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.
 A: 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)
A: $U=\{(x,y,z)\in\mathbb{R}^{3}\mid x^2+y^2+z^2\leq a^2,z\geq0\}$
$\vec F = (x,y,z+1)$
$U$ is union of disk $D: x^2 + y^2 \leq a^2, z = 0$ and hemisphere surface  $S: x^2 + y^2 + z^2 = a^2, z \geq 0$
i) Flux through $D$:
We parametrize the surface as $\vec r(t) = (r \cos t, r \sin t, 0), 0 \leq r \leq a, 0 \leq t \leq 2\pi$
Outward unit normal vector to the disk at $z = 0$ is $\hat n = (0, 0, - 1)$.
So $\vec F(\vec r(t)) = (r \cos t, r\sin t, 1)$
$\vec F \cdot n = ((r \cos t, r\sin t, 1) \cdot (0, 0, -1) = - 1$
So the flux through $D$ is $ - \pi a^2$ ...$(i)$
ii) Flux through $S$:
Now coming to your parametrization and finding flux directly through surface $S$, please note that divergence theorem will give you flux considering outward normal vector. So you need to make sure you consider outward normal vector when doing surface integral.
$\sigma_{v}\times\sigma_{u}=(a^2\cos u\sin^{2}v,a^{2}\sin u\sin^{2}v, a^{2}\sin v\cos v)$
$F(\sigma(u,v))=(a\cos u\sin v,a\sin u\sin v,a\cos v+1)$
$\vec F \cdot (\sigma_{v}\times\sigma_{u}) = a^3 \sin v + a^2 \sin v \cos v$
So flux through $S$ is,
$\displaystyle \int_0^{2\pi} \int_0^{\pi/2} (a^3 \sin v + a^2 \sin v \cos v) \ dv \ du = \pi (2a^3 + a^2)$ ...$(ii)$
$(i) + (ii) = 2 \pi a^3$, which matches with the result obtained using divergence theorem.
