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Surface integral for line current

  1. Nov 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculate the integral ##\oint_C \vec F \cdot d\vec S##, where ##C## is the closed curve constructed by the intersection of the surfaces ##z = \frac{x^2+y^2}{4a}## and ##x^2+y^2+z^2=9a^2##, and ##\vec F## is the field ##\vec F = F_0\left( \frac{a}{\rho}+\frac{\rho^2}{a^2} \right)\hat \phi##. (##a## and ##F_0## are constants.)

    (I'm guessing this fits better here than in the calculus forum since since the question is as preparation for EM-class)
    2. Relevant equations
    Line current with strength ##J##
    ##\vec F = \frac{J}{2\pi \rho} \hat \phi##.

    3. The attempt at a solution
    We see the first part of the force is a line current with strength ##2\pi F_0a## so it's total contribution to the integral over a closed loop is ##2\pi F_0a##.

    I'm left with ##\vec F_2 = F_0\frac{\rho^2}{a^2}\hat \phi##. We note from ##z=\frac{x^2+y^2}{4a}## that if ##a >0## ##z## has to be strictly positive and if ##a < 0## ##z## is strictly negative.
    Substituting into the second equation we have
    ##(z+2a)^2-4a^2 = 9a^2 \Longrightarrow z=(\pm \sqrt{13} -2)a##. We conclude that ##z## is constant and since ##\rho^2 = x^2+y^2##, ##\rho## is also constant.
    Calculating the curl we get ##\nabla \times \vec F_2 = \frac{3F_0}{a^2}\rho \hat z##.
    Using stokes theorem and since we know ##\rho## is constant and using ##\hat z## as the normal vector
    ##\int_C \vec F_2 \cdot d\vec r = \frac{3F_0}{a^2}\rho \int_S dS = \frac{3\pi F_0}{a^2}\rho^3 ##
    Let's assume that ##a>0## we then have that ##\rho = 2a\sqrt{\sqrt{13}-2}##. Using this we have
    ##\int_C \vec F_2\cdot \vec r = 24\pi F_0a\left( \sqrt{13}-2 \right)^{3/2}##.
    So the total value of integral should be
    ##2\pi F_0a\left(12(\sqrt{13}-2)^{3/2}+1\right)##.


    Another approach would be to use the line element ##d\vec r = d\rho \hat \rho + \rho d\phi \hat \phi + dz \hat z##
    ##\int_C \vec F_2 \cdot d\vec r = \int_C \frac{F_0\rho^3}{a^2}d\phi## And everything inside should be a constant. Since both ##z## and ##\rho## are constant we should be on a circle thats parallel to the ##xy##-plane and we should integrate from ##\phi =0 \to \phi = 2\pi##. So the integral becomes
    ##\int_C \vec F_2 \cdot d\vec r = \frac{2\pi F_0 \rho^3}{a^2} = 8\pi aF_0(\sqrt{13}-2)^{3/2}##. Giving us the value
    ##2\pi F_0a\left(8(\sqrt{13}-2)^{3/2}+1\right)##.

    According to the answer the second approach gave me the right answer and the first one is wrong but I can't see where I go wrong?
     
  2. jcsd
  3. Nov 15, 2015 #2

    RUber

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    Homework Helper

    Have you drawn a picture of what the contour looks like?
    This is the intersection of a cone and a sphere.
    Since your integral uses spherical coordinates, convert the contour to spherical coordinates as well. That should help.
     
  4. Nov 15, 2015 #3
    Sorry I should've clarified better. The field is given in cylindrical coordinates (the book uses ##\rho## for cylindrical.)
    Should be a paraboloid not a cone since we have ##z=\frac{x^2+y^2}{4a}## not ##z^2=\frac{x^2+y^2}{4a}## so the intersection should be a circle. I did make a drawing but I'm not too great at those. Do you think the second solution is correct or did I just happen upon the correct answer there? Be cause I feel I did most of the work similar for both approaches.
     
  5. Nov 15, 2015 #4

    RUber

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    Homework Helper

    Yes, you are z is growing with the square of the radius of the circle, not linearly with the radius like in a cone.
    Your contour is a circle parallel to the xy plane. So the second method leaves fewer opportunities to make a mistake.
     
  6. Nov 20, 2015 #5
    I realised what my error was here. When I calculated the surface integral i got from stokes theorem I had
    ##3F_0a^2\int_S \rho d\vec S##. S is a ##z##-surface so the surface element is given by ##d\vec S = \hat z \rho d\rho r\phi##. Using this we have (let ##\rho_0## be the radius of the circle as to not confuse them)
    ##3F_0a^2 \int_0^{\rho_0}\int_0^2\pi \rho^2 d\phi d\rho = 2\pi F_0\rho_0^3## yielding the correct answer in the first case as well. So the mistake I made was that I didn't accountfor that I have scale factors in curvilinear coordinates.
     
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