Blood Flow and Biology
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 2.5: “Physical Applications” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/2-5-physical-applications |
| Textbook used in class | Stewart, Calculus, Section 8.4: “Applications to Economics and Biology” |
Opening Scenario
Blood does not flow uniformly through an artery. It flows fastest in the center (where there is no friction from the vessel wall) and slowest near the wall (where viscosity slows it down). The velocity at a distance $r$ from the center of a vessel of radius $R$ follows a parabolic profile: $$v(r) = \frac{P}{4\eta L}(R^2 - r^2),$$ where $P$ is the pressure difference along the vessel, $\eta$ is the blood viscosity, and $L$ is the vessel length. To find the total volume of blood passing per unit time, you cannot just multiply velocity by area -- the velocity varies with $r$, so you integrate.
Quick Reference
Poiseuille’s Law (velocity profile): $$v(r) = \frac{P}{4\eta L}(R^2 - r^2), \quad 0 \leq r \leq R.$$
Total flow rate (volume per unit time): $$F = \int_0^R v(r)\cdot 2\pi r\,dr = \frac{\pi R^4 P}{8\eta L}.$$
This is one of the most important formulas in physiology. Flow rate depends on $R^4$: doubling the vessel radius increases flow by a factor of 16.
Key Concepts
1. The Parabolic Velocity Profile
Viscous fluid in a tube (Poiseuille flow) has velocity that depends on radial position. At the center ($r = 0$), velocity is maximum: $v(0) = \dfrac{PR^2}{4\eta L}$. At the wall ($r = R$), velocity is zero: $v(R) = 0$. The profile is a downward-opening parabola in $r$.
2. Flow Rate Requires Integrating Over Annuli
To find total flow, sum contributions from thin annular rings at radius $r$. A ring of thickness $dr$ at radius $r$ has cross-sectional area $2\pi r\,dr$. The flow through that ring is velocity times area: $$dF = v(r)\cdot 2\pi r\,dr.$$
Integrate from the center to the wall: $$F = \int_0^R v(r)\cdot 2\pi r\,dr = \frac{P}{4\eta L}\cdot 2\pi\int_0^R (R^2 - r^2)r\,dr.$$
3. Evaluating the Integral
$$\int_0^R (R^2 r - r^3)\,dr = \left[\frac{R^2 r^2}{2} - \frac{r^4}{4}\right]_0^R = \frac{R^4}{2} - \frac{R^4}{4} = \frac{R^4}{4}.$$
$$F = \frac{P}{4\eta L}\cdot 2\pi\cdot\frac{R^4}{4} = \frac{\pi R^4 P}{8\eta L}.$$
4. The $R^4$ Dependence
The flow rate $F \propto R^4$ is the most physiologically significant result. A 10% reduction in artery radius reduces flow by $(0.9)^4 \approx 0.66$, a 34% decrease. This explains why even partial blockage of an artery dramatically reduces blood supply.
Worked Example
An artery has radius $R = 0.5$ cm, length $L = 20$ cm, pressure difference $P = 1000$ dyn/cm$^2$, and blood viscosity $\eta = 0.027$ g/(cm$\cdot$s). Find the flow rate $F$.
$$F = \frac{\pi R^4 P}{8\eta L} = \frac{\pi (0.5)^4 (1000)}{8(0.027)(20)}.$$
Numerator: $\pi \cdot 0.0625 \cdot 1000 = 62.5\pi \approx 196.35$.
Denominator: $8 \cdot 0.027 \cdot 20 = 4.32$.
$$F \approx \frac{196.35}{4.32} \approx 45.5 \text{ cm}^3/\text{s}.$$
Boxed answer: $F \approx 45.5$ cm$^3$/s (about 45.5 mL per second).
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Multiplying velocity by total cross-section area | $F = v_{\text{max}} \cdot \pi R^2$ | Different rings move at different speeds; integrate $\int_0^R v(r)\cdot 2\pi r\,dr$ |
| Forgetting the $2\pi r$ factor for the annular area | $F = \int_0^R v(r)\,dr$ | The area element for a ring of width $dr$ at radius $r$ is $2\pi r\,dr$, not $dr$ |
| Using $r = R$ for the maximum velocity | $v_{\text{max}} = \frac{PR^2-R^2}{4\eta L}$ | Maximum velocity is at the center $r = 0$: $v_{\max} = \frac{PR^2}{4\eta L}$ |
Common Misconceptions
total blood flow rate equals the maximum velocity at the center times the cross-sectional area of the vessel.
This is the average-rate-as-arithmetic-mean error. Velocity is not constant across the cross-section; it varies from $v_{\max} = PR^2/(4\eta L)$ at the center to $0$ at the wall. Multiplying $v_{\max}$ by $\pi R^2$ treats all fluid as moving at peak speed, which overcounts the flow. The correct total flow $F = \pi R^4 P/(8\eta L)$ is exactly half of $v_{\max}\cdot\pi R^2$, because the parabolic profile averages to $v_{\max}/2$ over the cross-section. The factor of two difference is not a detail; it is the quantitative consequence of integrating the true velocity profile.
the area element for integrating over the circular cross-section is $dr$, giving $F = \int_0^R v(r)\,dr$.
This is the concept-image-conflicts-definition error. Integrating $v(r)$ with respect to $r$ alone sums velocity values along a radius, which has no direct physical meaning. The area of an annular ring at radius $r$ with width $dr$ is $2\pi r\,dr$, not $dr$. The flow through that ring is $v(r)\cdot 2\pi r\,dr$, and the total flow is $F = \int_0^R v(r)\cdot 2\pi r\,dr$. The factor $2\pi r$ is essential because outer rings have larger circumference and contribute more cross-sectional area per unit of radial width.
Leveled Practice
Level 1 -- Verify the Formula
Problem 1. Confirm that $\displaystyle\int_0^R (R^2 r - r^3)\,dr = \dfrac{R^4}{4}$.
Show answer
$\displaystyle\int_0^R (R^2 r - r^3)\,dr = \left[\frac{R^2 r^2}{2} - \frac{r^4}{4}\right]_0^R = \frac{R^4}{2} - \frac{R^4}{4} = \frac{R^4}{4}$.
Level 2 -- Effect of Radius Change
Problem 2. If the radius of an artery is reduced from $R$ to $0.8R$ (say, by partial blockage), by what factor does the flow rate decrease?
Show answer
$F \propto R^4$, so the ratio of new flow to old flow is $(0.8R)^4 / R^4 = (0.8)^4 = 0.4096$.
Flow decreases to about $41\%$ of its original value -- a decrease of $59\%$ from a $20\%$ reduction in radius.
Mastery Checklist
Mental Model
To find the total flow in a tube, do not use average velocity. Instead, think of the tube as a stack of thin cylindrical rings. Each ring contributes a small flow: (velocity of that ring) times (area of that ring) = $v(r) \cdot 2\pi r\,dr$. Integrating from $r = 0$ to $r = R$ adds up all contributions. The inner rings move faster but have smaller area; the outer rings are slower but span more area. The integral balances these two effects. The key lesson: whenever a quantity varies across a cross-section, integrate to find the total.
Connections
Looking back
- Definite integral (Chapter 4): The flow rate integral is a standard definite integral.
- Polar areas (not in MATH162, but conceptually related): Annular area elements arise in polar coordinates.
Looking ahead
- Consumer and producer surplus (Section 8.4): Another application of integrating a rate (willingness-to-pay curve) to find a total (economic surplus).
Back to Pappus’s Theorems | Next: Consumer and Producer Surplus