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Polar coordinate to compute the volume

  1. Nov 13, 2011 #1
    1. The problem statement, all variables and given/known data
    Use polar coordinates to compute the volume of the region defined by
    4 - [itex]x^{2}[/itex] - [itex]y^{2}[/itex] ≤ z ≤ 10 - 4[itex]x^{2}[/itex] - 4[itex]y^{2}[/itex]


    2. Relevant equations



    3. The attempt at a solution
    I got z = 2 so set up the equation

    V = [itex]f^{2pi}_{0}[/itex][itex]f^{5/2}_{2}[/itex][itex]f^{0}_{2}[/itex]r*dzdrdθ

    is the domain correct?
     
  2. jcsd
  3. Nov 13, 2011 #2

    SammyS

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    Those are cylindrical coordinates, not polar. (If you were to do the problem in cylindrical coords, your limits of integration for z would be incorrect.)

    The two surfaces intersect at z=2, but that's not particularly important. At what value of r do they intersect?
     
  4. Nov 13, 2011 #3
    setting 4 - [itex]x^{2}[/itex] - [itex]y^{2}[/itex] and 10 - 4[itex]x^{2}[/itex] - 4[itex]y^{2}[/itex] to equal, r = ±√2, but since r must be greater than 0, it's r = √2
     
  5. Nov 13, 2011 #4

    SammyS

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    The volume element is the area element, r dr dθ, times the height, which you get from zupper - zlower .

    z goes from 4 - r2 to 10 - 4r2 .
     
  6. Nov 14, 2011 #5
    thank you I got it.
    But how do I know if it's cylindrical?
     
  7. Nov 14, 2011 #6

    HallsofIvy

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    How do you know if what is cylindrical? If you mean the coordinate system, "cylindrical coordinates" are just polar coordinates for the xy-plane with the z coordinate.
     
  8. Nov 14, 2011 #7
    I mean just by looking at the equation
    Is it because it contains z?
     
  9. Nov 14, 2011 #8

    SammyS

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    If you're asking how did I know you were using cylindrical coordinates rather than polar; cylindrical coordinates are in 3 dimensions and use r, θ, and z. Polar coordinates are 2 dimensional using r and θ.
     
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