Divergence physics homework

  • Thread starter thatguy14
  • Start date
  • #1
45
0

Homework Statement


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.


Homework 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


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.
 

Attachments

  • Figure.png
    Figure.png
    1.6 KB · Views: 342
  • figure2.png
    figure2.png
    2 KB · Views: 357

Answers and Replies

  • #2
45
0
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.
 
  • #3
TSny
Homework Helper
Gold Member
13,068
3,393
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)##?
 
  • #4
45
0
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
 
  • #5
TSny
Homework Helper
Gold Member
13,068
3,393
Yes, that's correct.
 
  • #6
45
0
Okay thank you again!
 

Related Threads on Divergence physics homework

  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
4
Views
5K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
0
Views
725
  • Last Post
Replies
5
Views
984
Top