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Vector Calculus

  1. Feb 23, 2006 #1
    Hi, can someone help me out with the following question parts?

    a) Let W be a compact region in R^3 bounded by a piecewise smooth closed surface S. Let [itex]f:W \to R[/itex] be a C^1 scalar function. Prove that for all constant vectors [itex]\mathop c\limits^ \to [/itex],

    [tex]
    \int\limits_{}^{} {\int\limits_S^{} f \mathop c\limits^ \to \bullet \mathop n\limits^ \to dS} = \int\limits_{}^{} {\int\limits_{}^{} {\int\limits_W^{} {\mathop c\limits^ \to \bullet \nabla fdV} } }
    [/tex]

    where [itex]\mathop n\limits^ \to [/itex] is the outward unit normal to S.

    b) Let S be the surface of a tetrahedron in R^3, consisting of 4 triangular faces. Suppose that the ith face has area A_i and outward unit normal n_i. Show that

    [tex]
    \sum\limits_{i = 1}^4 {A_i \mathop {n_i }\limits^ \to } = \mathop 0\limits^ \to
    [/tex]

    [Hint: apply part (a) with a suitable choice of f.]

    I can do part (a) but I can't figure out how to do part (b). In fact, I'm really unsure about what kind of a problem it is. For example, I know that part (a) is a divergence theorem question but the second one doesn't look familar.

    Looking at what I need to show (the sum) I'm thinking that surface integrals might be involved. A surface integral is equal to the surface area over the appropriate region if the integrand is equal to one. I think I'm supposed to sum the 'vector' areas of the tetrahedron. I don't see how it can be zero unless the unit normals have opposite directions on opposing faces. But even that doesn't seem to lead anywhere because the area of each of the faces of the tetrahedron are not necessarily equal.

    I don't know how to proceed. Can someone help me out? Thanks.
     
    Last edited: Feb 23, 2006
  2. jcsd
  3. Feb 23, 2006 #2

    AKG

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    Suppose

    [tex](\forall j \in \{1, 2, 3, 4\})\left(\vec{n_j}\cdot\left (\sum _{i=1}^4A_i\vec{n_i}\right) = \vec{0}\right)[/tex]

    Then we can conclude that

    [tex]\sum _{i=1}^4A_i\vec{n_i}\right = \vec{0}[/tex]

    Hint: the choice of f is very very simple, and we essentially use part a) four times, each time with a different [itex]\vec{c}[/itex] but the same f.
     
  4. Feb 24, 2006 #3
    Thanks for your input AKG. I'm not really sure about your use of "Then." I wrote out the sum in full without the summation notation but I still don't see how what you said follows from the dot product being zero for all j.

    I still can't think of another way of doing this question apart from using the result that if the integrand of integrals of scalar functions over surfaces is equal to one, then the integral is equal to the surface area.

    [tex]
    \int\limits_{}^{} {\int\limits_S^{} {f\mathop c\limits^ \to \bullet \mathop n\limits^ \to dS} } = \int\limits_{}^{} {\int\limits_{}^{} {\int\limits_W^{} {\mathop c\limits^ \to } \bullet \nabla fdW} }
    [/tex]

    I want to get a 'scalar' surface integral. If I consider a face, say the 1st face then n is n_1. Let c = n_1 then c.n_1 = ||n_1|| since both vectors in the dot product are unit vectors. Also, set f = 1. From this, I get that the LHS of the above is equal to the surface area of the 1st face. But then the RHS has an integrand of zero which implies that the surface area of one of the faces is equal to zero which can't be right.
     
    Last edited: Feb 24, 2006
  5. Feb 24, 2006 #4

    AKG

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    Homework Helper

    Lets call that sum, X. Now if X.nj is zero for each of the four normal vectors, then X is zero, right? That's my "use of 'then'." And this is true because the normal vectors form a basis (how could they be coplanar or collinear?). So it suffices to show that X.nj is zero for each normal nj.
    This is on the right track, but the LHS isn't equal to the surface area of the 1st face. It is equal to the area of the 1st face PLUS n1.n2 x (Area of the second face) PLUS n1.n3 x (Area of the third) PLUS n1.n4 x (Area of the fourth). Using part a), you will get that this sum is zero. The fact that this sum is zero is not going to tell you anything alone. But you will find that if you replace c with n2, n3, or n4 you will again get this sum being zero. Now this shows that for each nj, the integral on the left with nj in place of c is zero. I said earlier that it suffices to prove that for each nj, X.nj is zero. In more suggestive words, it suffices to show that for each nj, the product X.c with nj in place of c is zero.
     
  6. Feb 24, 2006 #5
    Thanks for the help AKG, it should be enough for me to do this question.
     
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