Find the flux integral of $$ F=(\cos xyz, \tan xyz, 1+\arctan xyz) $$ across the surface $x^2+y^2=1$, $0\leq z \leq2$ and oriented by normal vectors pointing away from the z-axis.
I tried to do it in a straightforward way by parametrizing the surface $S=(\cos{t}, \sin{t}, s)$ and plugging this into $$ \iint_{S} (\cos xyz, \tan xyz, 1+\arctan xyz) \cdot n dS $$ $$ =\iint_{S} (\cos xyz, \tan xyz, 1+\arctan xyz) \cdot \frac{(2x,2y,0)}{\|(2x,2y,0)\|} dS$$ But it seems the effort will not be successful because the integral looks very complicated.
Let us compute this using divergence theorem $$ \int X\cdot ndS=\int \nabla \cdot X dV $$ Here the upper and lower pieces are $x^2+y^2=1,z=0$ and $x^2+y^2=1,z=2$. The volume integral is given by $$ \int \int \int (-yz \sin[xyz]+xz \sec^2[xyz]+\frac{xy}{1+(xyz)^2})dx dy dz $$ The first and second integral is odd $x$ and they vanishes. Similarly the last integral is symmetric and vanishes if we divide the circle into four quadrants. Thus the original integral is the negative of the sum of surface integral on the $z=0$ and $z=2$ plane. When $z=0$ we have $F=(1,0,1),n=(0,0,-1)$, thus the integral is $-\pi$.When $z=2$ we have $$ \int_{x^2+y^2\le 1} (1+\arctan[2 xy])dxdy=\pi $$ Thus the whole surface integral is $\pi+(-\pi)=0$. Hopefully there is no typo here.