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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?

    or should i be putting it in polar coordinates, in which case how do i set up the limit of integration
    [tex]\int\int_{\Omega} f(x,y) dA[/tex]
    [tex]\Omega = \left\{ (x,y) | 0<=x<=1, 0<=y<=2 \right\}[/tex] ? is this right?
     
  2. jcsd
  3. Sep 23, 2007 #2
    You're looking for the volume under the surface, within the bounds of the rectangle, right? Your first one is fine, though I think you meant to say f(x,y) = 4-x^2-y^2.
     
  4. Sep 23, 2007 #3

    HallsofIvy

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    I'm not sure why you would write that: z= 4- x2- y2, not x.

    Of course, your first calculation is correct. In fact, the second formula,
    [tex]\int\int_{\Omega} f(x,y) dA[/tex]
    is also correct- it's just a general statement of the formula for volume.
    But there is no reason to change to polar coordinates. Polar coordinates would be appropriate if you were finding the volume over a disk.
     
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