This excercise asks me to integrate $\omega=z dx \wedge dy$ over the 2-manifold $A = \{(x, y, z): x^2+z^2=y, y<4\}$ with orientation $o(x)$ such that $o^{31}(x)>0$ using Divergence's theorem. So I try to proceed as follow: $$D = \{(x, y, z): y>x^2+z^2, y<4\}$$ Then $$A^o = \partial{D}^o$$ with $$o(x) = \frac{2ze_{12}+e_{13}+2xe_{23}}{\sqrt{1+4z^2+4x^2}}$$ So by divergences theorem: $$\int_{A^o}\omega = \int_{A^+}\omega ·o=\int_{\partial{D}^+}\omega·o=\int_{{D}^+}d(\omega·o)$$ Where $d(\omega·o)=\dfrac{8z^3+\left(16x^2+4\right)z}{\left(4z^2+4x^2+1\right)^\frac{3}{2}}dx\wedge dy \wedge dz$
So in cylindrical coordinates ($y=y\space \space \space z=rsin(\theta)\space \space \space x=rcos(\theta)$): $$\int_0^{4}\int_0^{2\pi}\int_0^{y}\dfrac{8(r·sin(\theta))^3+\left(16(r·cos(\theta))^2+4\right)(r·sin(\theta))r}{\left(4r^2+1\right)^\frac{3}{2}}drd\theta dy=0$$ Which is an odd function on $\theta$ so its integral is $0$. But the book where the excercise appears says it should give me $-8\pi$. What am I missing? What did I do wrong? Thanks.
Update: I think I found where the error is (at least one of them). The integral of dr is from $0$ to $\sqrt{y}$ not from $0$ to $y$. I will do the calculations later and put them here in case someone wants to see it.
Update2: The function is still odd over $\theta$ so it's still $0$. So that last one error can't be the unique one. I would appreciate some help. Thanks.