Evaluate $\int_M(x-y^2+z^3)dS$ when $M$ is the part of the cylinder $x^2+y^2=a^2$ where $a>0$ which is between the two planes $x-z=0$ and $x+z=0$.
So I did not manage to use green/gauss/stocks, so I tried to solve it as a surface integral.
first to find $\|n\|$ we use the parameterisation $\phi(u,v)=(a\cos (u),a\sin (u),v)$
$\phi_u\times\phi_v=(a\cos(u),a\sin(u),0)$
So $\|n\|=a$
So the integral is $\iint (a \cos(u)-a^2\sin^2(u)+v^3)adudv$ but I can I find the limit of integration? I know that $u\in[0,2\pi]$ and as for $v$ is is bounded by $x$ and $-x$
P.S or I can say that $F=\nabla\cdot(\frac{x^2}{2},-\frac{y^3}{3},\frac{z^4}{4})$ and so I can use gauss?
On $\phi(r,t,v)=(r\cos t,r\sin t,v)$ where $\theta\in[0,2\pi],r\in[0,a],v\in[-r \cos \theta,\cos\theta]$
You have to compute this as a surface integral. Stokes' and Gauss' theorems deal with the flux of certain vector fields across a surface, but here a scalar function (representing, e.g., a temperature) is integrated over $M$.
Your parametrization is fine, and leads to the scalar surface element $${\rm d}S=|\phi_u\times\phi_v|\>{\rm d}(u,v)=a\>{\rm d}(u,v)\ .$$ In a sketch of the $(x,z)$-plane you can see that the part of the cylinder we are interested in is characterized by $|z|\leq |x|$, or $|v|\leq a|\cos u|$. Therefore we have to split the integral ($=:J$) into two parts corresponding to $-{\pi\over2}\leq u\leq{\pi\over2}$ and ${\pi\over2}\leq u\leq{3\pi\over2}$. In this way we obtain $$J=a\int_{-\pi/2}^{\pi/2}\int_{-a\cos u}^{a\cos u}\Psi(u,v)\>dv\>du+a\int_{\pi/2}^{3\pi/2}\int_{a\cos u}^{-a\cos u}\Psi(u,v)\>dv\>du\ .$$ Here $\Psi(u,v)$ denotes the pullback $$\Psi(u,v)=(x-y^2+z^3)\biggr|_{(x,y,z):=\phi(u,v)}=a\cos u-a^2\sin^2 u +v^3\ .$$ It becomes obvious that the contribution of the $v^3$ term is zero, by symmetry. I may leave the rest to you.