7. (20 points) Calculate the surface integral $\iint_{S} \mathbf{F} \cdot \mathbf{N} d S$, where $\mathbf{F}(x, y, z)=x \mathbf{i}-y^{3} \mathbf{j}+2 \mathbf{k}$ and $S$ is the surface with parameterization $\mathbf{r}(u, v)=\left\langle u v,-2 u, v^{3}\right\rangle, 0 \leq u \leq 3,0 \leq$ $v \leq 2$.

Step 1 ：First, we calculate the partial derivatives of the parameterization \(\mathbf{r}(u, v)=\left\langle u v,-2 u, v^{3}\right\rangle\) with respect to u and v. We get \(\mathbf{r}_u = \left\langle v, -2, 0\right\rangle\) and \(\mathbf{r}_v = \left\langle u, 0, 3v^2\right\rangle\).

Step 2 ：Next, we calculate the cross product of these two vectors to get the normal vector \(\mathbf{N}\). We find that \(\mathbf{N} = \left\langle -6v^2, -3v^3, 2u\right\rangle\).

Step 3 ：We then substitute the parameterization of the surface into the vector field \(\mathbf{F}(x, y, z)=x \mathbf{i}-y^{3} \mathbf{j}+2 \mathbf{k}\) to get \(\mathbf{F}\) in terms of u and v. We get \(\mathbf{F} = \left\langle u v, 8u^3, 2\right\rangle\).

Step 4 ：We calculate the dot product of \(\mathbf{F}\) and \(\mathbf{N}\). We find that \(\mathbf{F} \cdot \mathbf{N} = -24u^3v^3 - 6uv^3 + 4u\).

Step 5 ：Finally, we calculate the double integral of this dot product over the given range of u and v. We find that the surface integral \(\iint_{S} \mathbf{F} \cdot \mathbf{N} d S = -2016\).

Step 6 ：So, the final answer is \(\boxed{-2016}\).