Find the area of the region bounded

  1. Apr 7, 2005 #1
    Find the area of the region bounded by: [tex]r= 6-2sin(\theta)[/tex]

    here's what i did:

    [tex] 6-2sin(\theta) = 0[/tex]
    [tex] sin(\theta) = 1/3[/tex]

    so the bounds are from arcsin(-1/3) to arcsin(1/3) right?

    my integral:
    [tex]\int_{-.339}^{.339} 1/2*(6-2sin(\theta))^2[/tex]

    i get a answer of 0.6851040673*10^11, and it's wrong. all my steps seems to be correct, i cant figure out the problem.
     
  2. jcsd
  3. Apr 7, 2005 #2
    What is the answer that you should have got ???

    marlon
     
  4. Apr 7, 2005 #3
    [tex]r= 2sin(\theta)[/tex] is an ellipse so [tex]r= 6-2sin(\theta)[/tex] is just shifting and stretching it.

    Therefore the bounds on [tex] \theta [/tex] are [tex] 0 \leq \theta \leq 2 \pi [/tex]
     
  5. Apr 7, 2005 #4

    dextercioby

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    I agree.It's a shifted & stretched ellipse.Pay attention with the numbers...You can't get a big value for the area.It's ~100...

    Daniel.

    [itex] 38\pi [/itex] to be exact.
     
  6. Apr 7, 2005 #5
    can you explain it to me agian? i dont really understand it that well. are you saying if it's an ellipse, the bounds will always be from 0 ->2pi?
     
  7. Apr 7, 2005 #6

    dextercioby

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    Yes,it's like for a circle,or for any closed curve enclosing the origin inside it...

    Daniel.
     
  8. Apr 7, 2005 #7
    ah, i get it now. thank you
     
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