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Find the area inside r=3+2sin(theta) and r=2

  1. May 7, 2012 #1
    1. The problem statement, all variables and given/known data

    find the area inside r=3+2sin(theta) and r=2

    2. Relevant equations

    c above

    3. The attempt at a solution
    I knew how to do inside the polar curve and outside the circle, well, i guess everyone know how to solve what i knew. but not both inside to me.
    also, I am kind of stuck at the range of integral...
    ∏=7∏/6 and 11∏/6
     
  2. jcsd
  3. May 7, 2012 #2
    Start out by finding when [itex]2 > 3+2 \sin(\theta) [/itex] and use that to write the integration limits. You should be finding something like

    [tex] A = \int_{C_1} d\theta \int_2^{3+2 \sin(\theta)} dr + \int_{C_2} d\theta \int^2_{3+2 \sin(\theta)} dr [/tex]
     
  4. May 7, 2012 #3
    Okay, so you want the area inside [itex]r=3+2sinθ[/itex] and outside [itex]r=2[/itex]. You found the intersections by plugging [itex]r=2[/itex] into [itex]r=3+2sinθ[/itex]:

    [itex]2=3+2sinθ\rightarrow sinθ=-\frac{1}{2}\rightarrowθ=\frac{7π}{6},\frac{11π}{6}[/itex]

    But on the interval [itex](\frac{7π}{6},\frac{11π}{6})[/itex], [itex]r=3+2sinθ[/itex] is less than [itex]r=2[/itex]. For instance, [itex]3+2sin(\frac{3π}{2})=1[/itex], which is less than 2.

    So you can't use [itex]\frac{7π}{6}[/itex] as your smaller angle and [itex]\frac{11π}{6}[/itex] as your larger angle. Can you think of another way to represent one of these angles so that [itex]\frac{7π}{6}[/itex] is the larger angle?
     
  5. May 7, 2012 #4

    sharks

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    Gold Member

    Personally, i found the OP's username inappropriate and offensive. Reported.
     
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