If the velocity distribution of a fluid flowing through a pipe is known,
the flow rate Q can be calculated by $Q=\int vdA$, where v is the velocity and A is the area
pipeline cross section. Consider a circular pipe of $A=\pi r^2$ and $dA=2\pi r$
So, we have: $$Q=\int_0^r v(2\pi r)dr$$ Where r is a radial distance measured outward from the center of the pipe. If the distribution speed is given by: $$v=2(1-\frac{r}{r_0})^{\frac{1}{6}}$$
Where $r_0$ is the total radius (in this case 3 cm), calculate Q.
my attempt:
With the information I have, I wrote the integral as follows $$Q= \int_0^3 2(1-\frac{r}{3})^{\frac{1}{6}}2\pi r dr$$ Since the pipe is circular I wrote R = 3 cm and $r_0$ = 3cm. But I'm not sure if this is the correct way to write the integral, I don't know if I have to change the 3 cm to meters or if it doesn't matter. Any help that clarifies if I wrote the integral correctly, I will greatly appreciate it.