Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Test Stokes' Theorem for the function

  1. Sep 28, 2006 #1
    Test Stokes' Theorem for the function
    [itex] \vec{v} = (xy) \hat{x} + (2yz) \hat{y} + (3yz) \hat{z} /[/itex]

    for the triangular shaded region

    [tex] \int_{S} (\grad \times v) \dot da = \oint_{P} v \bullet dl [/tex]

    for the left hand side

    [tex] \int_{0}^{2} \int_{0}^{2} (-2y \hat{x} - 3z \hat{y} - x \hat{z}) \hat{x} dy dz = \int_{0}^{2} \int_{0}^{2} (-2y) dy dz = \left[ \frac{-y^2} \right]_{0}^{2} \int_{0}^{2} dz = (-4)(2) = -8 [/tex]

    so far so good??

    ok for hte right hand side

    i) x=0, z=0, [itex] v \bullet dl = 2y dy[/itex] integration limits 0 to 2

    ii) x=0, y=2, [itex] v \bullet dl = 0 dz [/itex] integration from 0 to 2

    i am not sute about the last one though....

    am i right with i and ii though??

    thank you for any advice!

    Attached Files:

    Last edited by a moderator: Sep 29, 2006
  2. jcsd
  3. Sep 29, 2006 #2


    User Avatar
    Science Advisor

    No, your [itex]\nabla \vec{v}[/itex] is wrong. The [itex]\hat{y}[/itex] component is 0.

    That should be [itex]v \bullet dl= 2yz dy= 0[/itex]

    for the last one, you can take y= t, z= 2- t integrating from t= 0 to 2.

    By the way, to get [itex]\nabla[/itex] use "\nabla", not "\del".
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook