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Area bounded by polar curves

  • Thread starter itzela
  • Start date
34
0
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).
 

Attachments

185
3
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.)
 
Last edited:

LeonhardEuler

Gold Member
858
1
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 [itex]\frac {\pi} {3}[/tex], but they don't really intersect at the origin. There [itex]\theta = \pi [/tex] for [itex] r=1+cos(\theta)[/tex], but [itex]\theta= \frac {\pi} {2}[/tex] for [itex] r=3cos(\theta)[/tex]. What you need to do is two seperate integrals with different limits. Find the whole area for [itex] r=1+cos(\theta)[/tex] and then subtract out the area for [itex] r=3cos(\theta)[/tex].
 
113
1
Find and double the area from [itex]\pi/2[/itex] to [itex]\pi[/itex] for the [itex]1+cos(\theta)[/itex] graph. Then, find the area from [itex]\pi/3[/itex] to [itex]\pi/2[/itex] for [itex][(1+cos(\theta))^2 - (3cos(\theta))^2][/itex]; double this as well. The total area is the sum of the two.
 
Last edited:
34
0
Thanks a bunch guys =)

What I did was:

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

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

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

... Does it sound on the right path?
 
113
1
Yup, that works.
 

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