# Problems with divergence

by Summer2442
Tags: divergence
 P: 8 Hello, I am new to calculus, and am having problems with divergence, I was reading something to explain the physical interpretation of divergence and i got stuck in the very first part. it says that if we have a small volume dxdydz at the origin, and that a fluid flowing into this volume from the positive x-direction per unit time, the the rate of flow in is = ρvx|x=0 = dy dz, where ρ is the density at (x, y, z), and vx is the velocity of the fluid in the x-direction, what does "|x=0 = dy dz" part mean. Thanks Alot.
 P: 345 "|x=0" probably means that the function, here ρvx, should be evaluauted at x=0. So they imagine the volume element as a cuboid with one vertex at the origin and the sides as dx, dy, and dz in the posotive directions. One face of the cuboid is then contained in the plane x=0. The flux into the volume through this face is then ρvxdydz, so something seems wrong in what you wrote anyway. Btw, you wrote "from the positive x-direction", but I assume that you meant "along the positive x-direction", so that positive vx is directed to the right.
 P: 8 yes what i meant is "along the positive x-direction" but about the flux into the volume, i agree that it should be 'ρvxdydz' but i am pretty sure this is what the book says 'ρvx|x=0 = dy dz', i think its a mistake. I have another question please, it then says that the flux out of the opposing face is ρvx|x=dx dydz which i understand but then it equates this equation with the following [ρvx + ∂ (ρvx)/∂x dx] dydz which i do not understand, how was the partial derivative introduced and why? thanks,
P: 345
Problems with divergence

 Quote by Summer2442 yes what i meant is "along the positive x-direction" but about the flux into the volume, i agree that it should be 'ρvxdydz' but i am pretty sure this is what the book says 'ρvx|x=0 = dy dz', i think its a mistake.
Yes, it is a typo. It should be ρvx|x=0dxdy.

 Quote by Summer2442 I have another question please, it then says that the flux out of the opposing face is ρvx|x=dx dydz which i understand but then it equates this equation with the following [ρvx + ∂ (ρvx)/∂x dx] dydz which i do not understand, how was the partial derivative introduced and why? thanks,
The second expression is a linear approximation of the first one.

Recall that (ρvx|x=dx - ρvx|x=0)/dx ≈ ∂(ρvx)/∂x|x=0.
 P: 41 i think it is talking about mass flow rate... density X velocity X area=flow rate..or more precisely mass flow rate..it says rho X V X x such that at x=0 area is dy dz..for differential element the area perpendicular to flow in x direction is dydz... the second thing is taylor's theorem is being applied here..which means when fluid has flowed a length dx its mass flowrate has changed depnding upon dx..that is why patial derivative is introduced here..
 P: 8 Ok I get it now, thanks guys.

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