Mathematical Tools for Physics

9—Vector Calculus 1 252

y

x

θk+1

nˆk

b

θk

ˆy

ˆx

y

b/ 2

(b/2) sinθ

The velocity field is the same as before,~v(x,y,z) =v 0 ˆxy/b, so the contribution to the flow rate through this
piece of the surface is

~vk.∆A~k=v 0

yk

b

ˆx.a

b

2

∆θknˆk

The value ofykat the angleθkis

yk=

b

2

+

b

2

sinθk, so

yk

b

=

1

2

[1 + sinθk]

Put the pieces together and you have

v 0

1

2

[

1 + sinθk

]

xˆ.a

b

2

∆θk

[

ˆxcosθk+yˆsinθk

]

=v 0

1

2

[

1 + sinθk

]

a

b

2

∆θkcosθk

The total flow is the sum of these overkand then the limit as∆θk→ 0.

lim

∆θk→ 0

∑

k

v 0

1

2

[

1 + sinθk

]

a

b

2

∆θkcosθk=

∫π/ 2

−π/ 2

v 0

1

2

[

1 + sinθ

]

a

b

2

dθcosθ

Finally you can do the two terms of the integral: Look at the second term first. You can of course start grinding
away and find the right trigonometric formula to do the integral, OR, you can sketch a graph of the integrand,
sinθcosθ, on the interval −π/ 2 < θ < π/ 2 and write the answer down by inspection. The first part of the
integral is

v 0

ab

4

∫π/ 2

−π/ 2

cosθ=v 0

ab

4

sinθ

∣

∣

∣

∣

π/ 2

−π/ 2

=v 0

ab

2