I have a question on the handling of gradients in coordinate transformations:
For a coordinate transformation in form of a rotation defined by the mapping $\Phi: \Omega \mapsto \Omega'$, $\psi \mapsto R\psi = x$ with an orthogonal matrix R, the following integral transformation can be made:
$\int_{\Omega'} u(x) dx = \int_\Omega u(R\psi) d\psi$,
as |det R| = 1.
Now I am wondering how this integral transformation looks if there is the gradient $\nabla_x u(x)$ on the left, as the derivatives also depend on the coordinate system (therefore the subscript x). I know that due to the chain rule $\nabla_\psi(u(R\psi)) = \nabla_\psi u(R\psi) R$, so my question is if I can just write:
$\int_{\Omega'} \nabla_x u(x) dx = \int_\Omega R \nabla_\psi u(R\psi) d\psi$