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Surface Integration Problem

  1. Apr 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex]D=\{(x,y,z)| z^{2}=1+x^{2}+y^{2} , 1<z<3\}[/itex] Compute[itex]\int\int_{D}zdS[/itex]

    2. Relevant equations
    From lectures I know;
    [itex]\int\int_{D}\delta dS=\int\int_{D}\delta\sqrt{(\frac{\partial f}{\partial x})^{2}+(\frac{\partial f}{\partial y})^{2}+1}dxdy[/itex]


    3. The attempt at a solution
    I'm not sure what I'm doing is correct my answer seems wrong;

    [itex]z^{2}=1+x^{2}+y^{2}[/itex]
    so

    [itex]z=\sqrt{1+x^{2}+y^{2}}[/itex]

    Taking partial derivatives of x and y and substituting into equation from 2 I get

    [itex]\int\int_{D}\sqrt{5x^{2}+5y^{2}+1}dxdy[/itex]

    making change of variables to cylindrical coordinates

    [itex]\int\int_{D}\sqrt{5r^{2}+1}r drdt[/itex]

    making substitution

    [itex]u=5r^{2}+1[/itex]

    I get

    [itex]\int\int_{D}\sqrt{u} dudt[/itex]

    I brought the limits through as well to have limits for u of 0 and 41 and limits of t of 0 and 2[itex]\pi[/itex]

    giving me a final answer of 1099.675108 which seems completely wrong;

    Any suggestions?
     
  2. jcsd
  3. Apr 17, 2012 #2
    I got 2 instead of 5. Really, [itex]\frac {\partial z}{\partial x} = \frac{x}{z}[/itex] and [itex]\frac {\partial z}{\partial y} = \frac{y}{z}[/itex]
    [itex]z \sqrt{1+\frac{x^2}{z^2} + \frac{y^2}{z^2}} = z \sqrt{ \frac{z^2 + x^2 + y^2}{z^2} } = \sqrt{1+2x^2+2y^2}[/itex]

    I then got 73.39 which is a lot more likely.
     
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