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Stoke's Theorem check

  1. Mar 5, 2006 #1
    JDGriffiths, 3rd ed, Prob. 1.33:

    Given v = xy x + 2yz y + 3xz z, check that Stoke's theorem is valid using the surface bounded by:
    x = 0, 0 <= y <= 2, and 0 <= z = -y + 2. (See attached image.)

    [​IMG]

    Area Integral Result:
    [tex] \int \nabla \times \mathbf{v} \cdot d\mathbf{a} = \int (-2y) (\frac{1}{2} dy dz) = -4 [/tex]

    Line Integral Result:
    [tex] \int \mathbf{v} \cdot d\mathbf{l} } = \int (xy dx + 2yz dy + 3 xz dz) = \frac{8}{3} [/tex]

    Stoke's theorem says they should be equal. What am I doing worng?
     

    Attached Files:

    Last edited: Mar 5, 2006
  2. jcsd
  3. Mar 5, 2006 #2
    check your integral over the surface, note that
    [tex] d\vec{a} = dy \, dz \hat{x} \Rightarrow \vec{\nabla} \wedge \vec{v}\cdot d\vec{a} = 2 \,y\,dy\,dz [/tex]
    and watch the limits of the integral for dz. you should obtain the same results as your line integral. hope this helps. sincerely, x
     
    Last edited: Mar 5, 2006
  4. Mar 5, 2006 #3

    HallsofIvy

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    Why "(1/2)dydz"? That area is in the yz-plane so the differential of area is just dydz.

    Be careful of the direction in which you are integrating. I presume you see that it is only the line z= 2- y that gives a non-zero integral.
     
  5. Mar 5, 2006 #4
    I checked and get the same result. So here is what I do in detail:

    [tex] \vec{\nabla} \times \vec{v} = \left| \begin{array}{ccc}
    \vec{x} & \vec{y} & \vec{z} \\
    \partial_{x} & \partial_{y} & \partial_{z} \\
    xy & 2yz & 3xz \\
    \end{array} \right| = (-2y) \hat{x} + ... [/tex]

    I still get a "-2y" for the curl since only the negative x component of the curl is non-zero and only that component is needed for the dot product with da.

    I think I see my mistake now that you both have mentioned it - my area integral is messed up. I have to review area integrals and their limits.

    Thanks for the help!
    -LD

    EDIT: I got it! I wasn't setting my limits correctly at all. I read Schaum's Outline Series on Advanced Calculus (Chapter 9, pg. 180) and solved it in seconds.

    [tex] \int_{y=0}^{y=2} \int_{z=0}^{z=-y+2} 2y dydz = \frac{8}{3} [/tex] :!!)

    thx again!
    -LD
     
    Last edited: Mar 5, 2006
  6. Mar 5, 2006 #5
    Oh, I am getting a negative sign when I do the curl. How are you getting a plus sign??

    [tex] (\vec{\nabla} \times \vec{v})_{x} = \left| \begin{array}{ccc}
    \vec{x} & \vec{y} & \vec{z} \\
    \partial_{x} & \partial_{y} & \partial_{z} \\
    xy & 2yz & 3xz \\
    \end{array} \right| = 0 - 2y = -2y
    [/tex]
     
    Last edited: Mar 5, 2006
  7. Mar 9, 2006 #6
    The minus sign dissappears from the dot product with the [tex] d\vec{a} = -yz \hat{x} [/tex] which corresponds to the direction we are transversing the loop for the line integral.
     
  8. Mar 9, 2006 #7
    The minus sign dissappears from the dot product with the [tex] d\vec{a} = -dydz \hat{x} [/tex] which corresponds to the direction we are transversing the loop for the line integral.
     
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