This is an extension for Linear Piola Transform, where we discuss the cases where the transformation from the reference element to the physical element is nonlinear such as a parametric element. The main tool here is actually from the exterior calculus framework, and will just be from the FEEC book by Arnold.

Let $\Omega, \hat \Omega$ by the physical element and reference element with an orientation-preserving, differentiable map $\phi: \hat \Omega \to \Omega$. Suppose $p: \Omega \to \mathbb{R}, v: \Omega \to \mathbb{R}^3$ are the functions on the physical element. The quantity $(p, \nabla \cdot v) = \int_\Omega p \nabla \cdot v \, dx$ is of importance.

The key observation here is that by letting $\nu = v^1 dx^2 \wedge dx^3 – v^2 dx^1 \wedge dx^3 + v^3 dx^1 \wedge dx^2$, we have $d\nu = v$ where $d$ is the exterior derivative. A key fact here is that with integral of differential forms is it preserves pullbacks (see second answer) hence \begin{align*} (p, \nabla \cdot v) = \int_\Omega p \nabla \cdot v \, dx = \int_\Omega p \wedge d\nu = \int_{\hat \Omega} \phi^*(p \wedge d\nu) \end{align*} where $\phi^*$ is the pullback of differential form $\phi$.

Now, it’s just a matter of algebra \begin{align*} \int_{\hat \Omega} \phi^*(p \wedge d\nu) = \int_{\hat \Omega} \phi^*p \wedge \phi^* d\nu = \int_{\hat \Omega} \phi^*p \wedge d\phi^*\nu. \end{align*} Now, we need to look up what the pullback does to a 0-form $p$ and on the 2-form $\nu$. Well, we can look this up or just algebraically do it (which I’ll do it sooner or later because I couldn’t find a good source online), we have $\phi^* p = p \cdot \phi$ and \begin{align*} \phi^* \nu = (\det \phi’ )(\phi’)^{-1} (v \cdot \phi): \hat \Omega \to \mathbb{R}^3 \end{align*} which is exactly the Piola transform.