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Homework Help: Another flux problem

  1. Apr 28, 2008 #1
    1. The problem statement, all variables and given/known data

    Calculate the outward flux of the vector field F = <2x, -3y, 3z> across the boundary of the first-octant unit cube with opposite vertices (0, 0, 0) and (1, 1, 1).

    2. The attempt at a solution

    The top and the bottom of the box is easy. The flux here is 3 and 0 respectively. But how do I calculate the other sides?
     
    Last edited: Apr 28, 2008
  2. jcsd
  3. Apr 28, 2008 #2

    HallsofIvy

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    I hate to be the one to break it to you but the flux over the bottom is NOT -3. The unit, outward, normal to the bottom, z= 0, is <0, 0, -1> so you are integrating [itex]<2x,-3y,2z>\cdot<0, 0, ->= <2x, -3y, 0>\cdot<0, 0, -1>= 0[/itex], since z= 0 there, over the xy-plane, 0< x< 1, 0< y, 1. That flux is 0. It is correct that the flux over the top, z= 1 is
    [tex]\int_{x=0}^1\int_{y= 0}^1 <x, y, 3>\cdot<0, 0, 1> dxdy= 3[/itex].

    The "left" face is x= 0 and an outward unit normal is <-1, 0, 0>. On that face, x= 0 so the integrand is [itex]<0, -3y, 3z>\cdot<-1, 0, 0>= 0[/itex] so the flux there is also 0.
    The "right" face is x= 1 and an outward unit normal is <1, 0, 0>. On that face, x= 1 so the integrand is [itex]<2, -3y, 3z>\cdot<1, 0, 0>[/itex] integrate that for 0<y<1, 0< z< 1.

    The "front" face is y= 0 and an outward unit normal is <0, -1, 0>. On that face, y= 0 so the integrand is [itex]<2x, 0, 3z>\cdot<0, -1, 0>[/itex].
    The "back" face is y= 1 and an outward unit normal is <0, 1, 0>. On that face, y= 1 so the integrand is [itex]<2x, -3, 3z>\cdot<0, 1, 0>[/itex].

    Of course, it would be easy to very easy to use the divergence theorem to get the answer.
     
  4. Apr 28, 2008 #3
    Yes, the divergenze theorem is part of the next chapter, but it would definitely be easier. How about this one:

    Calculate the outward flux of the vector field F = <0, 0, z^2> out of the boundary of te solid bounded by the paraboloids z= z^2 and z = 18 - z^2 - y^2

    I change to polar coordinates: z = 18 - r^2 and z = r^2. These functions intersect in the circle r=3.

    Then I use the divergence theorem, integrating 2z dz dr d(tetha) within the limits [r^2, 18-r^2], [0, 3] og [0, 2*pi] respectively. The answer I get is 1296*pi. The correct answer is 1458*pi. Where's my mistake?
     
  5. Apr 28, 2008 #4

    HallsofIvy

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    The differential of volume in cylindrical coordinates is [itex]r dzdrd\theta[/itex]. You forgot the "r".
     
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