What is the Volume Flux of Fluid Through a Surface?

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Homework Statement


The velocity components of a flow are given by:
u=-x v=y
Compute the volume of fluid flowing per unit time per unit area through a small surface at (1,2) whose normal makes an angle of 60 deg with the positive x-axis.

Homework Equations



V= u i + y j (velocity vector)

dq = V dot n dS (volume efflux)

The Attempt at a Solution



So for this problem V = -1 i + 2 j
then dq/dS = Vcos(60) = -cos(60) i + 2cos(60) j

i believe it wants the magnitude of dq/dS which is where i get confused.

is the magnitude this:
magnitude dq/dS = sqrt((-.5)^2 + (1)^2)

or is it this:
magnitude dq/dS = sqrt(-(.5^2) + (1^2))
 
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You have to compute the unit normal vector and take the dot product of this nornal with the velocity vector field.

You know that the surface is at 60 degrees from the positive x-axis. can you compute a normal from there?
 
would the normal vector be...

n = cos(60) i + sin(60) j

so then...

dq/dS = V dot n = -cos(60) i + 2sin(60) j

and...

magnitude of dq/dS = sqrt( (-cos(60))^2 + (2sin(60))^2)
 
Almost, the dot product of 2 vectors is a scalar, so:
<br /> \frac{dq}{dS}=\mathbf{V}\cdot\hat{\mathbf{n}}=u\cos \Bigg(\frac{\pi}{3}\Bigg) +v\sin \Bigg(\frac{\pi}{3}\Bigg)<br />
 
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