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Flux

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Recall that a unit normal vector to a surface can be given by

n=ru×rv|ru×rv|

There is another choice for the normal vector to the surface, namely the vector in the opposite direction, n.

By this point, you may have noticed the similarity between the formulas for the unit normal vector and the surface integral. This idea leads us to the definition of the Flux Integral. Consider a fluid flowing through a surface S. The Flux of the fluid across S measures the amount of fluid passing through the surface per unit time. If the fluid flow is represented by the vector field F, then for a small piece with area ΔS of the surface the flux will equal to

ΔFlux=FnΔS

Adding up all these together and taking a limit, we get

Definition: Flux Integral

Let F be a differentiable vector field on a surface S oriented by a unit normal vector n. The flux integral of F across n is given by

S(Fn)dσ.

Note that the method for choosing the value for dσ for flux is identical to doing it for the integrals described above. Also notice that the denominator of n and the formula for dS both involve |ru×rv|. Canceling, we get

ndS=ru×rvdvdu

for a surface that is defined by the function z=g(x,y), we get the nice formula

ndS=gx(x,y)ˆigy(x,y)ˆj+ˆk(oriented upward)

or

ndS=gx(x,y)ˆi+gy(x,y)ˆjˆk(oriented downward)

Example 1

Find the flux of F=yzˆj+z2ˆk outward through the surface S cut from the cylinder y2+z2=1,z0, by the planes x=0 and x=1.

Solution

First we calculate the outward normal field on S. This can be calulated by finding the gradient of g(x,y,z)=y2+z2 and dividing by its magnitude.

n=g|g|=2yˆj+2zˆk4y2+4z2=2yˆi+2zˆk21=yˆj+zˆk

Next, we calculate the value for dσ=g|gp|dA. Note the similarity to the value of n. All we need to do is find the value of p. Because our cylinder sits with its shadow region in the xy plane, the vector normal to that region is in the k direction. Thus,

dσ=g|gk|dA=22zdA=1zdA

Note: we dropped the absolute value bars in the last step there because the problems specifies z0 on S.

Now find the value of Fn.

Fn=(yzˆj+z2ˆk)(yˆj+zˆk)=y2z+z3=z(y2+z2)=zbecause the surfaceis defined as y2+z2=1

Once all the preliminary work is done, plug them into the integral:

SFndσ=S(z)(1zdA)=RxydA=area(Rxy)=2.

Example 2

Find the flux of F=yˆjzˆk through the paraboloid S=y=x2+z2,y1.

151166977515379.png

Solution

The flux can be described by SFndσ with n=2xˆiˆj+2zˆk1+4x2+4z2.

Take the dot product of F and n

(0ˆi+yˆjzˆk)2xˆiˆj+2zˆk1+4x2+4z2=11+4x2+4z2(y+2z2)

Substitute x2+z2=y to simplify n to 1+2z2y.

Thus, the integral for the flux is

1111(1+2z2y)dzdy=11z+2z33y|11dy=11(1+23y)(1+23y)dy=112dy=0

The total flux through the surface is 0.

Example 3

Find the flux of

F(x,y,z)=xˆi+2yˆj+zˆk

across the part of the surface

z=x+y2

with upward pointing normal that lies within the box

0x3 and 2y5

Solution

We compute

NdS=ˆi2yˆj+ˆkdydx

and

FNdS=x4y2+x+y2=3y2

The flux integral is

30523y2dydx=351


Flux is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Michael Rea & Larry Green.

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