I've got a 2-part problem to solve, in which there is some syntax that I don't recognize. It looks like this: Let $$ \Psi: [-1,1]^2 \mapsto \mathbb{R}^3, \Psi(u,v) := ( u-\frac{u^3}{3}+uv^2, v-\frac{v^3}{3}+vu^2, u^2-v^2 ) $$ 1) Calculate the Normal $N$, which I got to be:
$$ N = \frac{\partial \Psi}{\partial u} \times \frac{\partial \Psi}{\partial u} = (-2(u+u^3), -2v(u^2+v^2+1), -2u^2v^2-u^4-v^4+1) $$
The second part of the problem is:
2) Let $\omega = i_N$ det be the volume form. Calculate the surface $\int_{\Psi} i_N$ det.
What does $i_N$ mean in this context? The normal is a line in $\mathbb{R}^3$, but in order for the integral to result in a surface, I assume that $i_N$ det, also must be a line?
Here $\det = dx \wedge dy \wedge dz$ is the volume form of $\mathbb R^3.$ The name comes from the triple product formula $$(dx \wedge dy \wedge dz)(u,v,w) = (u \times v) \cdot w = \det \left[ u \; v \; w \right].$$
The interior product $i_N \det$ is the two-form resulting from plugging the vector $N$ in to this three-form. Thus we have $$(i_N \det)(v,w) = (v \times w) \cdot N;$$
so when you integrate the two-form over the surface $\Psi(U)$ you get
$$ \int_{\Psi(U)} i_N \det = \int_{U} (\Psi_u \times \Psi_v) \cdot N\,du\,dv = \int_U |\Psi_u \times \Psi_v|du\,dv$$ since $N$ is just the unit-length normalization of $\Psi_u \times \Psi_v.$ This is the area formula from multivariable calculus, which is hopefully familiar.
(You didn't include the normalization of $N$ in your question, but it is necessary if you want this to give you the area.)