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Vector calculus double integrals

  1. Sep 23, 2007 #1
    Find the volume V of the solid under the surface z=4-x^2-y^2 and over the rectangle R consisting of all points (x,y) such that 0<=x<=1 and 0<=y<=2.

    I have started, but am unsure if my approach is correct or not.

    x = 4-x^2-y^2
    [tex]\int^{2}_{0}\int^{1}_{0} 4-x^{2}-y^{2} dx dy[/tex]

    is this correct?
     
  2. jcsd
  3. Sep 23, 2007 #2

    Astronuc

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  4. Sep 23, 2007 #3

    HallsofIvy

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    I could have sworn I had answered this before! I particularly remember the peculiar "x= 4- x^2- y^2" term to which I noted that it should be "z=" rather than "x= ".

    But that post also asked about changing to polar coordinates! Rectangle, as here, uses "rectangular" coordinates. Disks and circles use polar coordinates.
     
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