Finding the area between two polar curves

[burt]

What is the area of the region that lies inside the cardioid \(r = 1 + \cos( θ)\) and outside the circle \(r = \cos (θ)\)?

In attempting to solve this problem, I reasoned that the area inside the cardioid but outside the circle is the area of the cardioid minus the area of the circle. This gave me the setup: \[\frac12\left(\int^{2\pi}_{0}\left(1+\cos(\theta)\right)^2-\cos^2(\theta)\ d\theta\right)\\=\frac12\left(\int^{2\pi}_{0}1+2\cos(\theta)+\cos^2(\theta)-\cos^2(\theta)\ d\theta\right)\\=\frac12\left(\int^{2\pi}_{0}1+2\cos(2\theta)\ d\theta\right)\\=\frac12\left(\theta+\sin(2\theta)\right)|^{2\pi}_0\\=\pi\]
Why doesn't this method work? Is there something wrong with my calculation or is it my logic that is not holding true?


**EDIT:**
As I've been working more, I see that this kind of method does not seem to work for this problem either:
>What is the area of the region that lies outside the circle \(r = \cos θ\) and inside the circle \(r = 2 \cos θ\)?

I do not seem to be able to simply subtract the area of the second circle from the area of the first. What is wrong with this method? Looking at the graphs it seems like this would work?

 

[MarkFL]

The issue is that the circle \(r=\cos(\theta)\) is traversed twice for one traversal of the cardioid.

 

[Cubist]

\[=\frac12\left(\int^{2\pi}_{0}1+2\cos(\theta)+\cos^2(\theta)-\cos^2(\theta)\ d\theta\right)\\=\frac12\left(\int^{2\pi}_{0}1+2\cos(2\theta)\ d\theta\right)\]

@MarkFL has pointed out the primary reason why your method could not work

However to help with your trig it might also be worth mentioning that the above two lines in your post aren't equivalent. There was no need to go from 2cos(θ) to 2cos(2θ). But maybe this was a simple typing mistake.

 

[burt]

However to help with your trig it might also be worth mentioning that the above two lines in your post aren't equivalent. There was no need to go from 2cos(θ) to 2cos(2θ). But maybe this was a simple typing mistake.

Yup, it was - but thanks for pointing that out!

 

[burt]

The issue is that the circle r=cos(θ)r=cos⁡(θ)r=\cos(\theta) is traversed twice for one traversal of the cardioid.

So this means that you cannot put the two integrals together even if they have the same limits of integration?

 

[Cubist]

So this means that you cannot put the two integrals together even if they have the same limits of integration?

The limits are different. The circle \(r=\cos(\theta)\) would be traversed once between 0 and π (not between 0 and 2π).

But you could make the limits match by simply adding a scaling factor, say k, like this \(r=\cos(k \cdot \theta)\). The value of k should be obvious. I don't know if this would make the integration easier than just working out the two separate integrals.

 

[MarkFL]

So this means that you cannot put the two integrals together even if they have the same limits of integration?

They wouldn't have the same limits. I would likely just find the area of the cardioid, and then subtract the area of a circle of radius 1/2.

 

[Cubist]

But you could make the limits match by simply adding a scaling factor, say k, like this \(r=\cos(k \cdot \theta)\). The value of k should be obvious. I don't know if this would make the integration easier than just working out the two separate integrals.

My suggestion above is incorrect. Adding in the scale factor and changing the limits gives a different answer for the circle's area. This is because the shape changes...



I'd stick to doing the integrals separately!

 

[Cubist]

It seems that changing the limits to go between 0 and π for r=cos(θ) still gives a wrong answer. I get an area of π/2 if I work this out.

But the circle, of radius 1/2, obviously has area π/4.

Does anyone know the reason for this?

 

[MarkFL]

My suggestion above is incorrect. Adding in the scale factor and changing the limits gives a different answer for the circle's area. This is because the shape changes...



I'd stick to doing the integrals separately!

Yes, I initially had the same thought you did, well if theta is moving twice as fast as we want, then we'll just cut it in half. Checked the Desmos plot and found it was no longer a circle. What we can do to write one integral with the same limits is:

[MATH]A=\frac{1}{4}\left(\int_0^{2\pi} 2(1+\cos(\theta))^2-\cos^2\left(\frac{\theta}{2}\right)\,d\theta\right)[/MATH]

 

[burt]

You basically halved the second one and kept the first the same. Can the same be done like this?
[MATH]A=\frac12\left(\int^{2\pi}_{0}(1+\cos(\theta))^2-\frac12(\cos^2\left(\frac{\theta}{2}\right))\ d\theta\right)[/MATH]

 

[Cubist]

It seems that changing the limits to go between 0 and π for r=cos(θ) still gives a wrong answer. I get an area of π/2 if I work this out.

But the circle, of radius 1/2, obviously has area π/4.

Does anyone know the reason for this?

I made a mistake in my integration - I now get area π/4 with the changed limits

 

[MarkFL]

You basically halved the second one and kept the first the same. Can the same be done like this?
[MATH]A=\frac12\left(\int^{2\pi}_{0}(1+\cos(\theta))^2-\frac12(\cos^2\left(\frac{\theta}{2}\right))\ d\theta\right)[/MATH]

Yes, what I did was I realized that when essentially letting:

[MATH]u=\frac{\theta}{2}\implies du=\frac{1}{2}\,d\theta[/MATH]
And so the area of the circle becomes:

[MATH]A_C=\frac{1}{2}\left(\frac{1}{2}\int_0^{2\pi} \cos^2(u)\,du\right)[/MATH]
Now that I actually write it out, I see that the correct result I got with the above formula was simply coincidental. Since the variable in a definite integral is a dummy variable, we could just as easily write:

[MATH]A_C=\frac{1}{2}\left(\frac{1}{2}\int_0^{2\pi} \cos^2(\theta)\,d\theta\right)[/MATH]
So, it would appear then that the "scaling factor" applies to the integral itself, not to the variable of integration. So, what I should have written is:

[MATH]A=\frac{1}{4}\left(\int_0^{2\pi} 2(1+\cos(\theta))^2-\cos^2(\theta)\,d\theta\right)[/MATH]

 

[Cubist]

So, it would appear then that the "scaling factor" applies to the integral itself, not to the variable of integration. So, what I should have written is:

In the range 0 to 2π the circle is drawn twice - so you effectively have two circles laid exactly on top of each other. Therefore it makes sense that the area of a single circle would be half of this "overlaid double circle".

That was a tricky question!

 

[Cubist]

@burt I've been thinking that the integral in your original question was very nice because the cos² term cancelled. And the answer that you obtained was "area of cardioid" minus "double the area of the circle." Therefore it would be valid to take the result from your post #1 and just add the area of ONE circle back to it (using π r²). This way the integral is kept much simpler.