1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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]

    Incorrect.
    The correct answer is

    [itex]\frac{14\pi}{3}[/itex]
     
  2. jcsd
  3. Apr 15, 2014 #2

    HallsofIvy

    User Avatar
    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted