intro) (aqa-alevel-maths-coordinate-geometry-depth / parametric-equations-intro) — the parametric curves whose areas we are now integrating.
Integration techniques (aqa-alevel-maths-pure-2 / integration-techniques) — substitution and standard integrals applied to the t-integral.
Applications of integration (aqa-alevel-maths-calculus-applications / applications-of-integration) — volumes of revolution and area-between-curves problems that often arrive in parametric form.

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 $V = \pi\int x^2\,dy = \pi\int [x(t)]^2\dfrac{dy}{dt}\,dt$ .

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)

Specimen Question (Modelled on the AQA Paper Format)

Worked Examination-Style Problem

Synoptic Links

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

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