Change of variables for integral involving curl So I'm working on a finite element problem. Usually, I have to deal with the integral of a gradient over a triangular region $K$. I map the integral to the reference triangle $\hat{K}$ with vertices at $(0,0)$, $(1,0)$ and $(0,1)$ with $F_{K}: \hat{K} \to K$:
$F_{K}(\mathbf{\hat{x}}) = B_{K}\mathbf{\hat{x}} + \mathbf{b}_{K}$
Then we have
$$\int_{K}\nabla\phi_{i}\cdot\nabla\phi_{j} = \frac{1}{\text{det} \;B_{K}}\int_{\hat{K}}B_{K}^{-T}(\nabla\hat{\phi_{i}}\circ F_{K}^{-1})\cdot B_{K}^{-T}(\nabla\hat{\phi}_{j}\circ F_{K}^{-1})$$
where $\phi_{i}$ is the global basis function and $\hat{\phi_{i}}$ is the local basis function on the reference triangle.
Now I have an integral of the following form:
$$\int_{K}(\nabla\times\varphi_{i}) \cdot (\nabla\times\varphi_{j})$$
where $\varphi_{i}$ is a global basis field. I'm at a bit of a loss as to how to map this integral to the reference triangle $\hat{K}$...
edit: also adding that the basis field has two components only. I edited the problem statement since in the end, the integrand should be scalar
 A: For it is in a 2 dimensional triangulation, the curl can be represented using gradient:
$$
\nabla \times \psi = \begin{vmatrix}\partial_x&\partial_y\newline \psi_1&\psi_2 \end{vmatrix} = \partial_x \psi_2  - \partial_y \psi_1 = \nabla \psi_2 \cdot (1,0) + \nabla\psi_1\cdot (0,-1). $$
Say your global basis field is $$\psi_i = (a\phi_{i,1}, b\phi_{i,2}),$$
i.e., it can be represent by scalar basis $\phi_i$ component-wise. Then 
$$
\nabla \times \psi_i = b\nabla \phi_{i,2}\cdot (1,0) + a\nabla\phi_{i,1}\cdot (0,-1),
$$
and
$$
\nabla \phi_i(\mathbf{x}) = \frac{1}{|\det B_K^T|} B_K^T (\nabla\hat{\phi_{i}}\circ F_{K}^{-1})(\mathbf{x}),
$$
hence
$$
\nabla \times \psi_i(\mathbf{x}) = \frac{b}{|\det B_K^T|} B_K^T (\nabla\hat{\phi}_{i,2}\circ F_{K}^{-1})(\mathbf{x})\cdot (1,0) +  \frac{a}{|\det B_K^T|} B_K^T (\nabla\hat{\phi}_{i,2}\circ F_{K}^{-1})(\mathbf{x})\cdot (0,-1).
$$
Now write the other $\psi_j =  (c\phi_{j,1}, d\phi_{j,2})$ and do the same.
BTW: your way of writing integral looks a bit confusion, don't forget $dxdy = |\det B_K^T|d\hat{x}d\hat{y} $.
