3 Attachment(s)

Using the definite integral

The heart pumps blood throughout the body, the arteries are the blood vessels carrying

blood away from the heart, and the veins return blood to the heart. Close to the heart, arteries

are very large. The smallest blood vessels are capillaries

Precise measurements demonstrate that the flux (rate of flow) of fluid through the capillary

wall is not constant over the length of the capillary. Fluids in and around the capillary are

subjected to two forces. The *hydrostatic pressure*, resulting from the heart’s pumping, pushes

fluid out of the capillary into the surrounding tissue. The *oncotic pressure *drives absorption

in the other direction. Along the length of

the capillary, the hydrostatic pressure decreases while the oncotic pressure is approximately

constant

Attachment 30059Attachment 30061

Along a cylindrical capillary of length L = 0.1 cm and radius r = 0.0004 cm, the hydrostatic

pressure, ph, varies from 35 mm Hg at the artery end to 15 mm Hg at the vein end. (mm

Hg, millimeters of mercury, is a unit of pressure.) The oncotic pressure, po, is approximately

23 mm Hg throughout the length of the capillary

**(a) **Find a formula for The *hydrostatic pressure* as a function of x, the distance in centimeters from the artery end of the capillary, assuming that The *hydrostatic pressure* is a linear function of x.

**(b) **Find a formula for p, the net outward pressure, as a function of x.

**(c) **The rate of movement, j, of fluid volume per capillary wall area across the capillary wall is

proportional to the net pressure.We have j = k*p where k, the hydraulic conductivity, has

value

Attachment 30060

Check that j has units of volume per time per area.

**(d) **Write and evaluate an integral for the net volume flow rate (volume per unit time) through

the wall of the entire capillary.

Re: Using the definite integral

For (a), you are given ph=35 at x=0 and ph=15 at x=0.1, so since the variation is linear, $\displaystyle \text{ph}(x)=35-200x$. The 200 is $\displaystyle \frac{35-15}{0.1}$, which is what it needs to be to get ph=15 at x=0.1.

In (b), is it correct to just subtract the oncotic pressure? If that's correct, then $\displaystyle p(x)=(35-200x)-23=12-200x$.

The units of k are the same as j/p, j is in cm/sec and p is in mm Hg, so k is cm/(sec mm Hg).

For part (d), you are integrating over the length of the capillary, x=0 to x=0.1. Call the volume flow rate Q. The volume flow rate through a small piece of the capillary with length dx is j times p(x) times the area. You get p(x) from part (b) and the area is the area of the side of a cylinder of radius r=0.0004cm and height dx, so it's $\displaystyle 2\pi{r}\,dx$. Putting it together gives $\displaystyle \int_0^{0.1}2\pi{r}jp(x)\,dx$, and since r and j are constants, $\displaystyle 2\pi{r}j\int_0^{0.1}p(x)\,dx$. I assume you can take it from there.

- Hollywood