Evaluate $\displaystyle \int\int \vec{F}.ds$ where $\vec{F}= -x\hat{i} + 2y\hat{j} -z\hat{k}$ and $S$ is the portion of the $y = 3x^2 + 3z^2$ that lies behind $y =6$ oriented positive $y -axis$ direction.
Now, Using Stoke's theorem this surface integral is equivalent to
$\displaystyle \int F.dr$ where $C$ is the portion of the curve $x^2 + z^2 = 2$, forming an outer boundary to the surface.
So, in the Line Integral $dy= 0$, and taking $x = \sqrt{2}\cos\theta$ ,$y = \sqrt{2}\sin\theta$,
I get my integral as :
$\displaystyle \int (-2 \cos\theta\sin \theta + 2\sin\theta\cos \theta)d \theta$, giving me the answer $0$.
However the answer given is $24 \pi$
I cannot understand where am I wrong in my solution ?
Can anyone please help me here ?
By Stokes theorem you may evaluate the flux through the disc $x^2+z^2\leq 2$ in the plane $y=6$. Such disc $D$ has the same outer boundary as the given surface $S$, but the flux is much simpler to compute. Thus we find $$\iint_D (-x,2y,z)\cdot (0,1,0)dS=\iint_D 2ydS=2\cdot 6\cdot |D|=12\cdot\sqrt{2}^2\pi=24\pi.$$
P.S. According to your approach, in the line integral you should use a vector field $G$ such that $\nabla\times G=F$.