Using Stokes' Theorem to find the line integral I am having a bit of trouble understanding line integrals. I've muddled my way through a lot of them, but I just can't understand their relation to Stokes' theorem. Here is a question that I've wrassled with for a while with little to no progress.
I'm supposed to use Stokes' Theorem to find the line integral of the vector field $ \langle xz, 7x+2yz, 3x^2 \rangle $ around the path $ C $ determined by the circle $ x^2 + y^2 = 9, z=3 $.
Here's what I've tried to do. I knew it was wrong as I was doing it, but I have no other idea of where to start. I computed the curl of my field as $ \langle -2y, 5x, 7 \rangle $ and my boundary as $ \langle 3\cos\theta, 3\sin\theta, 3 \rangle $. I put my vector field in terms of the boundary and got  $ \langle -6\sin\theta, 15\cos\theta, 3 \rangle $, computed the derivative of the boundary as  $ \langle -3\sin\theta, 3\cos\theta, 0 \rangle $, and took their cross product to be $ 18\sin^2\theta + 45\cos^2\theta $, which I integrated with $ 0 $ and $ 2\pi $ as my boundaries, yielding $ 36\pi $.
This isn't correct, but I don't know where to start this problem. Any help would be appreciated!

There are a couple of flaws.
First, the curl of the vector $\vec F(\vec r)=\hat x(xz)+\hat y(7x+2yz)+\hat z(3x^2)$ is given by
$$\nabla \times \vec F(\vec r)=\hat x(-2y)+\hat y(-5x)+\hat z(7)$$
Next, from Stokes' Theorem, we can evaluate the line integral of $\vec F(\vec r)$ around the contour $C$ described by $x^2+y^2=1$, $z=3$ as
$$\oint_C \vec F(\vec r)\cdot \,d\vec l=\int_S \hat n \cdot  \nabla \times \vec F(\vec r)$$
where $S$ is any surface bounded by $C$ and $\hat n$ is the unit normal to $S$ oriented to conform with the right-hand rule.  Then, we have
$$\begin{align}
\int_S \hat n \cdot  \nabla \times \vec F(\vec r)&=\int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \hat z\cdot \left(\hat x(-2y)+\hat y(-5x)+\hat z(7)\right)\,dx\,dy\\\\
&=\int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} (7)\,dx\,dy\\\\
&=7\pi
\end{align}$$
