P4 6.5 - Areas from parametric equations | Edexcel A-Level International Mathematics

Calculate area without eliminating a curve's parameter. Convert vertical-strip widths using the derivative of its x-coordinate, match parameter limits to the required boundaries, and keep the direction and sign of each contribution consistent.

Converting y dx to a parameter

If a curve is given by $x=t^2$ , $y=t+1$ , why is $\int(t+1)\,dt$ not its area under the curve? The height is correct, but an equal step in $t$ does not give an equal horizontal width.

For example, increasing $t$ from $1$ to $1.1$ changes $x$ from $1$ to $1.21$ . The strip width is $0.21$ , not $0.1$ . Infinitesimally, $dx/dt=2t$ , so $dx=2t\,dt$ . Multiply the height by this horizontal-width factor.

[DIAGRAM: asset_name: Parametric strips need horizontal width; asset_slug: parametric-strip-width; caption: The area strip has width delta x, which differs from the change in t.; description: The supplied curve x equals t squared and y equals t plus one runs from zero comma one to four comma three. The area under the curve is pale blue. A highlighted vertical strip between x equals one and one point two one corresponds to t increasing from one to one point one; its horizontal width is labelled delta x equals zero point two one, and the parameter change is delta t equals zero point one.]
The area strip has width delta x, which differs from the change in t.

The sketch is supplied to explain the strip; you do not need to sketch a curve from its parametric equations for this unit.

Parametric integration

For $x=x(t)$ , $y=y(t)$ ,

\int y\,dx=\int y(t)\frac{dx}{dt}\,dt.

When the curve is traversed once from left to right and stays above the axis, this gives its area above the axis. Choose parameter limits corresponding to the required x-boundaries.

Not in the formula booklet: learn it.

Substitution has changed the variable used to describe the width, while retaining the same geometric strips.

Worked example 1: a changing horizontal speed

A curve has $x=t^2$ , $y=t+1$ , $0\le t\le2$ . Find the area bounded by it, the x-axis, the y-axis and $x=4$ .

Plan. Use $dx/dt=2t$ , with $t=0$ at $x=0$ and $t=2$ at $x=4$ .

For this range, $x$ increases and $y>0$ , so

\begin{aligned} A&=\int_0^2(t+1)2t\,dt\\ &=\left[\frac23t^3+t^2\right]_0^2 =\frac{16}{3}+4=\frac{28}{3}. \end{aligned}

Answer. $28/3$ square units.

Check. Eliminating the parameter here gives $y=\sqrt{x}+1$ , so $\int_0^4(\sqrt{x}+1)\,dx=16/3+4=28/3$ , independently confirming the parametric method.

Change only $x=t^2$ to $x=3t^2$ , keeping the same parameter interval and heights. The region becomes three times as wide.

Worked example 2: a horizontal stretch

For $x=3t^2$ , $y=t+1$ , $0\le t\le2$ , find the corresponding area above the x-axis between the endpoint verticals.

Plan. Multiply each height by $dx/dt=6t$ ; the endpoint verticals are now $x=0$ and $x=12$ .

A=\int_0^2(t+1)6t\,dt =\left[2t^3+3t^2\right]_0^2 =16+12=28.

Answer. $28$ square units.

Check. Each horizontal coordinate is three times its previous value while each height is unchanged. The area is therefore $3(28/3)=28$ .

A zero $dx/dt$ at the endpoint $t=0$ causes no problem here. The curve still moves monotonically to the right on the full interval.

Watch out

$\int y(t)\,dt$ uses parameter width, not horizontal width. Include $dx/dt$ ; replacing it with $dy/dt$ computes a different integral.

Change only the horizontal scale again.

Parameter limits and direction

Increasing the parameter can move a curve leftwards. Match limits to the intended left-to-right area calculation, rather than automatically writing the smaller parameter first.

For $x=4-t^2$ , $y=t+1$ , $0\le t\le2$ , increasing $t$ moves from $(4,1)$ to $(0,3)$ , because $dx/dt=-2t$ . An integral in that direction is negative, although the region is above the axis. To add positive widths from $x=0$ to $x=4$ , let $t$ run from $2$ down to $0$ .

Worked example 3: decreasing parameter limits

For $x=4-t^2$ , $y=t+1$ , $0\le t\le2$ , find the area above the x-axis between $x=0$ and $x=4$ .

Plan. Use $t=2$ at the left endpoint and $t=0$ at the right endpoint, with $dx/dt=-2t$ .

\begin{aligned} A&=\int_2^0(t+1)(-2t)\,dt\\ &=\left[-\frac23t^3-t^2\right]_2^0 =0-\left(-\frac{16}{3}-4\right)=\frac{28}{3}. \end{aligned}

Answer. $28/3$ square units.

Check. Here $y=1+\sqrt{4-x}$ . Reflecting $y=1+\sqrt{x}$ horizontally preserves the area $28/3$ from Worked example 1. The integral with limits $0$ to $2$ would be its negative.

Change only the left boundary from $x=0$ to $x=1$ . Solve for its new parameter rather than using $t=1$ .

Worked example 4: convert a vertical boundary

For the same curve, find the area above the x-axis between $x=1$ and $x=4$ .

Plan. Solve $1=4-t^2$ in the allowed parameter interval, then integrate left to right.

The left endpoint has $t=\sqrt3$ ; the right endpoint still has $t=0$ . Thus

\begin{aligned} A&=\int_{\sqrt3}^0(t+1)(-2t)\,dt\\ &=\left[-\frac23t^3-t^2\right]_{\sqrt3}^0\\ &=2\sqrt3+3. \end{aligned}

Answer. $3+2\sqrt3$ square units.

Check. The interval width is $3$ , with heights between $1$ and $1+\sqrt3$ . The result lies between $3$ and $3+3\sqrt3$ . Direct integration of $1+\sqrt{4-x}$ from $1$ to $4$ gives the same value.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Pure Mathematics 1

Pure Mathematics 2

Pure Mathematics 3

Pure Mathematics 4

Mechanics 1

Mechanics 2

Statistics 1

Statistics 2

Decision Mathematics 1

P4 6.5 - Areas from parametric equations

Converting y dx to a parameter

Parameter limits and direction

Continue this lesson.

Learn faster with Nova