Why Do I Need to Sum All the Torques Along the Radius for Calculating Torque?

[EEristavi]

Homework Statement

uniform disc of radius R is spinned to the angular velocity
w and then carefully placed on a horizontal surface. How long will
the disc be rotating on the surface if the friction coefficient is equal
to k? The pressure exerted by the disc on the surface can be regarded
as uniform.

Relevant Equations

T = Ia

Only problem I have is in calculating Torque

I say:
dT = R dF = R k g dm
&
dm = $\frac m {\Pi R^2}$ R dr d$\theta$

However, in the solution I see that:
dT = r dF = r k g dm
&
dm = $\frac m {\Pi R^2}$ r dr d$\theta$I don't get it: when taking the whole T (when I integrate), why do I have to "Sum" all the Torques along the radius

 

[haruspex]

Problem Statement: uniform disc of radius R is spinned to the angular velocity
w and then carefully placed on a horizontal surface. How long will
the disc be rotating on the surface if the friction coefficient is equal
to k? The pressure exerted by the disc on the surface can be regarded
as uniform.
Relevant Equations: T = Ia

Only problem I have is in calculating Torque

I say:
dT = R dF = R k g dm
&
dm = $\frac m {\Pi R^2}$ R dr d$\theta$

However, in the solution I see that:
dT = r dF = r k g dm
&
dm = $\frac m {\Pi R^2}$ r dr d$\theta$I don't get it: when taking the whole T (when I integrate), why do I have to "Sum" all the Torques along the radius

It is not clear exactly what your variables mean. E.g. dT is the contribution to torque from what element?

Consider an element of area r dr dθ. The normal force is ρgr dr dθ, the frictional force kρgr dr dθ, and this acts tangentially. The torque it exerts about the centre is kρgr² dr dθ.

 

[EEristavi]

Found the mistake in my point of view - Thanks