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Flux through a section of a sphere

  1. Apr 15, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the flux of F=<y,-x,z> through the piece of ρ=2 that lies above z=1 and is oriented up.

    2. Relevant equations

    3. The attempt at a solution

    [itex]S = < x, y, \sqrt{4-x^{2}-y^{2}} >[/itex]

    Take Find Sx and Sy, cross them and end up with:

    [itex]dS = < \frac{x}{\sqrt{4-x^{2}-y^{2}}}, \frac{y}{\sqrt{4-x^{2}-y^{2}}}, 1 > [/itex]

    Z is positive, orientation is OK.

    F dot dS = [itex]\sqrt{4-x^{2}-y^{2}}[/itex]

    Therefore the integral should be

    [itex]\int^{2\pi}_{0}\int^{\sqrt{3}}_{0} r\sqrt{4-r^{2}}drd\theta[/itex]

    = [itex]\frac{4\pi}{3} (\sqrt{32} -1)[/itex]

    The correct answer is

  2. jcsd
  3. Apr 15, 2014 #2


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    Staff Emeritus
    Science Advisor

    All I can say is that you have clearly integrated wrong. Since you don't show how you did that integral, I cannot say more.
  4. Apr 15, 2014 #3
    [itex] \int^{2\pi}_{0} \int^{\sqrt{3}}_{0}r \sqrt{4-r^{2}}dr d\theta[/itex]

    = [itex]2\pi \int^{\sqrt{3}}_{0}r\sqrt{4-r^{2}}dr[/itex]

    =[itex]\pi \int^{4}_{1}u^{1/2}du[/itex]

    = [itex]\frac{2\pi}{3}(4^{3/2}-1)[/itex]

    I see what I did. Thanks.
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