Does this integral relation follow from Stokes' theorem?

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In Landau and Lifhsitz's Electrodynamics of continuous media there is an integral relation for the curl of magnetisation (just after equation 27.8) in the form

$\int\limits_{V}\mathbf{r}\times\left(\nabla\times\mathbf{M}\right)dV=-\oint\limits_{\partial V}\mathbf{r}\times(\mathbf{M}\times\mathbf{dS})-\int\limits_V(\mathbf{M}\times\nabla)\times\mathbf{r}\,dV$,

where $\mathbf{r}$ is the position vector and $\mathbf{M}$ is the magnetisation. The integral goes over the volume $V$, or its boundary $\partial V$.

I am having trouble proving this - does the relation follow from Stokes' theorem? Any suggestions would be much appreciated.

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Bring the rightmost integral to the left side; pull both under the same integral sign. Now the thing in the left integral looks like a derivative of a product, so you have $$ \int_V d(thing) = \int_{\partial V} thing $$ which looks like Stokes' to me.

Of course, this requires that $d(r \times (M \times dS))$ is indeed the thing under the (newly combined) left integral, and I don't know what magnetization is, nor any of its properties, nor do I know how to do integrals with $dS$ and $dV$ in them reliably -- I really only "get" the differential-forms version of Stokes' Theorem -- but this seems to be pretty clearly the way to go.