Mathematical Tools for Physics

9—Vector Calculus 1 251

Now to implement the calculation of the flow rate:
Divide the area intoN pieces of length∆kalong the slant. The length in and out isaso the piece of area is∆Ak=a∆k.
The unit normal isˆnk=ˆxcosφ−yˆsinφ. (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 ˆxyk/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

(

ˆxcosφ−ˆysinφ

)

=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 ˆnk=xˆcosθk+ˆysinθk (4)