How to find area under a parametric curve

Let’s take a look at the formulas we’ll use for calculating parametric area

Given a parametric curve where our function is defined by two equations, one for $x$ and one for $y$, and both of them in terms of a parameter $t$,

$x=f(t)$

$y=g(t)$

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we’ll find the area under the curve using the integral formula

$A=\int^{\beta}_{\alpha}y(t)x'(t)\ dt$

where $A$ is the area under the curve, $y(t)$ is $y=g(t)$, and $x'(t)$ is the derivative of $x=f(t)$.

Keep in mind as you’re working these kinds of problems that this area formula won’t give us a real-number answer. Instead, it’ll give us a function that represents the area under any part of the parametric curve. In order to find a number value for the area, we’ll have to use a definite integral by defining an interval for the area.

How to calculate the area under (enclosed by) a parametric curve

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Finding the function that defines the area under the parametric curve

Example

Find the function that defines the area under the parametric curve.

$x=2\theta-\cos{\theta}$

$y=2+\sin{\theta}$

Don’t be confused by the fact that the parameter is $\theta$ instead of $t$. It’s still a parameter value, because $x$ and $y$ are both defined in terms of $\theta$.

We’ve already been given $y(\theta)$, but we need to find $x\prime(\theta)$ before we can plug into the area formula.

$x=2\theta-\cos{\theta}$

$x'(\theta)=2+\sin{\theta}$

Instead, it’ll give us a function that represents the area under any part of the parametric curve.

Plugging $y(\theta)$ and $x'(\theta)$ into the area formula, we get

$A=\int(2+\sin{\theta})(2+\sin{\theta})\ d\theta$

$A=\int4+4\sin{\theta}+\sin^2{\theta}\ d\theta$

Using the formula

$\sin^2{\theta}=\frac12(1-\cos{2\theta})$

we’ll make a substitution for $\sin^2{\theta}$.

$A=\int4+4\sin{\theta}+\frac12(1-\cos{2\theta})\ d\theta$

$A=\int4+4\sin{\theta}+\frac12-\frac12\cos{2\theta}\ d\theta$

$A=\int\frac92+4\sin{\theta}-\frac12\cos{2\theta}\ d\theta$

$A=\int\frac92\ d\theta+\int4\sin{\theta}\ d\theta-\int\frac12\cos{2\theta}\ d\theta$

$A=\frac92\theta-4\cos{\theta}-\frac14\sin{2\theta}$

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