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I can't accept the solution manual's explanation

  1. Jul 2, 2015 #1
    1. The problem statement, all variables and given/known data

    It's a long winded problem, I'll post a picture and an imgur link

    5wvbqO2.jpg

    Imgur link: http://i.imgur.com/5wvbqO2.jpg

    2. Relevant equations

    Divergence Theorem

    [tex]\iint\limits_S \vec{F}\cdot d\vec{S} = \iiint\limits_E \nabla \cdot \vec{F} \ dV[/tex]

    3. The attempt at a solution

    I'll follow my own solution up to the point that it disagrees with the solution given in the solutions manual, then I'll explain why I cannot understand/accept what the solution manual is telling me.

    First off, the area of S(a) is

    [tex]a^2\int_{\theta_1}^{\theta_2}\int_{\phi_1}^{\phi_2} \sin{\phi} \ d\phi \ d\theta[/tex]

    Thus

    [tex]| \Omega (S) | = \int_{\theta_1}^{\theta_2}\int_{\phi_1}^{\phi_2} \sin{\phi} \ d\phi \ d\theta[/tex]

    I can see that for

    [tex]\vec{F} = \frac{\vec{r}}{|\vec{r}|^3}[/tex]

    [tex]\nabla \cdot \vec{F} = 0[/tex]

    Thus

    [tex]\iiint\limits_E \nabla \cdot \vec{F} \ dV = 0[/tex]

    I can see that for [itex]\partial S[/itex] there are 6 surfaces, but that the lateral polar rectangular surfaces up/around the sides have normal vectors that are orthogonal to the specified vector field at all points on said surfaces, and that therefore the flux across those surfaces will be zero.

    So

    [tex]\iiint\limits_E \nabla \cdot \vec{F} \ dV = 0 = \iint\limits_{S} \vec{F}\cdot d\vec{S} + \iint\limits_{S(a)} \vec{F}\cdot d\vec{S}(a)[/tex]

    Thus

    [tex]\iint\limits_{S} \vec{F}\cdot d\vec{S} = -\iint\limits_{S(a)} \vec{F}\cdot d\vec{S}(a)[/tex]

    Now, our S(a) is a portion of the surface of the sphere subtended by the surface S.

    The negative sign of the RHS indicates that we used a normal vector in the opposite direction to the normal vector used for the flux across S on the LHS, since we are interested in a quantity without direction here, let's change the direction of the normal vector, to give us

    [tex]\left|\iint\limits_{S} \vec{F}\cdot d\vec{S}\right| = \left|-\iint\limits_{S(a)} \vec{F}\cdot d\vec{S}(a)\right| = \iint\limits_{S(a)} \vec{F}\cdot d\vec{S}(a)[/tex]

    So if we show that

    [tex]\iint\limits_{S(a)} \vec{F}\cdot d\vec{S}(a) = |\Omega (S)| = \int_{\theta_1}^{\theta_2}\int_{\phi_1}^{\phi_2} \sin{\phi} \ d\phi \ d\theta[/tex]

    Then

    [tex]\iint\limits_{S} \vec{F}\cdot d\vec{S} = |\Omega (S)| [/tex]

    where [itex]\vec{F} = \frac{\vec{r}}{|\vec{r}|^3}[/itex], [itex]\vec{r}[/itex] being a vector from P to any point on S, which is what we set out to prove in the first place.

    I will omit some of my working now. I set out to compute the surface integral of S(a). I get to here.

    As S(a) is a portion of the surface of a sphere

    [tex]\vec{n} = \frac{1}{a}\langle x,y,z \rangle[/tex]

    Then

    [tex]\iint\limits_{S(a)} \vec{F} \cdot d\vec{S}(a) = a \int_{\theta_1}^{\theta_2} \int_{\phi_1}^{\phi_2} \frac{1}{|\vec{r}|} \sin{\phi} \ d\phi \ d\theta[/tex]

    The solutions manual takes a slightly different, but equivalent route, but is telling me, at roughly this point, that [itex]|\vec{r}| = a[/itex].

    I don't understand this. I can't accept it. [itex]\vec{r}[/itex] is the position vector from P to the surface S, not the position vector connecting P to S(a)...how can [itex]|\vec{r}| = a[/itex]?

    edit: wait

    in order to get to here

    [tex]\iint\limits_{S(a)} \vec{F} \cdot d\vec{S}(a) = a \int_{\theta_1}^{\theta_2} \int_{\phi_1}^{\phi_2} \frac{1}{|\vec{r}|} \sin{\phi} \ d\phi \ d\theta[/tex]

    I need to assume that the x,y,z that I use in my normal vector, are the same x,y,z that I use in [itex]\vec{r}[/itex] anyway...so that would mean that [itex]|\vec{r}| = a[/itex], but how can we justify this?
     
    Last edited: Jul 2, 2015
  2. jcsd
  3. Jul 2, 2015 #2

    vela

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    In the integral over S, the vector ##\vec{r}## goes from point P to a point on S, and in the integral over S(a), the vector ##\vec{r}## goes from the point P to a point on S(a). This is true simply because the integrals are being taken over those surfaces. The problem statement isn't saying that the symbol ##\vec{r}## always represents the vector from point P to a point on S in any context; it's saying this is the case for that specific integral. Note that you already used this fact when you evaluated the divergence of ##\vec{F}##. There, you took ##\vec{r}## to correspond to a point in the volume between S and S(a).
     
  4. Jul 2, 2015 #3
    Ok, but, to quote the problem itself "r is the radius vector from P to any point on S"...there is absolutely a context within which to understand it that way.

    I still don't get it. 'r' gives us our vector field. They really shouldn't write that r goes to any point on S if what they really mean is that r goes to any point on [itex]\partial S[/itex].

    "Note that you already used this fact when you evaluated the divergence of ##\vec{F}##. "

    Really?? I definitely thought the opposite right there. My reasoning went that for the surface S, x,y,z would be parametrized with phi and theta, BUT, that you could still just apply the vector differential operator to them first...

    So, the vector field is just a position vector to any point in the region E...ok.
     
  5. Jul 2, 2015 #4

    vela

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    Yes, that's what the problem statement says, but what I'm saying is that it's saying that's what ##\vec{r}## represents in that particular integral, which happens to be evaluated over the surface S. When you calculated the volume integral, you evaluated the divergence of ##\vec{F}## at points inside the volume, so there ##\vec{r}## corresponded to the vector from P to a point in the volume, not on S or S(a).
     
  6. Jul 3, 2015 #5
    So if we happened to get a parametrization of the surface S, [itex]x=x(\phi, \theta ), y = y(\phi, \theta ), z = z(\phi, \theta)[/itex] and chose a vector field [itex]\vec{r}(\phi, \theta) = \langle x(\phi, \theta ), y(\phi, \theta ), z(\phi, \theta) \rangle[/itex] and we were to evaluate that surface integral over some range of phi and theta, essentially evaluating the surface integral where the vector field IS the r vector we constructed from the parametrization, the result would be equal to the area of a subtended portion of sphere divided by it's own radius squared...giving us a measure of the angular subtend-ation that has occurred...

    And if we were to parametrize x,y,z so as to describe the surface of the subtended sphere, it would be a surface integral where the vector field was also just the r vector we constructed from the parametrization, then it would have the same quantity value as the first S surface that had x,y,z parametrized differently.

    Is there something about the fact that the vector field is a vector constructed from the parametrization of the surface itself that makes it not matter between the three cases of doing it on two different surfaces and then doing it in a volume...? What's making the r vector so flexible here?
     
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