Prove that there isn't such a 2-form $\omega$ on $\mathbb{R}^3$ that $\omega$ restricted to any surface $\Sigma$ gives its volume form.
Suppose that there is such a form:
$$\omega=f_3(x,y,z)dx \wedge dy + f_1(x,y,z)dy \wedge dz+ f_2(x,y,z) dz \wedge dx$$
Consider plane (x,y,c) for any $c$. Then $\omega$ restricted to this plane is:
$$f_3(x,y,c) dx \wedge dy$$
Volume form of plane (x,y,c) is $dx \wedge dy$, so we have:
$$f_3(x,y,c)=1$$
for any $x,y,c$, so $f_3 \equiv 1$. We can do the same with $f_1$ and $f_2$. So:
$$\omega=dx \wedge dy + dy \wedge dz+ dz \wedge dx$$
Now let's consider plane: $(x,y,x)$, $\omega$ restricted to this plane is:
$$dx \wedge dy + dy \wedge dx+ dx \wedge dx=0$$
But volume form of this plane is not $0$.
My question: is it correct? Is there any simplier way to prove that?
Your proof is correct. I don't think it can be done simpler, but perhaps the following is more "conceptual". The form $\omega$ is the Hodge-* of the $1$-form $f_1\,dx+f_2\,dy+f_3\,dz$. Hence, the volume element of its plane restriction is proportional to the dot product of $(f_1,f_2,f_3)$ with the normal vector to the plane. No matter what $(f_1,f_2,f_3)$ is, there are normal vectors for which the dot product is zero.