# How to choose the limits of integration to find the velocity field of a circular pipe from the Laplace operator?

I have a cylindrical pipe where the pressure gradient($$G$$), the kinematic viscosity ($$\nu$$), and density of liquid ($$\rho$$) are constants. The no-slip boundary condition may be taken at the wall of the pipe, so $$u(R) = 0$$. The velocity $$u(r)$$ satisfies the Navier-Stokes equation:

$$\nabla^2 u = \frac{1}{r}\frac{d}{dr}(r\frac{du}{dr}) = -\frac{G}{\nu \rho}$$

I now know that I should multiply by r and do a first integration from 0 to r.

$$\int d(r\frac{du}{dr}) = r(\frac{du}{dr}) = -\int_{0}^{r} r\frac{G}{\nu \rho} dr = -\frac{r^2}{2}\frac{G}{\nu \rho}$$

Next, I divide by r and integrate from r to R.

$$\int du = u(R) - u(r) = -\int_{r}^{R}\frac{r}{2}\frac{G}{\nu \rho} dr = -\frac{G}{4 \nu \rho}(R^2-r^2)$$

Applying the no-slip boundary condition.

$$u(r) = \frac{G}{4\nu\rho}(R^2-r^2) = \frac{GR^2}{4\nu\rho}\left (1-\left (\frac{r}{R}\right)^2 \right).$$

Question: How does one relate the physical dimensions of the pipe to the Laplace equation in $$r$$? How does one know to use $$0$$ to $$r$$ for the first integration and then $$r$$ to $$R$$ for the second integration?