Areas under polar curves—why these boundaries?

KrabLord

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So, the area of
[MATH] r = \cos(\theta) [/MATH]is found by taking
[MATH] \int_{0}^{\pi} \frac{1}{2} (\cos(\theta))^2 d\theta = \frac{\pi}{4} [/MATH]or by using the pi r^2 formula with r = 1/2.
So, and this is where my question begins, why is it that the integration boundaries are 0 and pi for cos(theta), yet for 1+cos(theta) we use 0 to 2pi?
 
Have you tried making the graphs yourself, to get a feel for how they work? If you vary theta from 0 to 2pi, the black curve is traced twice, while the red curve is traced only once.

I myself would probably use -pi/2 to pi/2 for the latter, just because it feels more natural.
 
Have you tried making the graphs yourself, to get a feel for how they work? If you vary theta from 0 to 2pi, the black curve is traced twice, while the red curve is traced only once.

I myself would probably use -pi/2 to pi/2 for the latter, just because it feels more natural.
I have not tried making them myself, as I didn’t see how that would be more beneficial than Desmos—but that was clearly naive! Thank you!
 
I think that the title should be area between two polar curves.

I think that you should answer the following in order to answer your question.

What should \(\displaystyle \theta\) go from to graph r=cos(\(\displaystyle \theta\))? Same question for r= 1+cos(\(\displaystyle \theta\))
 
I have not tried making them myself, as I didn’t see how that would be more beneficial than Desmos—but that was clearly naive! Thank you!
I should probably add that it isn't necessary to actually graph it entirely on your own; what's most valuable is to look at the graph and act out the process of drawing it, making sure you see why it is what it is.

But it's an interesting example where technology can be harmful if we rely on it to do the thinking for us. We need at least to think along with it. (Another example is when a program can't show holes in a graph.)
 
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