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Divergence physics homework

  1. Sep 24, 2013 #1
    1. The problem statement, all variables and given/known data
    In deriving the formula
    div v = [itex]\frac{∂v_{x}}{∂x}[/itex] + [itex]\frac{∂v_{y}}{∂y}[/itex] + [itex]\frac{∂v_{z}}{∂z}[/itex]
    we used a rectangular solid infinitesimal volume; however, any shape will do (although the calculation gets harder). To see an example, derive the same formula using the prism-shaped volume shown (figure attached). This could be a volume element next to an irregular surface.


    2. Relevant equations
    Definition of divergence

    div v = lim Δτ→0 [itex]\frac{\int v \cdot da}{Δτ}[/itex] where Δτ = small volume and the integral is a over a closed surface bound Δτ


    3. The attempt at a solution
    So I am having some issue with this question. A thing to note is that we are approximating the integral as just v[itex]\circ[/itex]da where v is at some point on the surface. This is due to it being an infinitesimal volume.

    The main problem I am having is I think I am making a mistake on the top face somehow. If you look at figure 2 which is my crude drawing to represent n it is then

    n = (cos[itex]\theta[/itex] jhat + sin[itex]\theta[/itex] khat)

    So then for the top face if h = the slanted side
    da = hΔx(cos[itex]\theta[/itex] jhat + sin[itex]\theta[/itex] khat)
    =Δx(Δz jhat + Δy khat) because h can be expressed in terms of sin and cos of the other sides.

    so then v dot da = v[itex]_{y}[/itex](x, y, z)ΔxΔz + v[itex]_{z}[/itex](x,y,z)ΔxΔy

    hopefully so far so good. Now for the other faces

    bottom

    da = ΔxΔy (-khat direction)

    v dot da = -v[itex]_{z}[/itex](x, y, z-[itex]\frac{1}{2}[/itex]Δz)ΔxΔy

    back(zx plane)

    da = ΔxΔz

    v dot da = -v[itex]_{y}[/itex](x, y-[itex]\frac{1}{2}[/itex]Δy, z)ΔxΔz

    so then when I put it all together and divide by Δτ = [itex]\frac{1}{2}[/itex]ΔxΔyΔz

    it almost looks like the definition of a derivative except I am left with a 1/2. What did I miss? Any help would be greatly appreciated and if you need me to clarify let me know.
     

    Attached Files:

  2. jcsd
  3. Sep 24, 2013 #2
    Hey guys, I would really appreciate some help so if there is something I was unclear about let me know and I will try and clarify further thanks.
     
  4. Sep 24, 2013 #3

    TSny

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    Homework Helper
    Gold Member

    Note ##f(x+a) \approx f(x) + f'(x)\:a## for small ##a##.

    Likewise ##f(x-a) \approx f(x) - f'(x)\:a##

    Thus, how would you approximate ##V_y(x, y-\frac{\Delta y}{2}, z)##?
     
  5. Sep 24, 2013 #4
    So

    =[itex]\frac{[V_{y}(x,y,z)-V_{y}(x, y-\frac{1}{2}Δy,z)]}{\frac{1}{2}Δy}[/itex]
    =V[itex]_{y}[/itex](x,y,z) - V[itex]_{y}[/itex](x,y,z) + [itex]\frac{∂v_{y}}{∂y}[/itex]([itex]\frac{1}{2}[/itex]Δy) / ([itex]\frac{1}{2}[/itex]Δy)
    =[itex]\frac{∂v_{y}}{∂y}[/itex]

    correct?

    Sorry for the poor formatting I am still getting used to this latex stuff
     
  6. Sep 24, 2013 #5

    TSny

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    Homework Helper
    Gold Member

    Yes, that's correct.
     
  7. Sep 24, 2013 #6
    Okay thank you again!
     
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