The Area Under A Parametric Curve

Suppose $x = f(\theta)$ and $y = g(\theta)$ are the parametric equations of a curve. Then the area bounded by the curve, the -axis and the ordinates $\theta = a$ and $\theta = b$ will be

Contents

EXAMPLE 1
SOLUTION

STEP 1
STEP 2

EXAMPLE 2
SOLUTION

STEP 1
STEP 2

Now I’ll solve some examples of that.

EXAMPLE 1

According to Stroud and Booth (2013)*, “Determine the area of one arch of the cycloid $x = \theta - \sin \theta, y = 1 - \cos \theta$ , i.e. find the area of the plane figure bounded by the curve and the -axis between $\theta = 0$ and $\theta = 2 \pi$ .”

SOLUTION

Now here the given parametric equations of the cycloid are

\[x = \theta - \sin \theta, y = 1 - \cos \theta.\]

And I have to find out the area of the plane figure bounded by the curve and the -axis between $\theta = 0$ and $\theta = 2 \pi$ . So I’ll start with the formula

STEP 1

First of all, I’ll find out the value of in terms of $d\theta$ . As I know the value of is

Next, I’ll differentiate with respect to $\theta$ to get

\[\frac{dx}{d\theta} = 1 - \cos \theta.\]

So that gives

Therefore the area bounded by the curve and the -axis between $\theta = 0$ and $\theta = 2 \pi$ is

\[\text{Area} = \int_0^{2\pi} (1 - \cos \theta) (1 - \cos \theta) d\theta.\]

Then I’ll simplify it to get

\begin{eqnarray*}\text{Area} &=& \int_0^{2\pi} (1 - \cos \theta)^2 d\theta\\ &=& \int_0^{2\pi} (1 - 2\cos \theta + \cos^2 \theta) d\theta.\end{eqnarray*}

As per the standard formulas in trigonometry,

\[\cos 2 \theta = 2 \cos^2 \theta - 1.\]

And that means,

\[\cos^2 \theta = \frac{1}{2}\left(1 + \cos 2 \theta \right).\]

So the area will be

\[\text{Area} = \int_0^{2\pi} \left[ 1 + \frac{1}{2}\left(1 + \cos 2 \theta \right) - 2 \cos \theta \right]d\theta.\]

Next, I’ll integrate it.

STEP 2

And for that, I’ll use the standard formulas in integration. Thus it will be

\[\text{Area} = \left[ \theta + \frac{1}{2}\left(\theta + \frac{\sin 2 \theta}{2}\right) - 2 \sin \theta \right]_0^{2\pi}.\]

Now I’ll substitute the limits to get

\begin{eqnarray*}\text{Area} &=& \left[ 2\pi + \frac{1}{2}\left(2\pi + \frac{\sin 4\pi}{2}\right) - 2 \sin 2\pi \right]\\&& - \left[ 0 + \frac{1}{2}\left(0 + \frac{\sin 0}{2}\right) - 2 \sin 0 \right].\end{eqnarray*}

As I know, $\sin 0 = \sin 2 \pi = \sin 4 \pi = 0$ . So I’ll put these values in the equation of the area and simplify it to get

\begin{eqnarray*}\text{Area} &=& \left[ 2\pi + \frac{1}{2}\left(2\pi +0\right) - 2 (0) \right]\\&& - \left[ \frac{1}{2}\left( \frac{0}{2}\right) - 0 \right]\\ &=& [2 \pi + \pi - 0]-[0]\\ &=& 3 \pi .\end{eqnarray*}

Therefore the area bounded by the curve and the -axis between $\theta = 0$ and $\theta = 2 \pi$ is $3 \pi$ .

Hence I can conclude that this is the answer to the given example.

Now I’ll give another example.

EXAMPLE 2

According to Stroud and Booth (2013)*, “The parametric equations of a curve are $y = 2 \sin \cfrac{\pi}{10}t, x = 2 + 2t - 2 \cos \cfrac{\pi}{10} t.$ Find the area under the curve between t = 0 and t = 10 .”

SOLUTION

Now here the parametric equations of a curve are $y = 2 \sin \cfrac{\pi}{10}t, x = 2 + 2t - 2 \cos \cfrac{\pi}{10} t.$ And I have to find the area under the curve between t = 0 and t = 10 . So I’ll start with the formula