<s>[INST] Surface delimited by an elliptical cylinder and z+y = 9. We're covering Stokes' Theorem in class right now, and I can't understand anything. I'm struggling to solve even the most trivial examples. Here's one I can't solve: Calculate the surface integral of $\mathbf{F}(x,y,z) = (z,x,y)$ where the surface $S$ is the surface delimited by $\frac{x^2}{4}+\frac{y^2}{9} = 1$ and $z+y=9$. What do I do? I feel like I should solve for $x$ and $y$ and then find a parametric curve in $z$. But then what's the surface associated with it? Should I solve $\frac{x^2}{4}+\frac{y^2}{9} - 1 = z + y - 9$ instead? Would that be a surface? I have no idea what is going on. A hint towards a general approach to solve these types of problems would help also.
 [/INST] I am going to interpret the question as computing the line integral of $\mathbf{F}(x, y, z)=(z, x, y)$ over the closed curve given by the intersection of the elliptical cylinder $\frac{x^2}{4}+\frac{y^2}{9} = 1$ and the plane $y+z=9$. The reason I interpret it like this is twofold: the elliptical cylinder and the plane do not "delimit" any finite surface, and the alternative interpretation of computing the surface integral over some ellipse requires computation of an "anti-curl", which is less straightforward. In any case, the alternative interpretation can be solved in the same way by noting that $(xz, xy, yz)$ is an anti-curl of $\mathbf{F}$.
The intersection of the elliptical cylinder and the plane is a closed curve $C$ that can be parameterized as
$$(x, y, z)=(2\cos\theta, 3\sin\theta, 9-3\sin\theta),\qquad \theta\in[-\pi,\pi]$$
Calculation using Stokes' theorem
Stokes' theorem says
$$\oint_C\mathbf{F}(x, y, z)\cdot d\boldsymbol{\ell}=\iint_S \left(\nabla\times\mathbf{F}(x, y, z)\right)\cdot d\mathbf{S}$$
where $S$ is any surface whose boundary is $C$.
The curl of $\mathbf{F}$ is
$$\nabla\times\mathbf{F}(x, y, z)=(1, 1, 1)$$
We choose $S$ to be the surface in the plane $y+z=9$ that is delimited by the curve $C$. A parameterization of $S$ is
$$(x, y, 9-y),\qquad x\in[-2, 2], y\in\left[-\frac{1}{2}\sqrt{36-9x^2},\frac{1}{2}\sqrt{36-9x^2}\right]$$
and the differential of surface is
$$d\mathbf{S}=(0, 1, 1)\:dy\:dx$$
Note that the norm of $d\mathbf{S}$ must be the area of the surface element, which is $\sqrt{2}$ in this case.
We compute the surface integral:
\begin{align*}
\iint_S \left(\nabla\times\mathbf{F}(x, y, z)\right)\cdot d\mathbf{S} & = \int_{x=-2}^{2} \int_{y=-\frac{3}{2}\sqrt{4-x^2}}^{\frac{3}{2}\sqrt{4-x^2}} (1, 1, 1) \cdot (0, 1, 1) \:dy\: dx \\
& = \int_{x=-2}^{2} \int_{y=-\frac{3}{2}\sqrt{4-x^2}}^{\frac{3}{2}\sqrt{4-x^2}} 2 \:dy\: dx \\
& = \int_{x=-2}^{2} 6\sqrt{4-x^2} \:dy\: dx \\
& = 12\pi.
\end{align*}
Direct computation of the line integral
The line integral can also be computed directly. The differential of length is
$$d\boldsymbol{\ell}=(-2\sin\theta, 3\cos\theta, -3\cos\theta)\:d\theta$$
so the line integral is
\begin{align*}
\oint_C\mathbf{F}(x, y, z)\cdot d\boldsymbol{\ell} & = \int_{-\pi}^{\pi} \mathbf{F}(2\cos\theta, 3\sin\theta, 9-3\sin\theta)\cdot (-2\sin\theta, 3\cos\theta, -3\cos\theta)\:d\theta \\
& = \int_{-\pi}^{\pi} (9-3\sin\theta, 2\cos\theta, 3\sin\theta)\cdot (-2\sin\theta, 3\cos\theta, -3\cos\theta)\:d\theta \\
& = \int_{-\pi}^{\pi} \left[-18\sin\theta + 6\sin^2\theta + 6 \cos^2\theta - 9\sin\theta\cos\theta  \right]\:d\theta \\
& = 12\pi.
\end{align*}
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