Flux
( \newcommand{\kernel}{\mathrm{null}\,}\)
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=F⋅nΔS
Adding up all these together and taking a limit, we get
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(F⋅n)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)ˆi−gy(x,y)ˆj+ˆk(oriented upward)
or
ndS=gx(x,y)ˆi+gy(x,y)ˆj−ˆk(oriented downward)
Find the flux of F=yzˆj+z2ˆk outward through the surface S cut from the cylinder y2+z2=1,z≥0, 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+2zˆk√4y2+4z2=2yˆi+2zˆk2√1=yˆj+zˆk
Next, we calculate the value for dσ=∇g|∇g⋅p|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|∇g⋅k|dA=22zdA=1zdA
Note: we dropped the absolute value bars in the last step there because the problems specifies z≥0 on S.
Now find the value of F⋅n.
F⋅n=(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:
∬SF⋅ndσ=∬S(z)(1zdA)=∬RxydA=area(Rxy)=2.
Find the flux of F=yˆj−zˆk through the paraboloid S=y=x2+z2,y≤1.
Solution
The flux can be described by ∬SF⋅ndσ with n=2xˆi−ˆj+2zˆk√1+4x2+4z2.
Take the dot product of F and n
(0ˆi+yˆj−zˆk)⋅2xˆi−ˆj+2zˆk√1+4x2+4z2=1√1+4x2+4z2(−y+2z2)
Substitute x2+z2=y to simplify n to −1+2z2y.
Thus, the integral for the flux is
∫1−1∫1−1(−1+2z2y)dzdy=∫1−1−z+2z33y|1−1dy=∫1−1(−1+23y)−(1+−23y)dy=∫1−1−2dy=0
The total flux through the surface is 0.
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
0≤x≤3 and 2≤y≤5
Solution
We compute
NdS=−ˆi−2yˆj+ˆkdydx
and
F⋅NdS=−x−4y2+x+y2=−3y2
The flux integral is
∫30∫52−3y2dydx=−351