$\iint_A\nabla\times\textbf{u}\cdot \textbf{n}\ dS$ with $\nabla\times\textbf{u}$ known. 
Let $$\textbf{v}=\operatorname{curl}\textbf{u}= (-2+43x-86y+43z, 3-2x+4y-2z, 2-47x+94y-47z)$$ and let $A$ be the part of $z(x,y)=9-x^2-9y^2$ that is above the plane $z(x,y)=x+45y+\frac{113}{2}$. Now calculate $$\iint_A\nabla\times\textbf{u}\cdot \textbf{n}\ dS$$

A hint is given to use Stoke's theorem twice.
So my take is: $$\iint_A\nabla\times\textbf{u}\cdot \textbf{n}\ dS=\iint_A\textbf{v}\cdot \textbf{n}\ dS=\int_{\partial A} \nabla\times \textbf{v} \cdot d\textbf{r}$$
So I get $d\textbf{r}$ as the boundary to an ellipse that satisfies $1=(\frac{x+\frac{1}{2}}{3})^2+(y+\frac{5}{2})^2$, so we have $\textbf{r}=(x,y,(\frac{x+\frac{1}{2}}{3})^2+(y+\frac{5}{2})^2)$ and
$\nabla\times\textbf{v}= (96,90,84)$
I am not sure  how to proceed from here. Should I parametrize $\textbf{r}$ in terms of trig functions?c
 A: The hint essentially means that you should translate this to another surface (with the same boundary curve) than the one you currently have. In other words,
$\displaystyle \iint_{S} (\nabla\times\textbf{u})\cdot \textbf{n}\ dS = \int_{C} \textbf{u}\cdot d\textbf{r} = \iint_{S_1} (\nabla\times\textbf{u})\cdot \textbf{n}\ dS$.
The boundary curve of the elliptic paraboloid surface, as you correctly found, is -
$\displaystyle (\frac{x+0.5}{3})^2+(y+2.5)^2 = 1, z = 9 - x^2 - 9y^2$.
Now using Stokes' Theorem a second time, we can equate line integral of the vector field along the boundary curve to the surface integral of the curl of $\textbf{u}$ over elliptic disk given by,
$\displaystyle (\frac{x+0.5}{3})^2+(y+2.5)^2 \leq 1, z = x + 45y + \frac{113}{2}$.
$\nabla \times \textbf{u}= (-2+43x-86y+43z, 3-2x+4y-2z, 2-47x+94y-47z)$
For the dot product, use the fact that normal vector to the plane is $(-1, -45, 1)$ and the dot product is simply $-131$.
So your task is just finding the area of the ellipse. You can, Either use the formula $\pi ab$ for area of the ellipse and multiply by $131$ which gives you answer of $-393 \pi$. Or you can do the integral as below.
Use substitution, $x = - 0.5 + 3r \cos\theta, y = -2.5 + r \sin\theta$ for the surface integral, $0 \leq r \leq 1, 0 \leq \theta \leq 2\pi$ and plug in Jacobian $3r$ instead of usual $r$. That is $3$ times the area of a unit circle( = $3 \pi$).
