This problem and the next one illustrate how some integrals can be difficult to evaluate numerically. Watch for how a function that changes dramatically over a small domain causes the difficulty.

The flow rate $Q$ in a cylindrical pipe can be determined by integrating the velocity at each position over the cross sectional area of the pipe. It is convenient to use cylindrical coordinates, so the velocity is a function of the radial position $r,$ with $r=0$ in the center of the pipe. This leads to

$Q={\displaystyle \underset{0}{\overset{r_0}{}}}v(2r)dr$

When the flow is smooth $("$laminar$"),$ the flow rate can be computed analytically. You will probably do this in CHE $348.$ When the flow is turbulent, one approximation for the velocity profile is

$v(r)=10(1-\frac{r}{r_0}{)}^{\frac{1}{n}}$

with $n=7.$ The maximum velocity here is $10c\frac{m}{s},$ at the center of the pipe. The minimum velocity is zero at the inner radius of the pipe, where $r_0=0\mathrm{.}75cm.$

Gauss$-$Legendre integration provides an easy way to compute an integral that spans from $-1$ to $1,$

${\displaystyle \underset{-1}{\overset{1}{}}}f(x)dx$~~${\displaystyle \underset{i=0}{\overset{n-1}{}}}{c}_{i}f(x_i)$

Integrals over other domains must be converted to spanning from $-1$ to $1.$

$($a$)$ Determine a substitution $R=??$ such that $R=-1$ at $r=0$ and $R=1$ at $r=r_0=\frac{3}{4}cm.$

$($b$)$ Evaluate the flow rate by using Gauss$-$Legendre quadrature using $n=2$ points

$($c$)$ Evaluate the flow rate by using Gauss$-$Legendre quadrature using $n=3$ points