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Divergence Theorem - Confused :s (2 problems)

  1. Apr 13, 2007 #1
    Question

    Evaluate both sides of the divergence theorem for

    V =(x)i +(y)j

    over a circle of radius R

    Correct answer

    The answer should be 2(pi)(R^2)

    My Answer

    the divergence theorem is

    **integral** (V . n ) d(sigma) = **double intergral** DivV d(tau)

    in 2D. Where (sigma) is the curve bounding the area (tau.) n is the unit normal vector to the curve.

    I did the RHS of the divergence theorem fine and got 2(pi)(R^2)

    *******
    For the LHS:

    First I need to work out n, the unit normal vector to the circle.

    The equation of the circle is y=(1 - (x^2)^1/2)

    To get the unit normal you use the nabla (gradient) function to get a vector that points away from the line and then normalize it.

    First problem: In the solution they took grad(x^2 + y^2.) You end up with n (normalised) as (x/R)i + (y/R)j. . But i dont understand why you take grad(x^2 + y^2) when the line has the function f(x)=(1 - (x^2)^1/2)

    Pretending i understood that bit and moving on you get V.n = (x^2/R) + (y^2R) = R^2/R =R

    So I need to work out *integral* R d(sigma)

    i presume d(sigma) (the elemental boundary length) is d(theta.) Then you do the line integral from 0 to 2(pi) . R is a constant so you can take it out then you just get 2(pi) for the integral

    => *integral* (V.n) d(sigma) = 2(pi)R

    which is wrong (missing an R)

    ******

    Sorry for the length of that. Some help on those two problems would be awesome. :wink:
     
  2. jcsd
  3. Apr 13, 2007 #2
    Well, the big problem with the first part you're having trouble with is that f(x) = (1 - x^2)^(1/2) only represents the top part of the circle.

    The second problem is that d(sigma) isn't d(theta).
     
  4. Apr 13, 2007 #3
    Yes! its R d(theta.) The length of the arc subtended. Which would provide the extra R. Excellent, thankyou :biggrin:

    Okay point taken. Im not sure how else to write the function though. f(x) = +/-(1 - x^2)^(1/2) would give the top and the bottom but i can't do anything with that.

    (just noticed something, that should be a an R not a 1. Doesnt affect anything)

    Since in the solution you do grad(x^2 + y^2.) you must be taking grad f(x,y) where

    f(x,y)= x^2 + y^2

    but thats the equation of a 3D surface (i dont know what surface, its not a sphere.)

    How does it represent my circle of radius R? (Theres no R in it for a start)
     
  5. Apr 13, 2007 #4
    Technically, what you should do is find a tangent vector for your circle, then get a vector perpendicular to that to find the normal. But, as far as I remember, taking the gradient is essentially a cheat.

    Basically, what you're doing is saying that every level set of f(x,y) = x^2 + y^2 represents a circle of some radius. So what you do is find the outward normal of the surface (which is (fx, fy, -1), then project it into the plane (level set) where z = r^2, and that gives you the outward normal for the circle in that plane.

    But don't quote me on that. I never got a really rigorous treatment of that kind of stuff because the few times it was covered in my classes, I was swamped with other work. I just used it if it worked.
     
  6. Apr 13, 2007 #5
    okay got it. So i can think of f(x,y) as like a contour map with a load of concentric circles each representing a value of f(x,y.)

    So you workout grad f(x,y) and normalize it then set (x^2 + y^2) to R^2... and it works!

    thankyou for all the help mystic. Youre a legend
     
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