I cannot fin the flux of
$$F(x,y,z)=(y^2cos(xz),x^3e^{yz},-e^{-xyz})$$
through the portion of sphere
$$\Sigma = \{x^2+y^2+(z-2)^2=8, z\ge0 \}$$
I think Stokes th. must be used, so in spherical coordinates I get $$x=\sqrt{8}cos\theta\sin\phi$$ $$y=\sqrt{8}sin\theta\sin\phi$$ $$z=2+\sqrt{8}cos\phi$$ and I try to calculate the integral on the curve $$\gamma=(\sqrt{8}cost,\sqrt{8}sint,2)$$ but the I cannot solve the integral which should be $$8\sqrt{8}\int_{[0,2\pi]}sin^3tcos(2\sqrt{8}cost)+\sqrt{8}cos^3t*e^{2\sqrt{8}cost}dt$$ Where is my error?
Thanks a lot for your help.
KB
In general, if $S_1$ and $S_2$ be oriented surfaces with the same oriented boundary curve $C$ and both satisfy the hypotheses of Stokes’ Theorem, then $$\iint\limits_{{{S}_{1}}}{\operatorname{curl}\,F.n\,\,d{{S}_{1}}}=\oint\limits_{C}{F.dr}=\iint\limits_{{{S}_{2}}}{\operatorname{curl}\,F.n\,\,d{{S}_{2}}}$$ This fact is useful when it is difficult to integrate over one surface but easy to integrate over the other. Anyway, let $$S_1 = \{(x,y,z) | x^2+y^2+(z-2)^2=8\,, z\ge0 \}$$ $$S_2 = \{(x,y,z) | x^2+y^2\le 4\,, z=0 \}$$ $$n=\overrightarrow{k}=(0,0,1)$$ we have $$\iint\limits_{{{S}_{2}}}{\operatorname{curl}\,F.n\,\,d{{S}_{2}}}=\iint\limits_{A}{\left( \frac{\partial {{F}_{2}}}{\partial x}-\frac{\partial {{F}_{1}}}{\partial y} \right)}\,dA=\iint\limits_{A}{(3{{x}^{2}}-2y)}\,dA$$ note $z=0$ on $S_2$ . Now set \begin{align} & x=r\cos \theta \\ & y=r\sin \theta \\ \end{align} $$\oint\limits_{C}{F.dr}=\int_{0}^{2\pi }{\int_{0}^{2}{(3{{r}^{2}}{{\cos }^{2}}\theta -2{r}\sin \theta }})\, r\,drd\theta$$