This is a rather simple problem, but one that I'm struggling with nonetheless. I'm given a 2-Form, $\beta = zdx \wedge dy-x^2dy \wedge dz$ that I need to integrate over the surface S : $z=4-x^2-y^2 \ge 0 $.
Parametrising the surface, I have:
$G(x,y) = < x,y,4-x^2-y^2>$ and the Jacobian: $$ \begin{bmatrix} DG \end{bmatrix} = \begin {bmatrix}1 & 0 \\0 & 1 \\ -2x & -2y \end{bmatrix}$$
Evaluating the 2-Form, I have $\int_S\beta= \iint_{z=4-x^2-y^2} (4-x^2-y^2) \begin{vmatrix}1&0 \\ 0&1 \end {vmatrix} - x^2 \begin{vmatrix} 0 & 1\\ -2x & -2y \end{vmatrix}dxdy$
This simplifies to: $\int_S\beta = \iint_{z=4-x^2-y^2} 4 -x^2-y^2-2x^3dxdy$
In determining the region of integration, though, I get stuck. My gut is telling me to go with polar coordinates. If this were the case, I know $0 \le \theta \le 2{\pi}$, but I'm uncertain of the limits of integration for $r$. Would it be $0 \le r \le 2$? I don't think this is correct but am uncertain of any other way you'd do it.