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Finding the area enclosed by r=3sin theta

  1. Dec 15, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the area enclosed inside r=3 sin (theta)

    2. Relevant equations

    integral?


    3. The attempt at a solution

    basically, I took [tex]\int3sin\Theta[/tex] from 0 to 2pi, then pulled the 3 out to get

    [tex]3\int sin\Theta[/tex] from 0 to 2pi and then

    [tex]3[-cos(\Theta)][/tex] evaluated from 0 to 2pi.

    that seems too easy. what am I missing?
     
    Last edited: Dec 15, 2008
  2. jcsd
  3. Dec 15, 2008 #2

    Dick

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    Who said everything has to be super hard? But if you work that out, you'll get 0 for the area. Is that right? It might help to draw a picture. And, hey, area in polar coordinates isn't the integral of r dtheta, is it? Would you look up the right formula for area?
     
    Last edited: Dec 15, 2008
  4. Dec 15, 2008 #3
    ah. that's what it was. I forgot about the formula for area. : (

    Thanks!
     
  5. Dec 15, 2008 #4

    HallsofIvy

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    Well, actually, Dick, area in polar coordinates is r dtheta! You didn't say quite what you meant to, did you?

     
  6. Dec 15, 2008 #5

    Dick

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    Are you SURE?
     
  7. Dec 15, 2008 #6

    Dick

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    The integral of r*dtheta*dr is the area. Not the integral of r*dtheta. I missed it at my first reading as well.
     
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