Area under one arc or loop of a parametric curve

 
 
Area under one arc or loop of a parametric curve blog post.jpeg
 
 
 

Finding area under one loop of a parametric curve

Sometimes we need to find the area under just one arc or loop of a parametric curve. In order to do it, we’ll use the area formula

A=aby(t)x(t) dtA=\int^b_a{y(t)}x'(t)\ dt

where [a,b][a,b] is the interval that contains the loop, and x(t)x'(t) is the derivative of x(t)x(t).

Krista King Math.jpg

Hi! I'm krista.

I create online courses to help you rock your math class. Read more.

 

So when we’re given equations for xx and yy, we simply plug yy and the derivative of xx into the formula for area.

In order to find the bounds [a,b][a,b], we want to look at the values of xx and yy as the parameter traces out its arc. When the parameter is equal to 00, that will correspond to a particular coordinate point (x,y)(x,y).

Once the arc arrives back at the same starting point (x,y)(x,y), you’ve closed one loop of the curve, and therefore you’ll use the limits of integration that correspond to those parameter values.

 
 

To isolate the area of just one loop of the parametric curve, we’ll use bounds on the integral that define the limits of the loop


 
Krista King Math Signup.png
 
Calculus 2 course.png

Take the course

Want to learn more about Calculus 2? I have a step-by-step course for that. :)

 
 

 
 

Building a table of values to find the bounds that enclose a single loop of the parametric curve

Example

Find the area under the parametric curve over the interval 0θ2π0\le\theta\le2\pi.

x=3+sinθx=3+\sin{\theta}

y=3+cosθy=3+\cos{\theta}

The first thing we need to do is find the limits of integration. We’ll set up a chart for θ\theta starting at 00, and then fill in the corresponding values of xx and yy.

a chart showing the values of x and y that correspond to theta

Because we started at the point (3,4)(3,4) and didn’t return to the same point until the parameter value reached 2π2\pi, the bounds for the integral will be [0,2π][0,2\pi].

Before we can plug everything into our area formula, we’ll need to find the derivative of x(θ)x(\theta).

x(θ)=cosθx'(\theta)=\cos{\theta}

Area under one arc or loop of a parametric curve for Calculus 2.jpg

Once the arc arrives back at the same starting point (x,y), you’ve closed one loop of the curve, and therefore you’ll use the limits of integration that correspond to those parameter values.

Plugging everything into the area formula, we get

A=02π(3+cosθ)(cosθ) dθA=\int^{2\pi}_0\left(3+\cos{\theta}\right)\left(\cos{\theta}\right)\ d\theta

A=02π3cosθ+cos2θ dθA=\int^{2\pi}_03\cos{\theta}+\cos^2{\theta}\ d\theta

Before we can integrate, we need to do a substitution for cos2θ\cos^2{\theta} using the formula

cos2θ=12(1+cos2θ)\cos^2{\theta}=\frac12(1+\cos{2\theta})

We’ll make the substitution.

A=02π3cosθ+12(1+cos2θ) dθA=\int^{2\pi}_03\cos{\theta}+\frac12(1+\cos{2\theta})\ d\theta

A=02π3cosθ+12+12cos2θ dθA=\int^{2\pi}_03\cos{\theta}+\frac12+\frac12\cos{2\theta}\ d\theta

A=3sinθ+12θ+14sin2θ02πA=3\sin{\theta}+\frac12\theta+\frac14\sin{2\theta}\Big|^{2\pi}_0

A=3sin(2π)+12(2π)+14sin(2(2π))[3sin(0)+12(0)+14sin(2(0))]A=3\sin{(2\pi)}+\frac12(2\pi)+\frac14\sin{(2(2\pi))}-\left[3\sin{(0)}+\frac12(0)+\frac14\sin{(2(0))}\right]

A=3(0)+π+14(0)[3(0)+12(0)+14(0)]A=3(0)+\pi+\frac14(0)-\left[3(0)+\frac12(0)+\frac14(0)\right]

A=πA=\pi

 
Krista King.png
 

Get access to the complete Calculus 2 course