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Iterated integral in polar coordinates

  1. Nov 10, 2014 #1
    1. The problem statement, all variables and given/known data
    Use polar coordinates to find the volume of the solid inside the hemisphere z=sqrt(16-x^2-y^2) and inside the cylinder x^2+y^2-4x=0

    2. Relevant equations
    z=sqrt(16-x2-y2)
    x2+y2-4x=0
    x=rcos(Θ)
    y=rsin(Θ)

    z=√(16-r2)
    3. The attempt at a solution

    ∫∫ r√(16-r2) dr dΘ

    The problem is the bounds; because the circle isn't centered it's throwing me off. Would dr be from 2 to 4? That's the start and end of the radius as it's a circle centered at (2,0) with a radius of 2. Of course I'm assuming that dΘ is from 0 to 2pi. I tried integrating with dr from 0 to 2 and from 2 to 4, but both times the answer was different than Wolfram's.
     
  2. jcsd
  3. Nov 11, 2014 #2

    haruspex

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    Do you understand what the object looks like? Can you post a decent picture, or describe it well in words?
    The first trick is to get the order of integration right, and you have done that (z first here). But you need r to be a radius of the circle. How can you arrange that?
     
  4. Nov 11, 2014 #3
    It's an offset vertical cylinder under a hemisphere. I don't follow what else you're saying.
     
  5. Nov 11, 2014 #4

    LCKurtz

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    So what do you get for the polar coordinate equation of that cylinder? You need that.
     
  6. Nov 11, 2014 #5

    haruspex

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    Yes, it's offset, but can you be more precise about how the cylinder fits within the hemisphere? Draw a diagram for z=0.
    In principle, you have a triple integral, ∫∫∫dxdydz. In what you posted, you have already done the dz integral, reducing it to ∫∫zdxdy. That was good. So now you can think of z as just a function of x and y, and forget the reality of the hemisphere.
    In your conversion to polar, you have set r = √(x2+y2). That represents a radius of the hemisphere, which is no longer interesting. What will work better?
     
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