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Double Integral

  1. Oct 28, 2009 #1
    I need to find the area of the hyperbolic paraboloid z=xy contained within the cylinder x^2+y^2=1. I know I need to take a double integral but am having real difficulty finding the correct limits, so far I've got that;

    [tex]\int dx[/tex][tex]\int dy[/tex]

    With the x limits being 1 and -1 and the upper y limit to be sqrt(1-x^2) I'm having trouble finding the lower y limit. Although to be honest I'm not completely sure about the other three limits! Sorry about my awful attempt at Latex-ing I don't know how to do it so couldn't write the limits of the integrals on the integral. Any help would be great! Thanks.
     
  2. jcsd
  3. Oct 29, 2009 #2
    the problem is symmetric by pi/2 so I'd just stick to the (+,+) quadrant and your lower limit is y=0. Then your x limits would be 1 and 0.

    The area in the entire domain is then four times the area in a single quadrant.
     
  4. Oct 29, 2009 #3

    HallsofIvy

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    Staff Emeritus
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    I have no idea what you mean by
    Shouldn't there be some function to be integrated in that? And it probably is NOT
    [tex]\int dx\int dy[/tex]
    but rather
    [tex]\int f(x,y) dxdy[/tex]
    Even ignoring the "f(x,y)" the two separate integrals implies that the two coordinates can be separated- which is not the case here- at least not in Cartesian coordinates.

    The surface area of z= f(x,y) is given by
    [tex]\int\int \sqrt{1+ \left(\frac{\partial f}{\partial x}\right)^2+ \left(\frac{\partial f}{\partial y}\right)^2} dA[/tex]
    where dA is the differential of area in whatever coordinate system you are using, in the xy-plane. Because of the circular symmetry I would recommend changing to polar coordinates- where the two coordinate variables can be separated.
     
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