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Surface integral

  1. Feb 3, 2008 #1
    1. The problem statement, all variables and given/known data
    I'm not sure how to convert this surface integral into a double integral for evaluation.

    [tex]\iint_S \frac{1}{1 + 4(x^2 + y^2)} dS[/tex]

    S is the portion of the paraboloid [itex]z = x^2 + y^2[/itex] between z = 0 and z = 1.

    3. The attempt at a solution
    How do you project this onto the xy-plane? From z = 0 to z = 1, we get circles of increasing radius. So do I just take z = 1, which is a circle of radius 1, and make the region:

    [tex]D = \{(r, \theta) | 0 <= r <= 1, 0 <= \theta <= 2 \pi\}[/tex]

    Also, I know:

    [tex]dS = \sqrt{1 + (2x)^2 + (2y)^2} dA = \sqrt{4r^2 + 1} dA[/tex]

    [tex]\iint_S \frac{1}{1 + 4(x^2 + y^2)} dS = \iint_D \frac{r}{1 + 4r^2} \sqrt{4r^2 + 1} dA[/tex]

    I just don't know what the limits should be. The region I chose gives me the right answer but I'm not sure exactly why the projection was when z = 1. What if it was the paraboloid between z = 1 and z = 2 then? Do I have to split up the surface integral to be the circle at z = 1 and the paraboloid surface? I'm really confused about this projection stuff.
  2. jcsd
  3. Feb 3, 2008 #2


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    The plane z=a cuts the paraboloid in the circle a=x^2+y^2. So the domain in the x,y plane is x^2+y^2<=a or r^2<=a. If the limits were z=1 to z=2, then the r limits would be r=1 to r=sqrt(2).
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