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Math Help - notation in vector calculus

  1. #1
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    notation in vector calculus

    Just a quick question about notation.

    I was given the vector field

    F = r + grad(1/bar(r)) where r= (x)i+(y)j+(z)k.
    grad is just written as the upside down delta (gradient) and the bar I wrote in the above equation looks like an absolute value around just the r (although I don't think it is absolute value).

    What would be a simplification of this vector field so that I can solve the rest of the problem?

    I want to find its flux across the surface of a sphere.
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    Quote Originally Posted by PvtBillPilgrim View Post
    Just a quick question about notation.

    I was given the vector field

    F = r + grad(1/bar(r)) where r= (x)i+(y)j+(z)k.
    No. I do not think that is the gradient, that is the Del Operator.
    Because the gradient transforms a scalar function into a vector function, not the other way around like you have it here.
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    It is the del operator, but then I take the gradient of (1/(r)).

    I think this whole thing simplifies to:

    3 + xi+yj+zk, but I'm not sure.
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    How can you preform the operation 1/r
    Vectors cannot be divided.
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    I don't know. I specifically have the problem:

    F= r + F= r + 1/double bar(r)
    The first part is easy enough, but what is 1/double bar (r) if r=xi+yj+zk?

    The double bar above is literally just two bars one each side surrounding r.

    This is my problem word for word:
    Consider the vector field
    F= r + grad(1/double bar(r)).
    Compute the flux (double integral F ndS) of F across the surface of the sphere x^2+y^2+z^2=a^2 where a>0. ndS is the vector element of surface with n the unit normal which here is assumed to point away from the enclosed volume.

    If I break it into two parts, I get F=r and F= grad(1/double bar(r))

    F=r just has flux 4*pi*a^3 since F=xi+yj+zk and F*N=a where * is the dot product. Note that N=(x/a)i+(y/a)j+(z/a)k. I know this just by looking at it, but how would I set up this double integral to get 4*pi*a^3?

    Also, I don't get the second part. What is grad(1/double bar(r))?
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    Could it be that ||r|| is simply the magnitude of the vector r? Then grad(1/||r||) is a vector.

    -Dan
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    Quote Originally Posted by PvtBillPilgrim View Post
    I don't know. I specifically have the problem:

    F= r + F= r + 1/double bar(r)
    The first part is easy enough, but what is 1/double bar (r) if r=xi+yj+zk?
    Okay I understand now, the double bar is the norm.

    Thus, ||r||=\sqrt{x^2+y^2+z^2}

    Thus, 1/||r||=1/sqrt(x^2+y^2+z^2)

    To find the gradient you need to find the partial derivatives, which are all symettric.

    (-x)/(x^2+y^2+z^2)^{3/2}-y/(x^2+y^2+z^2)^{3/2}j-z/(x^2+y^2+z^2)^{3/2}k
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    How do I find the flux of this mess then?

    I can simplify it by using a^2=x^2+y^2+z^2 and use polar coordinates, but how do I do this?

    I get that grad(1/magnitude(r))=
    [(-x)/a^(3/2)]i + [(-y)/a^(3/2)]j + [(-z)/a^(3/2)]k

    I know that the unit normal N=(x/a)i + (y/a)j + (z/a)k.

    Thus the dot product of grad(1/magnitude(r)) and the unit normal=
    (-x^2)/a^(5/2) + (-y^2)/a^(5/2) + (-z^2)/a^(5/2) =
    (-a^2)/(a^5/2)= -1/a^(3/2).

    Now how do I integrate this to get the flux of the surface?

    Is it,
    integral from 0 to 4pi, integral from 0 to a^2 of (-a^(-3/2))dr dtheta?
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    Is it easier to use spherical coordinates?

    OK, does an answer of (4*pi*a^3)-(4*pi*a^(-1/2)) make sense for the final answer to the problem posted above?
    Last edited by PvtBillPilgrim; February 12th 2007 at 03:30 PM.
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    Quote Originally Posted by ThePerfectHacker View Post

    (-x)/(x^2+y^2+z^2)^{3/2}-y/(x^2+y^2+z^2)^{3/2}j-z/(x^2+y^2+z^2)^{3/2}k
    Where does the i, j, and k go? Is the (-x), (-y), and (-z) included?
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  11. #11
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by PvtBillPilgrim View Post
    Where does the i, j, and k go? Is the (-x), (-y), and (-z) included?
    Think of ||r|| as sqrt(r dot r) where "dot" is the Euclidean dot product.

    -Dan
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