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Multivariable Cal, (Polar Coordinates)

  1. Aug 9, 2011 #1
    1. The problem statement, all variables and given/known data

    Evaluate by changining to polar coordinates :
    #19 in this picture http://i52.tinypic.com/2ngbt5z.jpg

    2. Relevant equations

    x = cos θ
    y = sin θ

    r^2 = x^2 + y^2 + z^2

    ∫∫∫w f(x,y) dxdy
    = ∫from θ1 to θ2 ∫from r1 to r2 f(cosθ, sinθ) (r dr dθ)


    3. The attempt at a solution

    The first thing I did is sketch y = (2x-x^2)^(1/2) in xy plane.

    y = (2x-x^2)^(1/2)
    y^2 = 2x-x^2
    0 = y^2 - 2x + x^2
    0 = y^2 + (x^2-2x+1) -1
    1 = y^2 + (x-1)^2 ====> circle with radius one, centered around (1,0).

    I also graphed y=0, x=1, and x=2. We know that we want the region between x=1 and x=2, and it is in postive y-axis due to y = 0. so we have a quarter of a circle that we need integral. I am stuck here, I do not what the bounds for θ and r are.

    I think the bounds for θ is 0 ≤ θ ≤ pi/4 but not too sure.
    I have no clue how to find the r for the problem.... I know it cannot be 1 ≤ r ≤ 2.

    I know how to change the given function into polar but I need help finding the bounds, if someone can help me out, please, thanks in advance.
     
  2. jcsd
  3. Aug 9, 2011 #2
    Im wondering if you should do a change of variables on that circle so you can get it centered on the origin, Because that radius will be tricky to deal with.
     
  4. Aug 9, 2011 #3
    I got the bounds for this, just a min ago.
    it will be 0 ≤ θ ≤ pi/4 and sec θ ≤ r ≤ 2cosθ.
    thanks anways.
     
  5. Aug 9, 2011 #4
    oh ok, ya those bounds look good thats what I got too .
     
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