How do I properly bound the area between polar curves?

[itzela]

I am having trouble finding the area between 2 polar curves... I have the procedure down, but the bounds are throwing me off. Any help with understanding how to bound would be great appreciated!

I have attatched one problem that I am having hard time with and the work I have done. I know that I am doing something wrong because I am getting a negative number for an area (which shouldn't be).

 

[qbert]

At Pi/2 your lower limit changes to 0. (Otherwise
you're integrating from the bottom half of the circle
up to the 1 + cos(theta) curve.)

 

[LeonhardEuler]

That was tricky, but I figured out what the problem is. The two graphs don't intersect where they appear to intersect. They do intersect at \frac {\pi} {3}[/tex], but they don't really intersect at the origin. There \theta = \pi [/tex] for r=1+cos(\theta)[/tex], but \theta= \frac {\pi} {2}[/tex] for r=3cos(\theta)[/tex]. What you need to do is two separate integrals with different limits. Find the whole area for r=1+cos(\theta)[/tex] and then subtract out the area for r=3cos(\theta)[/tex].

 

[Knavish]

Find and double the area from \pi/2 to \pi for the 1+cos(\theta) graph. Then, find the area from \pi/3 to \pi/2 for [(1+cos(\theta))^2 - (3cos(\theta))^2]; double this as well. The total area is the sum of the two.

 

[itzela]

Thanks a bunch guys =)

What I did was:

1. found the area from \pi to \pi/3 for the graph of 1+cos(\theta) -- doubled it

2. found the area from \pi/3 to \pi/2 for the graph 3cos(\theta) -- doubled it

3. Substracted the value i got in #2 from #1... and that was my answer.

... Does it sound on the right path?

 

[Knavish]

Yup, that works.