If the net force on a current loop in a magnetic field is zero, why is torque independent of choice of origin? Im trying to show that the integral over a closed loop of a crossproduct stays the same if I choose a different origin with $\overrightarrow{r}=\overrightarrow{r}\prime+\overrightarrow{r_0}$ and $\oint{d\overrightarrow{F}}=0$
$$ \overrightarrow{N} = \oint{\overrightarrow{r} \times d\overrightarrow{F}} = \oint{(\overrightarrow{r}\prime+\overrightarrow{r_0} )\times d\overrightarrow{F}} = \oint{\overrightarrow{r}\prime \times d\overrightarrow{F}} + \oint{\overrightarrow{r_0} \times d\overrightarrow{F}}$$
How do I show $\oint{\overrightarrow{r_0} \times d\overrightarrow{F}}= 0 $, or why can I say $\oint{\overrightarrow{r_0} \times d\overrightarrow{F}} = \overrightarrow{r_0} \times\oint{ d\overrightarrow{F}}$?
To clarify, $\overrightarrow{dF}$ is the force on the current $I \overrightarrow{dl}$ in a loop in a uniform and constant magnetic field. So $\overrightarrow{dF} = I \overrightarrow{dl} \times \overrightarrow{B}$.
 A: You are on the right way with your line of thought. The only part of the puzzle you are missing is, as you have correctly identified, the explanation of $$\oint{\overrightarrow{r_0} \times d\overrightarrow{F}} = \overrightarrow{r_0} \times\oint{ d\overrightarrow{F}}$$ It is actually very simple: $\overrightarrow{r_0}$ is a constant vector (it doesn't depend on coordinates) and as such you can factor it out of the integral. Rewriting the vectors in their component elements does nothing for you.
I am sure you know of the fact that you can factor constants out of integrals, but just for completeness, here is a link to the wikipedia entry for this rule called the Constant factor rule in integration.
A: You can extract the constant term out of integral directly, however keep the operation unchanged. See:
$$\oint \vec{r}_0\times \mathrm{d}\vec{F}=\oint\epsilon_{ijk}r_{0i}\mathrm{d}F_j\hat{e}_k=\epsilon_{ijk}r_{0j}\oint\mathrm{d} F_j \hat{e}_k=\vec{r}_0\times\oint\mathrm{d}F$$
Similarly:
$$\int \vec{B_0}\cdot \mathrm{d}\vec{S}= \vec{B_0}\cdot \int\mathrm{d}\vec{S}$$
