Compute the integral: $\int_{S} F \cdot N d \sigma$ where $S$ is the truncated cone $y=2 \sqrt{x^{2}+z^{2}}, 2 \leq y \leq 4$ with $N$ pointing outward, and $F(x, y, z)=(x,-2 y, z)$
Attempt. If $R_{y}=x$ and $Q_{z}=0$ then $R=x y+f(x, z)$ and $Q=g(x, y) .$ Thus $R_{x}=y+f_{x}$ and we may set $f=0$ i.e., $R=x y .$ since $P_{z}=-y$ implies $P=-y z+h(x, y)$ and $P_{y}=-z+h_{y},$ we may set $Q=0=h,$ i.e., $P=-y z$ Now $\partial S$ has two picces: $C_{1}$ given by $\phi(t)=(\sin t, 2, \cos t), t \in[0,2 \pi],$ and $C_{2}$ given by $\psi(t)=(2 \cos t, 4,2 \sin t)$ $t \in[0,2 \pi] .$ Thus
$$(P, Q, R) \cdot \phi^{\prime}(t)=(-2 \cos t, 0,2 \sin t) \cdot(\cos t, 0,-\sin t)=-2 \cos ^{2} t-2 \sin ^{2} t=-2$$
and $$ (P, Q, R) \cdot \psi^{\prime}(t)=(-8 \sin t, 0,8 \cos t) \cdot(-2 \sin t, 0,2 \cos t)=16 \cos ^{2} t+16 \sin ^{2} t=16 $$ Therefore by Stokes' Theorem that $$ \iint_{S} F \cdot \boldsymbol{n} d \sigma=\int_{0}^{2 \pi}(16-2) d t=28 \pi. $$
Can you check the attemp, if false then could you help? Thanks...