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Using Stroke's Theorem

  1. Jun 1, 2014 #1
    1. The problem statement, all variables and given/known data
    Use Stokes' Theorem to evaluate [itex]\int\int curl \vec{F}\bullet d\vec{S} [/itex] where [itex]\vec{F}(x,y,z) = <e^{z^{2}},4z-y,8xsin(y)> [/itex] and S is the portion of the paraboloid
    [itex]z = 4-x^{2}-y^{2}[/itex] above the xy plane.


    2. Relevant equations
    Stokes Thm:[itex]\int\int curl \vec{F}\bullet d\vec{S} = \int \vec{F}\bullet d\vec{r}[/itex]

    [itex]\vec{F}(x,y,z) = <e^{z^{2}},4z-y,8xsin(y)> [/itex]

    S: [itex]z = 4-x^{2}-y^{2}[/itex] above the z = 0.

    3. The attempt at a solution
    C: [itex]\vec{r}(t) = <2cos(t), 2sin(t), 0>[/itex] where [itex] 0\leq t\leq2\pi [/itex]
    [itex]\vec{r}'(t) = <-2sin(t), 2cos(t), 0> [/itex]

    [itex]\vec{F}(\vec{r}(t)) = <e^{0}, (4(0) - 2sin(t), 8(2(cos(t))sin(2cos(t))>[/itex]
    [itex]\vec{F}(\vec{r}(t)) = <1, -2sin(t), 16cos(t)sin(2cos(t))>[/itex]

    [itex]\int <1, -2sin(t), 16cos(t)sin(2cos(t))> \bullet <-2sin(t), 2cos(t), 0> dt [/itex] from 0 to 2pi
    [itex]=\int -2sin(t)-2sin(t)2cos(t) dt [/itex] from 0 to 2pi
    [itex]=0 [/itex]
     
    Last edited: Jun 1, 2014
  2. jcsd
  3. Jun 1, 2014 #2

    LCKurtz

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    Some latex pointers. If you will put a backslash in front of any function in tex it prints in a much nicer font. Also you can use \cdot instead of \bullet and \nabla\times for curl. I have done that in my quote so you can see the difference. Also you can put the limits on the integral as I illustrate in one line above.

    A problem like this should give some orientation for the surface, although with an answer of zero it doesn't matter much. Your work looks correct unless I have overlooked something.
     
  4. Jun 1, 2014 #3
    Thank you, I will remember those hints and tips for next time!

    And thanks for looking at my work, I think it is correct too; I just can't afford to miss any points on any problem, its dead week and I need all the points I can get.
     
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