Torque due to continuous force distribution / pressure In my fluid mechanics course, I was exposed to some cases where I need to calculate the torque due to the pressure and all solutions manuals or online tutorials take it as a known fact that $d\tau=rdF$ ($r$ is perpendicular to $F$). If I apply the normal differential rules to it, it should be 
$$d\tau = r dF + F  \mathrm{d}r.$$
I'm trying to understand why the second term is cancelled out. 
The best answer I thought of was that because the force is due to a pressure so at each point it would have a value of $p\cdot \mathrm{d}\mathbf{A}$ (with $p$ the pressure and $\mathbf{A}$ the area) which is an infinitely small force and thus F is 0. Is that correct? 
What if instead the force was defined as a function of r (continuous force distribution), so it does have a value at each point; for example $F = kr^2$?
Here is an example of such a case:

 A: I really don't like the way your book explained this, and it is definitely not the way I would have done it.  My problem is with the equation $dF=\tau_wdA=\tau_w(2\pi r)dr$.  Force is supposed to be a vector quantity (with a single specified direction), and yet the shear stress $\tau_w$ varies with direction around the circumference, so their expression for dF can't represent a force. Here is how I would have done it, which I hope makes more sense:
The differential force acting on a differential area of the disk $dA=rd\theta dr$ is $$\mathbf{dF}=\tau_w rd\theta dr \mathbf{i_{\theta}}$$where $\mathbf{i_{\theta}}$ is the unit vector in the tangential ($\theta$) direction.  The differential torque acting on this area of the disk is obtained by taking the cross product of the radius vector ($\mathbf{r}=r\mathbf{i_r}$) with the differential force such that $$d\mathbf{T}=\mathbf{r}x\mathbf{dF}=\tau_wr^2d\theta dr(\mathbf{i_r}x\mathbf{i_{\theta}})=\tau_wr^2d\theta dr\mathbf{i_z}$$where $\mathbf{i_r}$ is the unit vector in the radial direction and $\mathbf{i_z}$ is the unit vector in the axial direction.  The total torque on the disk is obtained by integrating this between $\theta = 0$ and $\theta = 2\pi$ and between the inner and outer radii (also taking into account the fact that $\tau_w$ is proportional to r, given by $\tau_w=\frac{r\Delta \omega}{h}$).
