Here is some starting work:

$\displaystyle z=9-x^2-y^2=5\quad\mbox{when}\quad x^2+y^2=2^2$

$\displaystyle \begin{aligned}

\mathbf{r}(t)&=2\cos t\mathbf{i}+2\sin t\mathbf{j}+5\mathbf{k}\\

\mathbf{r'}(t)&=-2\sin t\mathbf{i}+2\cos t\mathbf{j}\\

\mathbf{F}(\mathbf{r}(t))&=(2\cos t-10\sin t)\mathbf{i}+10\cos t\mathbf{j}\\

\oint_C\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r'}(t )\,dt&=\int_0^{2\pi}(-4\sin t\cos t+20\sin^2 t+20\cos^2 t)\,dt\\

&=\int_0^{2\pi}(20-4\sin t\cos t)\,dt.

\end{aligned}$

Also,

$\displaystyle \int\int_S \mbox{curl } \mathbf{F}\cdot\mathbf{n}\,dS=\int\int_R\left|

\begin{array}{ccc}

\mathbf{i} & \mathbf{j} & \mathbf{k}\\

\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\

x-yz & xz & 0

\end{array}\right|

\cdot\left(-\frac{\partial}{\partial x}\mathbf{i}-\frac{\partial}{\partial y}\mathbf{j}+k\right)\,dA$

$\displaystyle =\int\int_R (-x\mathbf{i}-y\mathbf{j}+2z\mathbf{k})\cdot(2x\mathbf{i}+2y\mat hbf{j}+k)\,dA$

$\displaystyle \int\int_R (-2x^2-2y^2+2(9-x^2-y^2))\,dA$

$\displaystyle \int_0^{2\pi}\int_0^2 (18-4r^2)\,r\,dr\,d\theta.$

We obtain the same answer for each.