Given $$z=10-x^2-y^2,\quad1\leq z\leq9,$$ I have to verify the Stokes' Theorem with $$F(x,y,z)=(3z,4x,2y)$$
Since $\nabla\times F=(2,3,4)$ and as we can parametrize the surface as follows $$\sigma(r,\theta)=(r\cos\theta,r\sin\theta,10-r^2)$$ with $1\leq r\leq 3,0\leq\theta\leq2\pi$, then $$\sigma_{r}=(\cos\theta,\sin\theta,-2r),\qquad\sigma_{\theta}=(-r\sin\theta,r\cos\theta,0)$$ So, $$\sigma_{r}\times\sigma_{\theta}=(2r^2\cos\theta,2r^2\sin\theta,r)$$ Therefore, $$\iint_{R}\nabla\times FdS=\int_{0}^{2\pi}\int_{1}^{3}(2,3,4)(2r^2\cos\theta,2r^2\sin\theta,r)drd\theta=32\pi$$
On the other hand, the boundary of the surface has two curves $$\gamma_{1}(t)=(\cos(t),\sin(t),1)$$ and $$\gamma_{2}(t)=(\cos(t),\sin(t),9)$$ for $0\leq t\leq2\pi$. Then $\gamma=\gamma_{2}\cup\gamma_{1}$, i.e., $$\int_{\gamma}Fdr=\int_{\gamma_{2}}Fdr-\int_{\gamma_{1}}Fdr$$ I know $\gamma_{1}'(t)=\gamma_{2}(t)=(-\sin(t),\cos(t),0)$. Therefore, $$\int_{\gamma_{1}}Fdr=\int_{0}^{2\pi}(3,4\cos(t),2\sin(t))(-\sin(t),\cos(t),0)dt=4\pi$$ Also, $$\int_{\gamma_{2}}Fdr=\int_{0}^{2\pi}(27,4\cos(t),2\sin(t))(-\sin(t),\cos(t),0)dt=4\pi$$ Hence, $$\int_{\gamma}Fdr=4\pi-4\pi=0\neq32\pi$$ So, Stokes' Theorem is not verified. Why? What is wrong?