Calculus 2nd Edition Rogawski

308 CHAPTER 6 APPLICATIONS OF THE INTEGRAL
Why is this formula true? Let’s fix an observation point P in the tube and ask: Which fluid
particles flow past P in a 1-min interval? A particle travels v centimeters each minute, so
it flows past P during this minute if it is located not more than v centimeters to the left of
P (assuming the fluid flows from left to right). Therefore, the column of fluid flowing past
P in a 1-min interval is a cylinder of radius R, length v, and volume πR 2 v (Figure 9).
v cm
FIGURE 9 The column of fluid flowing past
P in one unit of time is a cylinder of
volume πR 2 v.
P
R
In reality, the fluid particles do not all travel at the same velocity because of friction.
However, for a slowly moving fluid, the flow is laminar, by which we mean that the
velocity v(r) depends only on the distance r from the center of the tube. The particles at
the center of the tube travel most quickly, and the velocity tapers off to zero near the walls
of the tube (Figure 10).
FIGURE 10 Laminar flow: Velocity of fluid
increases toward the center of the tube.
r
P
R
v(r i )
Ring of
width r,
Area ≈ 2πr i r
If the flow is laminar, we can express the flow rate Q as an integral. We divide
the circular cross-section of the tube into N thin concentric rings of width r = R/N
(Figure 11). The area of the ith ring is approximately 2πr i r and the fluid particles
flowing past this ring have velocity that is nearly constant with value v(r i ). Therefore, we
can approximate the flow rate Q i through the ith ring by
FIGURE 11 In a laminar flow, the fluid
particles passing through a thin ring at
distance r i from the center all travel at
nearly the same velocity v(r i ).
r i
Q i ≈ cross-sectional area × velocity ≈ (2πr i r)v(r i )
We obtain
N∑
N∑
Q = Q i ≈ 2π r i v(r i )r
i=1
i=1
The sum on the right is a right-endpoint approximation to the integral 2π
Once again, we let N tend to ∞ to obtain the formula
∫ R
0
rv(r) dr.
Flow rate Q = 2π
∫ R
0
rv(r) dr 5
Note the similarity of this formula and its derivation to that of population with a radial
density function.
The French physician Jean Poiseuille
(1799–1869) discovered the law of
laminar flow that cardiologists use to study
blood flow in humans. Poiseuille’s Law
highlights the danger of cholesterol buildup
in blood vessels: The flow rate through a
blood vessel of radius R is proportional to
R 4 , so if R is reduced by one-half, the flow
is reduced by a factor of 16.
EXAMPLE 6 Laminar Flow According to Poiseuille’s Law, the velocity of blood
flowing in a blood vessel of radius R cm is v(r) = k(R 2 − r 2 ), where r is the distance
from the center of the vessel (in centimeters) and k is a constant. Calculate the flow rate
Q as function of R, assuming that k = 0.5 (cm-s) −1 .
Solution By Eq. (5),
Q = 2π
∫ R
0
(0.5)r(R 2 − r 2 )dr = π
) ∣ (R 2 r2
2 − r4 ∣∣∣
R
= π 4 0 4 R4 cm 3 /s
Note that Q is proportional to R 4 (this is true for any value of k).