Parametric Integration

Finding areas under curves defined by parametric equations

SPEC 8.3 • INTEGRATION

Why Parametric Integration?

We've learned how to find areas under curves when we have

y = f (x)

. But what happens when a curve is defined parametrically?

Recall

A curve defined parametrically has equations:

x = f (t) and y = g (t)

where

t

is the parameter.

The Challenge

To find area, we need

\int y d x

, but we don't have

y

as a function of

x

directly. We need a new approach!

The Parametric Integration Formula

Starting from the standard area formula:

Standard Formula

Area = \int_{a}^{b} y d x

Using the chain rule, we can write:

d x = \frac{d x}{d t} \cdot d t

Substituting this into our integral gives us the key formula:

Parametric Integration

Area = \int_{t_{1}}^{t_{2}} y \cdot \frac{d x}{d t} d t

where

t_{1}

and

t_{2}

are the parameter values at the boundaries

Converting Limits

When given limits in terms of

x

, you must convert them to

t

-values:

Method

If given

x = a

as a limit, solve

f (t) = a

to find the corresponding

t

value

If given

x = b

as a limit, solve

f (t) = b

to find the corresponding

t

value

Use these

t

-values as your new limits of integration

Important

The limits MUST be in terms of

t

when integrating with respect to

t

. Using

x

-limits with a

t

-integral is a common error!

Tip

Sometimes a question will give you the

t

-limits directly (e.g., "from

t = 0

to

t = π

"). In these cases, no conversion is needed.

Worked Example01

A curve has parametric equations $x = t^{2}$ and $y = t^{3}$ . Find the area under the curve from $t = 0$ to $t = 2$ .

1

Write the formula:

Area = \int_{t_{1}}^{t_{2}} y \cdot \frac{d x}{d t} d t

2

Find

\frac{d x}{d t}

: Since

x = t^{2}

, we have

\frac{d x}{d t} = 2 t

3

Set up the integral:

Area = \int_{0}^{2} t^{3} \cdot 2 t d t = \int_{0}^{2} 2 t^{4} d t

4

Integrate:

= [\frac{2 t ^{5}}{5}]_{0}^{2} = \frac{2 ( 32 )}{5} - 0 = \frac{64}{5}

Answer: Area = $\frac{64}{5}$ or 12.8 square units

Worked Example02

A curve is defined by $x = sin t$ and $y = cos^{2} t$ for $0 \leq t \leq \frac{π}{2}$ . Find the area bounded by the curve and the x-axis.

1

Find

\frac{d x}{d t}

: Since

x = sin t

, we have

\frac{d x}{d t} = cos t

2

Set up:

Area = \int_{0}^{π /2} cos^{2} t \cdot cos t d t = \int_{0}^{π /2} cos^{3} t d t

3

Rewrite

cos^{3} t = cos t (1 - sin^{2} t) = cos t - cos t sin^{2} t

4

Integrate:

= [sin t - \frac{sin ^{3} t}{3}]_{0}^{π /2} = (1 - \frac{1}{3}) - 0 = \frac{2}{3}

Answer: Area = $\frac{2}{3}$ square units

Worked Example03

A curve has parametric equations $x = e^{t}$ and $y = t e^{t}$ . Find the area under the curve from $x = 1$ to $x = e$ .

1

Convert x-limits to t-limits:

When

x = 1

:

e^{t} = 1 \Rightarrow t = 0

When

x = e

:

e^{t} = e \Rightarrow t = 1

2

Find

\frac{d x}{d t} = e^{t}

3

Set up:

Area = \int_{0}^{1} t e^{t} \cdot e^{t} d t = \int_{0}^{1} t e^{2 t} d t

4

Use integration by parts: Let

u = t

,

d v = e^{2 t} d t

Then

d u = d t

,

v = \frac{1}{2} e^{2 t}

= [\frac{t \cdot e ^{2 t}}{2}]_{0}^{1} - \int_{0}^{1} \frac{e ^{2 t}}{2} d t = \frac{e ^{2}}{2} - [\frac{e ^{2 t}}{4}]_{0}^{1}

5

= \frac{e ^{2}}{2} - \frac{e ^{2}}{4} + \frac{1}{4} = \frac{e ^{2}}{4} + \frac{1}{4} = \frac{e ^{2} + 1}{4}

Answer: Area = $\frac{e ^{2} + 1}{4}$ square units

Method Summary

Step-by-Step Method

Identify the limits: Are they given in terms of

x

or

t

?

Convert if needed: If x-limits are given, solve

x = f (t)

to find t-limits

Find dx/dt: Differentiate the x-equation with respect to t

Set up the integral:

\int_{t_{1}}^{t_{2}} y \cdot \frac{d x}{d t} d t

Simplify and integrate: May need trig identities or other techniques

Check the sign: Take absolute value if finding area

Exam Tip

Always show clearly that you're changing from

d x

to

d t

. Write: "Since

x = ...

,

\frac{d x}{d t} = ...

, so

d x = ... d t

"

Practice

Question 1: A curve has parametric equations

x = t^{2}

and

y = t^{2} - t

. Find the area under the curve from

t = 1

to

t = 3

.

\frac{d x}{d t} = 2 t

Area = \int_{1}^{3} (t^{2} - t) \cdot 2 t d t = \int_{1}^{3} (2 t^{3} - 2 t^{2}) d t

= [\frac{t ^{4}}{2} - \frac{2 t ^{3}}{3}]_{1}^{3} = (\frac{81}{2} - 18) - (\frac{1}{2} - \frac{2}{3})

= \frac{45}{2} - \frac{1}{2} + \frac{2}{3} = 22 + \frac{2}{3} = \frac{68}{3}

Answer: $\frac{68}{3}$ square units

Practice

Question 2: A curve is given by

x = cos t

and

y = sin 2 t

for

0 \leq t \leq \frac{π}{2}

. Find the area between the curve and the x-axis.

\frac{d x}{d t} = - sin t

Area = \int_{0}^{π /2} sin 2 t \cdot (- sin t) d t = - \int_{0}^{π /2} 2 sin t cos t \cdot sin t d t

= - 2 \int_{0}^{π /2} sin^{2} t cos t d t

Let

u = sin t

,

d u = cos t d t

. When

t = 0

,

u = 0

; when

t = \frac{π}{2}

,

u = 1

= - 2 \int_{0}^{1} u^{2} d u = - 2 [\frac{u ^{3}}{3}]_{0}^{1} = - \frac{2}{3}

Answer: Area = $\frac{2}{3}$ square units (taking absolute value)

Practice

Question 3: A curve has parametric equations

x = 4 t - t^{2}

and

y = t^{2}

. The curve crosses the x-axis at the origin and at another point P. Find the area of the region enclosed between the curve and the x-axis.

First find when the curve crosses the x-axis:

y = 0 \Rightarrow t^{2} = 0 \Rightarrow t = 0

Also check x-values: At

t = 0

,

x = 0

; At

t = 4

,

x = 16 - 16 = 0

So the curve returns to the origin when

t = 4

. We integrate from

t = 0

to

t = 4

.

\frac{d x}{d t} = 4 - 2 t

Area = \int_{0}^{4} t^{2} (4 - 2 t) d t = \int_{0}^{4} (4 t^{2} - 2 t^{3}) d t

= [\frac{4 t ^{3}}{3} - \frac{t ^{4}}{2}]_{0}^{4} = \frac{256}{3} - 128 = \frac{256 - 384}{3} = - \frac{128}{3}

Answer: Area = $\frac{128}{3}$ square units (taking absolute value)

Summary

Key Formula

Area = \int_{t_{1}}^{t_{2}} y \cdot \frac{d x}{d t} d t

StepAction

1Identify or convert limits to t-values

2Find dx/dt from the x-equation

3Set up

\int y \cdot \frac{d x}{d t} d t

4Simplify and integrate

5Check sign (take |answer| for area)

Exam Skills

Always convert x-limits to t-limits before integrating

Show your substitution clearly:

d x = \frac{d x}{d t} \cdot d t

Use trigonometric identities when needed

Be prepared to use integration by parts or substitution