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}???
