Using Gauss theorem in $F(x,y,z)=(x,y,z+1)$ I have to verify the Gauss' theorem in the vector field
$$F(x,y,z)=(x,y,z+1)$$
over the surface
$$U=\{(x,y,z)\in\mathbb{R}^{3}\mid x^2+y^2+z^2\leq a^2,z\geq0\}$$
Since $div(F)=3$ then, in spherical coordinates:
$$\iiint_{U}div(F)dV=\int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{0}^{a}3\rho^{2}\sin\phi d\rho d\phi d\theta=2\pi a^{3}$$
Now, using the following parametrization of the surface
$$\sigma(u,v)=(a\cos u\sin v, a\sin u\sin v, a\cos v)$$
with $0\leq u\leq2\pi,0\leq v\leq\pi/2$, I get
$$\sigma_{u}\times\sigma_{v}=(-a^2\cos u\sin^{2}v,-a^{2}\sin u\sin^{2}v,-a^{2}\sin v\cos v)$$
and
$$F(\sigma(u,v))=(a\cos u\sin v,a\sin u\sin v,a\cos v+1)$$
Hence
$$\iint_{S}F\cdot dS=\int_{0}^{\pi/2}\int_{0}^{2\pi}(-a^{3}\sin v-a^{2}\sin v\cos v)dudv=\pi(a^2-2a^3)$$
Therefore, I can't use Gauss' theorem here. My question is: why? For me, all the hypothesis of the theorem are ok.

Gauss' theorem works here. The problem is that the surface you parametrized is not the boundary of $U$, but rather the sphere of radius $a$ centered at the origin. Remember that $U=\overline{B}((0,0,0),a)\cap (\mathbb{R}\times\mathbb{R}\times[0,\infty))$, and therefore its boundary is the semi-sphere ($z\geq 0$) of radius $a$ union the circle of radius $a$ centered at the origin in the $XY$ plane.
(Edit: Here, $\overline{B}((0,0,0),a)=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2+z^2\leq a^2\}$ is the closed ball of radius $a$ centered at the origin)
