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A vector identity and surface integral

  1. Dec 23, 2005 #1
    Hi, can someone give me some assistance with the following questions?

    1. Let f(x,y,z), g(x,y,z) and h(x,y,z) be any C^2 scalar functions. Prove that [tex]\nabla \bullet \left( {f\nabla g \times \nabla h} \right) = \nabla f \bullet \left( {\nabla g \times \nabla h} \right)[/tex] .

    2. Let S be the part of the ellipsoid 2x^2 + y^2 + (z-1)^2 = 5 for z <=0 and [tex]\mathop F\limits^ \to = \left( {e^{y + z} + 3y,xe^{y + z} ,\cos \left( {xyz} \right) + z^3 } \right)[/tex] .

    Evaluate [tex]\int\limits_{}^{} {\int\limits_S^{} {\left( {\nabla \times \mathop F\limits^ \to } \right)} \bullet d\mathop S\limits^ \to } [/tex].

    (Use the normal to the surface pointing downwards.)

    My working:

    1. I will use [tex]
    \nabla \bullet \left( {\mathop A\limits^ \to \times \mathop B\limits^ \to } \right) = \mathop B\limits^ \to \bullet \left( {\nabla \times \mathop A\limits^ \to } \right) - \mathop A\limits^ \to \left( {\nabla \times \mathop B\limits^ \to } \right)
    [/tex] and [tex]\nabla \times \left( {f\mathop F\limits^ \to } \right) = f\nabla \times \mathop F\limits^ \to + \nabla f \times \mathop F\limits^ \to [/tex].

    \nabla \bullet \left( {f\nabla g \times \nabla h} \right)

    = \nabla h \bullet \left( {\nabla \times f\nabla g} \right) - f\nabla g \bullet \left( {\nabla \times \nabla h} \right)
    [/tex]...from the identies above .The second bracket is zero since curl(grad(h)) = 0 vector.

    = \nabla h \bullet \left( {\nabla \times f\nabla g} \right)

    = \nabla h \bullet \left( {f\nabla \times \nabla g + \nabla f \times \nabla g} \right)
    [/tex]...using identities listed above

    = \nabla h \bullet \left( {\nabla f \times \nabla g} \right)
    [/tex] since curl(grad(g)) = 0 vector.

    This is as far as I get in the first question.

    For the surface integral I calculated [tex]\nabla \times \mathop F\limits^ \to = \left( { - xz\sin \left( {xyz} \right) - xe^{y + z} ,e^{y + z} + yz\sin \left( {xyz} \right), - 3} \right)[/tex].

    I would normally parameterise the surface to find a normal to the surface. I'm not sure how to that or if I need to do that in this question.

    Any help would be good thanks.
    Last edited: Dec 23, 2005
  2. jcsd
  3. Dec 23, 2005 #2
    For question #2 have you learned Stokes's theorem yet, that might help.
  4. Dec 23, 2005 #3


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    Homework Helper

    For the first one, there is the vector identity:

    [tex]\vec A \cdot (\vec B \times \vec C) = \vec B \cdot (\vec C \times \vec A) =\vec C \cdot (\vec A \times \vec B) [/tex]

    This operation is called the triple product. For the second, use Stoke's theorem to turn the surface integral into a contour integral.
  5. Dec 24, 2005 #4
    Thanks for the help guys.

    The thing is, Stokes' theorem was what I thought about using but the question says something about using a normal so I figured that I need to evaluate it as a surface integral. In any case if I was to use Stoke's theorem, I would need to parameterise an ellipse. Perhaps x = acos(t) and y = bsin(t) where a and b are suitable constants? But if sub those into my expression for curl(F) it looks like it'll get quite messy even with z = 0 getting rid of the trig functions.
  6. Dec 24, 2005 #5
    But you wouldn't put them into curl(F) you would plug them into F because stokes' theorm says that the line integral of F dotted with dr is the same as the surface integral of curl(F) blah blah blah.
  7. Dec 24, 2005 #6
    Oh ok...I forgot about that.:redface:
  8. Dec 24, 2005 #7
    It's worth noting that this identity is a little trickier when you start throwing [tex]\nabla[/tex] into the mix. However, there is an identity like it that you might find useful.
  9. Dec 25, 2005 #8
    When you 'del' a scalar function you get a vector so it seems reasonable to suggest that it works. However, I haven't seen/(can't recall) any complete derivations of the cross product so I'm not sure if it matters if the vector has variable components.
  10. Dec 25, 2005 #9
    The problem is that del is a differential operator, a one-form, not a vector in the traditional sense, so there's an issue with ordering and such. For instance, what is [tex]\mathbf{b} \cdot (\mathbf{c} \times \nabla)[/tex]
  11. Dec 25, 2005 #10


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    Homework Helper

    That's true, but I was referring to:

    [tex]\nabla f \cdot (\nabla g \times \nabla h )[/tex]

    as it appears in his problem, and these are all vectors.
  12. Dec 25, 2005 #11
    Well first apply cyclic permutation to get the nabla out of the brackets. Then nabla must work on both b and c, so you will get a sum of two parts. The clue always is to write this formula in a sum of terms in which each nabla operates on ONE vector using the vector identities in the first post of this thread.

    Ofcourse it can happen that you have this
    [tex]\nabla \cdot \vec{b}[/tex]


    [tex]\vec{b} \cdot \nabla[/tex]

    The first formula is easy, right ? Just calculate the dot product and you will get a scalar.

    The second formula actually defines a new scalar operator

    Last edited: Dec 25, 2005
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