Mathematical Tools for Physics - Department of Physics - University

9—Vector Calculus 1 215

y

x

φ

ˆnk b

Now to implement the calculation of the flow rate:

Divide the area intoNpieces of length∆`kalong the slant.

The length in and out isaso the piece of area is∆Ak=a∆`k.

The unit normal isˆnk=ˆxcosφ−ˆysinφ. (It happens to be independent of the indexk, but that’s

special to this example.)

The velocity vector at the position of this area is~v=v 0 xyˆ k/b.

Put these together and you have the piece of the flow rate contributed by this area.

∆flowk=~v.∆A~k=v 0

yk

b

ˆx.a∆`k

(

xˆcosφ−yˆsinφ

)

=v 0

yk

b

a∆`kcosφ=v 0

`kcosφ

b

a∆`kcosφ

In the last line I put all the variables in terms of, usingy=cosφ.

Now sum over all these areas and take a limit.

lim

∆`k→ 0

∑N

k=1

v 0

`kcosφ

b

a∆`kcosφ=

∫b/cosφ

0

d`v 0

a

b

`cos^2 φ=v 0

a

b

`^2

2

cos^2 φ

∣∣

∣∣

b/cosφ

0

=v 0

a

2 b

(

b

cosφ

) 2

cos^2 φ=v 0

ab

2

This turns out to be independent of the orientation of the plane; the parameterφis absent from the

result. If you think of two planes, at anglesφ 1 andφ 2 , what flows into one flows out of the other.

Nothing is lost in between.

Another Flow Calculation
Take the same sort of fluid flow in a pipe, but make it a little more complicated. Instead of a flat
surface, make it a cylinder. The axis of the cylinder is in and out in the picture and its radius is half

the width of the pipe. Describe the coordinates on the surface by the angleθas measured from the

midline. That means that−π/ 2 < θ < π/ 2. Divide the surface into pieces that are rectangular strips

of lengtha(in and out in the picture) and widthb∆θk/ 2. (The radius of the cylinder isb/ 2 .)

∆Ak=a

b

2

∆θk, and nˆk=xˆcosθk+ˆysinθk (9.4)

y

x

θk+1

ˆnk

b

θk

ˆy

xˆ

y

b/ 2

(b/2) sinθ

The velocity field is the same as before,~v(x,y,z) =v 0 xy/bˆ , so the contribution to the flow rate