Verifying the flux transport theorem

  • #1
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Homework Statement:

Let ##S_t## be a uniformly expanding hemisphere described by ##x^2+y^2+z^2=(vt)^2, (z\ge0)##
Verify the flux transport theorem in this case

Relevant Equations:

flux transport theorem: $$\frac{d\phi}{dt} =\int\int_{S_t}\left(\frac{\partial \textbf{F}}{\partial t} + (\nabla \cdot\textbf{F})\textbf{v} + \nabla \times(\textbf{F}\times\textbf{v})\right)\cdot d\textbf{S}$$ where ##\textbf{F}(\textbf{R},t)=\textbf{R}t=(xt,yt,zt)## (I think?)
Let ##S_t## be a uniformly expanding hemisphere described by ##x^2+y^2+z^2=(vt)^2, (z\ge0)##

I assume by verify they just want me to calculate this for the surface. I guess that ##\textbf{v}=(x/t,y/t,z/t)## because ##v=\frac{\sqrt{x^2+y^2+z^2}}{t}##. The three terms in the parentheses evaluate quite nicely. We end up getting $$2\int\int_{S_t}(x,y,z)\cdot d\textbf{S}$$ Here is where I don't understand. So I am thinking to change to spherical coordinates, but I do not know what the ##d\textbf{S}## vector looks like in spherical coordinates. I also don't know what to integrate over. Don't we have 3 bounds that are changing? ##\theta, \phi## and ##r##? And what would the upper bound of r even look like? Would it be ##rt##?
 

Answers and Replies

  • #3
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You can actually figure the surface element intuitively. Imagine ##\theta## changing by ##d\theta##, that gives you ##dl_1=rd\theta##, then imagine ##\varphi## changing by ##d\varphi##, that gives you ##dl_2=(r\sin\theta)d\varphi##. We then have ##|d\vec S|=dl_1\times dl_2##.
Thinking infinitesimally, the lengths are too small to be considered arcs, so we think of them as straight lines.
Kugelkoord-lokale-Basis-s.svg.png
 
  • #4
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I should have clarified. I know ##dS= r^2sin\theta d\theta d\phi##. I want to be able to "interpret" it as ##d\textbf S=(dr e_r,rd\theta e_\theta,r\sin\theta d\phi e_\phi)##, or some equivalent form (I know that is not how you interpret it as it is not a Cartesian vector), so that I can evaluate ##(x,y,z)\cdot d\textbf{S} ##
 
  • #5
607
185
I should have clarified. I know ##dS= r^2sin\theta d\theta d\phi##. I want to be able to "interpret" it as ##d\textbf S=(dr e_r,rd\theta e_\theta,r\sin\theta d\phi e_\phi)##, or some equivalent form (I know that is not how you interpret it as it is not a Cartesian vector), so that I can evaluate ##(x,y,z)\cdot d\textbf{S} ##
If you find a way to parametrize your surface with ##\vec r(u,\,v)##, then you'll have:
$$d\vec S=\left(\frac{\partial\vec r}{\partial u}\times\frac{\partial\vec r}{\partial v}\right)du\,dv$$
https://en.wikipedia.org/wiki/Surface_integral#Surface_integrals_of_vector_fields
 
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