# How to convert this derivation of Poiseuille's law into the standard one? I am trying to derive Po

How to convert this derivation of Poiseuille's law into the standard one?
I am trying to derive Poiseuille's law. I have reached a point in the derivation where I have:
$V=\frac{\left(p1-p2\right)\left({R}^{2}\right)}{4lu}$
Where $l$ is length, u is viscosity, p is pressure, v is flow velocity and R is radius. What I am stuck on is shifting this to the volumetric flow rate:
$V\pi {R}^{2}=Q=\frac{\left(p1-p2\right)\left({R}^{4}\right)}{4lu}.$
$V\pi {R}^{2}=Q=\frac{\left(p1-p2\right)\left({R}^{4}\right)}{4lu}.$
However this is incorrect as Poiseuille's law is divided by 8. I know that I am probably missing something obvious, but I can't think of a reason as to why the whole equation needs to be halved. Any help understanding why (or whether my original derivation for V was inaccurate) would be appreciated.
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Eliezer Olson
The expression for the flow speed as a function of distance from the pipe center is:
$V\left(r\right)=\frac{\left({p}_{1}-{p}_{2}\right)\left({r}^{2}-{R}^{2}\right)}{4l\mu }$
The average velocity of the flow can be calculated by performing an integration over the pipe cross section, and then dividing by the cross-sectional area:
${V}_{avg}=\frac{{\int }_{0}^{R}{\int }_{0}^{2\pi }\frac{\left({p}_{1}-{p}_{2}\right)\left({r}^{2}-{R}^{2}\right)}{4l\mu }rd\theta dr}{\pi {R}^{2}}$
The extra factor of $r$ comes from the fact we are doing everything in polar coordinates. Since the function is independent of $\theta$,
${V}_{avg}=\frac{{\int }_{0}^{R}{\int }_{0}^{2\pi }\frac{\left({p}_{1}-{p}_{2}\right)\left({r}^{2}-{R}^{2}\right)}{4l\mu }2\pi rdr}{\pi {R}^{2}}=\frac{\left({p}_{1}-{p}_{2}\right){R}^{2}}{8l\mu }$
The factor of $8$ comes from the combination of integrating the ${r}^{3}$ term, resulting in a $\frac{1}{4}$ term, multiplied by the $2$ from the angular integration, and multiplied again by the original $\frac{1}{4}$ in the $V\left(r\right)$ expression to yield a factor of $\frac{1}{8}$