Use a parameterization to find the flux $\iint_{S} F \cdot n d \sigma$ of $F=z^{2} i+x j-3 z k$ in the outward direction (normal away from the $x$-axis) across the surface cut from the parabolic cylinder $z=4-y^{2}$ by the planes $x=0, x=1$, and $z=0$.
The flux is $\square$.
Answer
The flux is \(\boxed{-31.68}\).
Step 1 ：Parameterize the surface using \(y\) and \(x\) as parameters with bounds \(0 \leq x \leq 1\) and \(-2 \leq y \leq 2\).
Step 2 ：Find the normal vector by taking the cross product of the partial derivatives of the parameterization with respect to \(x\) and \(y\).
Step 3 ：The normal vector will have a positive \(x\)-component, as it is in the direction away from the \(x\)-axis.
Step 4 ：Calculate the surface integral of the vector field \(F = z^{2} i + x j - 3 z k\) dotted with the normal vector over the parameterized surface.
Step 5 ：The flux is \(\boxed{-31.68}\).
