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Area bounded by polar curves

  1. Jul 15, 2005 #1
    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).
     

    Attached Files:

  2. jcsd
  3. Jul 15, 2005 #2
    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: Jul 15, 2005
  4. Jul 15, 2005 #3

    LeonhardEuler

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    Gold Member

    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].
     
  5. Jul 15, 2005 #4
    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: Jul 15, 2005
  6. Jul 15, 2005 #5
    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?
     
  7. Jul 15, 2005 #6
    Yup, that works.
     
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