C3, H3, H5PureStretch

Areas Under Parametric Curves

When x and y are both functions of a parameter, you can still find area by integrating y with respect to x. Replace dx with the derivative of x times dt, then use parameter limits that trace the intended region.

40 min Integration

The big picture

A narrow strip has area approximately y times its horizontal width. The horizontal width is not a small change in t: it is x-prime(t) times that change. This is why the derivative of x appears in the integral. The direction of tracing matters too. A curve followed right to left gives a negative signed integral even if it lies above the axis.

What you'll be able to do

Derive and use $\int y\,dx=\int y(t)x'(t)\,dt$
Convert x-boundaries to parameter values
Distinguish signed integrals from geometric area
Split regions at zeros or turning points and find areas between a curve and a line

Change the horizontal variable

With $x=f(t)$ and $y=g(t)$ , the substitution rule gives $dx=f'(t)\,dt$ . Therefore $\int y\,dx=\int g(t)f'(t)\,dt$ . Do not integrate $y(t)$ alone unless $x=t$ .

First identify the region on a sketch or from sign information. Find parameter values at the boundaries, and check whether x increases throughout the chosen interval. A formula cannot decide which region the question means.

\int_{x(a)}^{x(b)}y\,dx=\int_a^b y(t)\frac{dx}{dt}\,dt

This gives a signed integral along the parameter interval.

Example 1Find the area under

x=t^2

y=t+1

, between

t=1

and

t=2

dx/dt=2t

. For

1\leq t\leq2

, both

y>0

and

dx/dt>0

, so the integral is a positive area.

A=\int_1^2(t+1)(2t)\,dt=\int_1^2(2t^2+2t)\,dt

A=[\tfrac23t^3+t^2]_1^2=(\tfrac{16}{3}+4)-(\tfrac23+1)

Answer

A=\dfrac{23}{3}

square units.

Use parameter limits, not x-limits

If boundaries are given as $x=A$ and $x=B$ , solve $f(t)=A$ and $f(t)=B$ to obtain the limits in $t$ . When several values are possible, select the branch specified by the curve's domain.

Using x-values as t-limits changes the region. Write a small limits table when the two sets of values differ, and keep any restriction on the parameter visible.

Example 1For

x=t^2+1

y=t

, with

t\geq0

, find the area under the curve from

x=2

x=5

1At

x=2

t=1

; at

x=5

t=2

. The negative square roots are excluded by

t\geq0

dx/dt=2t

, so

A=\int_1^2t(2t)\,dt

A=[\tfrac23t^3]_1^2=\tfrac23(8-1)

Answer

A=\dfrac{14}{3}

square units.

Tracing direction and signs

If $dx/dt<0$ , increasing $t$ traces the curve right to left. Above the axis this makes $\int y\,dx$ negative. Reverse the limits, or take the magnitude after confirming that the interval is one region with no cancellation.

If $y$ changes sign, or $x'(t)$ changes sign, split the calculation at the relevant parameter values. Taking the absolute value of one total can hide cancellation. A parameter interval that retraces part of a curve can also count that part more than once.

Example 1The curve

x=\cos t

y=\sin t

is traced for

0\leq t\leq\pi/2

. Find the area of the first-quadrant quarter-disc.

1As

t

increases,

x

decreases from 1 to 0.

dx/dt=-\sin t

2The oriented integral is

\int_0^{\pi/2}-\sin^2t\,dt

3Use

\sin^2t=(1-\cos2t)/2

to get

[-t/2+\sin2t/4]_0^{\pi/2}=-\pi/4

4The region is above the axis and traced right to left, so its geometric area is the positive magnitude.

Answer

A=\pi/4

square units.

Tip — Do not insert a minus sign simply because a trig parameter appears. Find the sign of dx/dt and identify the tracing direction.

Between a parametric curve and a line

For a curve above a line $y=h(x)$ , integrate $[y(t)-h(x(t))]x'(t)$ over a parameter interval where $x$ increases. Find intersections first by substituting $x(t)$ and $y(t)$ into the line equation.

Check which graph is upper on each interval. At intersections the order can swap. If the curve is traced in the opposite direction, reverse the limits or handle orientation explicitly.

Example 1Find the area between

x=t^2

y=t

0\leq t\leq1

, and the line

y=x

1Intersections satisfy

t=t^2

, giving

t=0,1

2For

0<t<1

t>t^2

, so the parametric curve is upper. Also

x'(t)=2t\geq0

A=\int_0^1(t-t^2)(2t)\,dt=\int_0^1(2t^2-2t^3)\,dt

A=\tfrac23-\tfrac12=\tfrac16

Answer

A=1/6

square units. Eliminating the parameter gives

y=\sqrt x

, which confirms the sketch.

Think like an examiner

Show dx/dt and transform both limits.
Justify the sign using the region and tracing direction.
Split before taking magnitudes when cancellation can occur.

Common misconceptions

Integrating

y(t)

directly with respect to

t

Multiply by

dx/dt

to obtain

y\,dx

Mixing x-limits with a t-integral.

Solve for the corresponding parameter values.

Taking the magnitude of a total that cancels several regions.

Split at crossings and changes of tracing direction first.

Key results to remember

A=\int_a^b\big(y(t)-h(x(t))\big)x'(t)\,dt

Upper minus lower, with consistent direction and parameter limits.

Stretch yourself

For $x=2t$ , $y=t^2-1$ , find the total area between the curve and the x-axis from $t=0$ to $t=2$ .

Hint — Split where $t^2-1=0$ , and use $dx=2\,dt$ .

Key takeaways

Replace $dx$ with $(dx/dt)\,dt$ .
Use parameter limits corresponding to the specified boundaries.
Check x-direction, y-sign and repeated tracing.
Subtract the lower boundary before integrating between graphs.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

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