Consider the sphere with radius $\sqrt{2}$ and centre the origin. Let $S'$ be the portion of the sphere that is above the curve $C$ (lies in the region $z \geq 1$) and has $C$ as a boundary. Evaluate the flux of $\nabla \times F$ through $S_0$. Specify which orientation you are using for $S'$.
$$F = (z-y,0,y)$$
(curve $x^2+y^2=1$ lying in the plane $z=1$).
So my thoughts are can we use the Divergence Theorem? If so when we use the $\text{curl} = \langle 1,1,1\rangle$?
Then when we apply the Divergence Theorem we can just say that the flux of this would be $0$? Or is this incorrect?
Method1:
Parametrize the surface as $(\sqrt2 \cos\theta \sin \phi, \sqrt2 \sin\theta \sin \phi,\sqrt2 \cos \phi)$
Outward unit normal vector $\hat{n} = (\cos\theta \sin \phi, \sin\theta \sin \phi,\cos \phi)$
At $z = 1, \phi = \frac{\pi}{4}$
So use $0 \leq \phi \leq \frac{\pi}{4}, 0 \leq \theta \leq 2\pi$ to set up your double integral.
$\displaystyle \iint_R (\nabla \times \vec{F}) \cdot \hat{n} \ (\sqrt2)^2 \ d\phi \ d\theta$
Or Method 2
Take boundary C which is a circle of radius $1$ and parametrize as $r(t) = (\cos t, \sin t, 1), 0 \leq t \leq 2\pi$.
$\displaystyle \int_C \vec{F} \cdot r'(t) \ dt$
Please make sure the orientation matches up for both integral.