$$\oint_S e^xdy\,dz-ye^xdz\,dx+3z\,dx\,dy, \quad S:x^2+y^2\le a^2,\quad 0\le z\le h$$
I tried to solve this integral by Gauss and the answer was $(3\pi a^2h)$ but it is required to verify Gauss i tried to integrate over each side but the resulting exponential function can't be integrated by usual methods. Can you please tell me how to do it by direct method. Thanks.
I don't think I have ever seen that perfectly dreadful notation for a surface integral before. I think it stands for the flux out of the closed surface $S$ $$\unicode{x222F}_S\vec F\cdot d^2\vec A$$ of the vector field $\vec F=\langle e^x,-ye^x,3z\rangle$. Assuming that is the case, we break $S$ into $3$ parts: the top $S_1$ where $z=h$, the curvy part $S_2$ where $\sqrt{x^2+y^2}=r=a$, and the bottom $S_3$ where $z=0$.
On $S_2$, $\vec r=\langle x,y,z\rangle=\langle a\cos\theta,a\sin\theta,z\rangle$, so the differential $$d\vec r=\langle-a\sin\theta,a\cos\theta,0\rangle d\theta+\langle0,0,1\rangle dz$$ Since the $d\theta$ term is the vector step you take along $S_2$ when $\theta$ goes from $\theta$ to $\theta+d\theta$ and the $dz$ term is the vector step you take along $S_2$ when $z$ goes from $z$ to $z+dz$, it follows that $$d^2\vec A=\pm\langle-a\sin\theta,a\cos\theta,0\rangle d\theta\times\langle0,0,1\rangle dz=\langle a\cos\theta,a\sin\theta,0\rangle d\theta\, dz$$ having the area of the parallelogram with these two vector steps as edges and normal to $S_2$, is the required vector areal element $d^2\vec A$. We took the $(+)$ sign so as to get the outward normal. Along the surface $S_2$, $$\vec F=\langle e^{a\cos\theta},-a\sin\theta e^{a\cos\theta},3z\rangle$$ So $$\unicode{x222F}_{S_2}\vec F\cdot d^2\vec A=\int_0^h\int_0^{2\pi}\left(a\cos\theta e^{a\cos\theta}-a^2\sin^2\theta e^{a\cos\theta}\right)d\theta\, dz$$ Now, it looks like we are in trouble here, but we can integrate by parts to get $$\begin{align}\int_0^{2\pi}a\cos\theta e^{a\cos\theta}d\theta&=\left.a\sin\theta e^{a\cos\theta}\right|_0^{2\pi}-\int_0^{2\pi}a\sin\theta e^{a\cos\theta}\left(-a\sin\theta\right)d\theta\\ &=\int_0^{2\pi}a^2\sin^2\theta e^{a\cos\theta}d\theta\end{align}$$ So $$\unicode{x222F}_{S_2}\vec F\cdot d^2\vec A=\int_0^h\int_0^{2\pi}\left(a^2\sin^2\theta e^{a\cos\theta}-a^2\sin^2\theta e^{a\cos\theta}\right)d\theta\, dz=0$$ The top is easy: along $S_1$, $\vec r=\langle r\cos\theta,r\sin\theta,h\rangle$, $$d\vec r=\langle \cos\theta,\sin\theta,0\rangle dr+\langle-r\sin\theta,r\cos\theta,0\rangle d\theta$$ $$d^2\vec A=\pm\langle \cos\theta,\sin\theta,0\rangle dr\times\langle-r\sin\theta,r\cos\theta,0\rangle d\theta=\langle0,0,r\rangle dr\, d\theta$$ Where the (+) sign is again outward. Then along $S_1$, $$\vec F=\langle e^{r\cos\theta},r\sin\theta e^{r\cos\theta},3h\rangle$$ so $$\unicode{x222F}_{S_1}\vec F\cdot d^2\vec A=\int_0^a\int_0^{2\pi}3hrd\theta dr=(3h)(2\pi)\left(\frac12a^2\right)$$ On the bottom along $S_3$, $\vec r=\langle r\cos\theta,r\sin\theta,0\rangle$, $$d\vec r=\langle \cos\theta,\sin\theta,0\rangle dr+\langle-r\sin\theta,r\cos\theta,0\rangle d\theta$$ $$d^2\vec A=\pm\langle \cos\theta,\sin\theta,0\rangle dr\times\langle-r\sin\theta,r\cos\theta,0\rangle d\theta=-\langle0,0,r\rangle drd\theta$$ Where now the (-) sign is outward. Then along $S_3$, $$\vec F=\langle e^{r\cos\theta},r\sin\theta e^{r\cos\theta},0\rangle$$ so $$\unicode{x222F}_{S_3}\vec F\cdot d^2\vec A=\int_0^a\int_0^{2\pi}0\,d\theta\, dr=0$$ Adding up the results, $$\unicode{x222F}_S\vec F\cdot d^2\vec A=\unicode{x222F}_{S_1}\vec F\cdot d^2\vec A+\unicode{x222F}_{S_2}\vec F\cdot d^2\vec A+\unicode{x222F}_{S_3}\vec F\cdot d^2\vec A=3\pi a^2h+0+0=3\pi a^2h$$ The divergence $$\vec{\nabla}\cdot\vec F=\langle\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\rangle\cdot\langle e^x,-ye^x,3z\rangle=e^x-e^x+3=3$$ So $$\int\int\int_V\vec{\nabla}\cdot\vec Fd^3V=3V=3(\pi a^2h)$$ Because the integrand is constant within the volume $V$. Thus the divergence theorem is verified.