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Homework Help: Polar Coordinates problem area of region

  1. Jul 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the area of the region inside: r = 9 sinθ but outside: r = 1


    2. Relevant equations



    3. The attempt at a solution
    r = 9 sinθ is a circle with center at (0, 4/2) and radius 4/2 while r= 1 is a circle with center at (0, 0) and radius 1. The two curves intersect where sin(θ)= 1/9. For 0 ≤ θ ≤ π that is satified by θ=sin^-1(1/9) and θ=π-sin^-1(1/9).
    This is where i get lost setting up the differential for the area?
     
  2. jcsd
  3. Jul 21, 2010 #2
    r = 9 sinθ is not what you say. I don't know where you went astray in your thinking, but if polar coordinates are giving you trouble, then think of it this way:

    r = 9 sinθ is the same thing as r^2 = 9r sinθ, which is equivalent to x^2 + y^2 = 9y, or...
    x^2 + (y-9/2)^2 = (9/2)^2. That's a circle centered at (0,9/2) with a radius of 9/2 units.

    Can you finish the rest by yourself? I hope this helps.

    EDIT: There, I fixed it :D
     
    Last edited: Jul 21, 2010
  4. Jul 21, 2010 #3
    Also, how did you arrive at this conclusion? Using that logic, we could let n = sin^-1(1/9) - k, where k is any real number. Then, using your logic, θ would also equal k in all these cases, i.e., sin^-1(1/9) has infinitely many solutions on a quite finite interval.
     
  5. Jul 21, 2010 #4
    It is actually a pi sign but a really bad one sorry for the confusion
     
  6. Jul 22, 2010 #5

    HallsofIvy

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    Using Latex, [itex]\theta= sin^{-1}(1/9)[/itex] and [itex]\theta= \pi- sin^{-1}(1/9)[/itex].

    Of course, for any [itex]\alpha[/itex], [itex]sin(\alpha)= sin(\pi- \alpha)[/itex].

    You should have learned when you first learned integration in polar coordinates that "dxdy" becomes "[itex]r dr d\theta[/itex]".
     
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