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Find the net flux thru Box

  1. Sep 16, 2011 #1
    1. The problem statement, all variables and given/known data
    Vector V = <x2yz2, xy3z>
    0<=x <=a
    0 <= y <= b
    0<= z <= c

    2. Relevant equations

    [itex]\int[/itex]V [itex]\bullet[/itex] [itex]\hat{n}[/itex]dS

    In our case, this amounts [itex]\int\int[/itex]V(either x or y component) dot [itex]\hat{i}[/itex] or [itex]\hat{j}[/itex] dA.


    3. The attempt at a solution

    Okay, I have tried to break apart the problem by looking for the flux through each side. My first difficulty is that there is always a free variable. I don't know how to get the equation into 2 equations. Second, ignoring that difficulty it seems that the answer would be zero. Because the only sides with any flux are going all the sides that are not the top and bottom. But each side will cancel the other side out.

    Now, I've tried to do the problem via the Divergence theorem, and I don't get zero.
     
  2. jcsd
  3. Sep 16, 2011 #2

    dynamicsolo

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    Is your vector field V missing a component? As it stands now, the field is confined to the xy-plane, but the "box" is in the "first octant" (in which case, the flux through the box would be zero)...
     
  4. Sep 16, 2011 #3
    No, it is not missing a component.

    If it is zero, if you integrate with the divergence theorem you don't get 0.

    For the Div of V = 2xyz^2 + 3xy^2z.

    Integrate that, and you'll wind up with a non-zero answer. I get (abc)^2 (c/6 +b/4).

    And why does it matter that the box is in the first octant? You make it sound as if because the box is in the first octant the flux must be zero because of that fact.
     
  5. Sep 16, 2011 #4
    Perhaps, it would be helpful to show me how in the world to use the divergence theorem to answer this question.
     
  6. Sep 17, 2011 #5
    i thought the vector V should have 3 components? or do we assume the 'k' direction is 0?
     
  7. Sep 17, 2011 #6

    dynamicsolo

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    As you've written this field, I am interpreting it to mean that it has no z-component (the vector as written is two-dimensional), and so lies only in the xy-plane. This would mean that the field is only on one face of the box and does not enter it.

    I am assuming you are interpreting this to be the description of the vector field as a function of x and y at every value of z. This doesn't seem to be notated correctly...

    In any case, the form of the vector tells us that V = 0 on the x = 0 , y = 0 , and z = 0 faces of the box, so there is no flux there. Since there is no z-component of this field, V is perpendicular to the normal on the z = c face, so there is also no flux there. Hence, there are only two faces to consider, x = a and y = b.

    The surface integral on the x = a face is (the dot product only brings in the x-component of V) [tex] \int_{0}^{c} \int_{0}^{b} a^{2}yz^{2} dy dz = \frac{1}{6} a^{2}b^{2}c^{3} . [/tex]

    Similarly, the surface integral on the y = b face is (we only have the y-component this time) [tex] \int_{0}^{c} \int_{0}^{a} xb^{3}z dx dz = \frac{1}{4} a^{2}b^{3}c^{2} . [/tex]

    So the sum of these fluxes through the two faces does agree with your result from the Divergence Theorem. It occurs to me now that the field should be notated as V = < x2yz2 , xy3z , 0 >. Now this looks right...
     
  8. Sep 17, 2011 #7
    "As you've written this field, I am interpreting it to mean that it has no z-component (the vector as written is two-dimensional), and so lies only in the xy-plane. This would mean that the field is only on one face of the box and does not enter it."

    I have no idea what you are trying to say here.

    "I am assuming you are interpreting this to be the description of the vector field as a function of x and y at every value of z. This doesn't seem to be notated correctly..."

    mmm... I don't understand. If you have a vector <x,y,0> it doesn't mean that there is no field at, say, point 3,2,1. It only means that the force isn't itself directed upwards at any point on the x,y,z. I'm not sure if this is what you were going for with the comment "It occurs to me now that the field should be notated as V = < x^2yz^2 , xy^3z , 0 >. Now this looks right..."

    "In any case, the form of the vector tells us that V = 0 on the x = 0 , y = 0 , and z = 0 faces of the box, so there is no flux there. Since there is no z-component of this field, V is perpendicular to the normal on the z = c face, so there is also no flux there. Hence, there are only two faces to consider, x = a and y = b."

    Are you trying to say here that the back plane and the leftmost plane of the box have a net flux of 0?
     
  9. Sep 17, 2011 #8
    i think he meant that the x=0 , y=0 and z=0 planes have no flux through it, because if you sub these values into the vector field expression given in the question, V=0.

    that means you only have to consider the other 3 faces which is the planes x=a, y=b, z=c

    but since the expression of the vector field has no component in the z-direction(k) ( since you only put two components in it, he assumed that they are the i and j components), so,, it would mean there is no flux going through the z=c plane.
     
  10. Sep 17, 2011 #9

    dynamicsolo

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    A vector written as <Vx, Vy> has only two components and means that it only lies in the xy-plane. If the third (z) component is zero, you do not simply omit it from the notation: you must write it as <Vx, Vy, 0 > , so that it is clear that you are describing a vector field in three dimensions. That's why what you were writing was causing confusion. What you are saying in your last sentence here is correct, but that wasn't what you were saying in the statement of your problem.

    That's true if you write a third component as zero, not if you don't write a third component at all.

    That is just what I'm getting at...

    This, of course, has no effect on the result for the divergence of V, but it does change the meaning of the problem. Correct notation is important in communicating the description of a physical situation: a value set to zero is not the same thing in mathematics or physics as saying that a quantity has no value. A vector in three dimensions with a z-component of zero is not the same entity as a two-dimensional vector in an important sense, even though they function the same way in certain calculations.

    Yes. The vector field V is < 0, 0, 0 > on the three faces of the "box" that lie in the coordinate planes because either x, y, or z are zero there. A zero vector on those surfaces means zero flux through those surfaces.

    Since the z-component of V is always zero, it is perpendicular to the k vector, so its "dot product" with the normal to the z=c face is also zero. Thus, there are only two faces of the box which require surface integration.
     
    Last edited: Sep 17, 2011
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