Torque of a distributed load: Integrating with dF vs using resultant force

Question

We have a rod that rotates around an fixed point A (where we fix the cartesian coordinate system origin), which coincides with the end of the rod. The mass distribution along the rod is uniform. We know that the torque generated by the force field at point A is:

$\int_c \vec{r} \times \vec{dF}$

I know that from reading different texts, at least when dF is constant throughout the rod and the mass distribution is homogenous, that:

$\int_c \vec{r} \times \vec{dF}$ = $\int_c \vec{r} \times \int_c\vec{dF}$ = $(\int_c \vec{r}) \times \vec{R}$ = $\vec{G} \times \vec{R}$

Vector $\vec{r}$ is the generic position vector of points throughout the rod, $\vec{dF}$ is the force exerted by the force field at each point of the rod, C is the rod that we integrate over, $\vec{R}$ is the sum of every $\vec{dF}$ across the rod (resultant force), and $\vec{G}$ is the position vector of the center of mass.

My questions are as follows:

1. Is $\int_c \vec{r} \times \vec{dF}$ = $\vec{G} \times \vec{R}$ under all circumstances, and if not, what are the conditions that allow this identity to be true?
2. Could you derive the above identity? I do not understand when $\int_c \vec{r} \times \vec{dF}$ = $\int_c \vec{r} \times \int_c\vec{dF}$ is allowed (I believe this is not always true, when is it?). I also do not understand how $\int_c \vec{r}$ = $\vec{G}$.

Any help will be appreciated.

Answer 1

The derivation above does not make sense.

The answer to 1. is only true when the force is constant and the density is constant. Accidentally, it might be true in other cases. Here is the correct derivation:

If the force is constant, and the total force is $\vec{R}$, then the force on each piece of size $|\vec{dr}|$ is $$\vec{dF}=\frac{\vec R|\vec{dr}|}{\int_c |\vec{dr}|}$$ The torque is then $$\int_c\vec r \times\vec{dF}=\frac{\int_c \vec r|\vec{dr}|}{\int_c |\vec{dr}|}\times \vec R $$ If the density is constant, then we can write it as $m/\int_c |\vec{dr}|$, where $m$ is the mass of the rod. The mass of any piece of the rod is $$dm=\frac{m|\vec{dr}|}{\int_c |\vec{dr}|}$$ The position of the center of the mass is then $$\vec G=\frac{\int_c \vec r dm}{m}$$