I need help. Consider $\int_S \nabla v(x) \cdot \nabla v(x)\;dx$. Here $x = (x_1, ..., x_n)$.
Use the substitution $x = \Phi(y)$, where $\Phi:T \to S$ is injective and $C^1$ and $y = (y_1, ..., y_n)$. So the integral becomes $\int_T \nabla v(\Phi(y)) \cdot \nabla v(\Phi(y)) |\det D\Phi|\;dy\tag{1}$ where $D\Phi$ is the matrix representing the derivative. How can I get this to the following form: $\int_T \nabla v(D\Phi)^{-1}(D\Phi)^{-T}\nabla v|\det D\Phi|\;dy$? I don't know how to get the inverse matrix there nor the transpose.. obviously I should apply the chain rule to the grad terms in (1) but not sure how. Thanks.