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Stoke's Theorem vector field

  1. Oct 5, 2013 #1
    1. The problem statement, all variables and given/known data
    Assume a vector field:[tex]\textbf{F} = \widehat{r} 2r sin\phi + \widehat{\phi} r^2 cos\phi[/tex]

    a) Verify the Stokes's theorem over the ABCD contour shown in Fig. 1 .
    b) Can F be expressed as the gradient of a scalar? Explain

    fig1.jpg

    My problems results in not being able to verify Stoke's Theorem

    2. Relevant equations
    Stoke's Theorem
    [tex]\oint_{C} \textbf{F}\cdot \overrightarrow{dl} = \int \int_{S} (\nabla\times\textbf{F})\cdot \overrightarrow{dS}[/tex]


    3. The attempt at a solution
    We can see that the line integral and surface integrals can be dealt with in cylindrical coordinates.
    for line integrals:
    [tex]\int _{DA}+\int _{AB}+\int _{BC}+\int _{CD} =-\int_{r=2}^{1}2rsin(0)dr + \int_{\phi=0}^{\pi/3}cos(\phi)d\phi + \int_{r=1}^{2}2rsin(\pi/3)dr + \int_{\phi=\pi/3}^{0}4cos(\phi)d\phi[/tex]
    [tex]= \frac{\sqrt3}{2} +\frac{\sqrt3}{2}(4-1) -4\frac{\sqrt3}{2}=0 [/tex]

    I also obtain that [tex]\nabla\cdot \textbf{F}=\widehat{z}cos\phi(3r-2)[/tex]
    And surface integral is evaluated as [tex]-\int_{r=1}^{2}\int _{\phi=0}^{\phi/3}cos\phi(3r-2)rdrd\phi=-(8-4)\frac{\sqrt3}{2}=-4\frac{\sqrt3}{2}[/tex]
    The negative is due to the fact that ds is in the -z direction.
    Am I doing something wrong while integrating?

    Thanks in advance!
     
  2. jcsd
  3. Oct 5, 2013 #2

    SteamKing

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    In your line integrals for segments AB and CD, d[itex]\phi[/itex] represents a change in angle, which s not the same as a change in arc length, dl.
     
  4. Oct 5, 2013 #3
    True, but isn't [tex]\overrightarrow{dl} = \widehat{\phi}r d\phi [/tex] for path AB and [tex]\overrightarrow{dl} = -\widehat{\phi} r d\phi[/tex] for path CD? The after dot product, I obtain the integrals previously?
     
  5. Oct 5, 2013 #4

    HallsofIvy

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    SteamKings point is that, in the integral from C to D, [itex]ds= 2d\phi[/itex]. You need another factor of 2.
     
  6. Oct 5, 2013 #5
    Ahh I see now I was missing r = 2 in the the CD line integral, now Stoke's theorem can be verified. Thank you very much guys!
     
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