Divergence theorem -- statement and direct verification

The formula of Ostrogradsky-Gauss relates the triple integral over a volume $V$ to the double integral over the surface $\sigma$ bounding it:

where $\overrightarrow {n}$ is the unit outward normal vector to $\sigma$, and $P$, $Q$, $R$ are the projections of the field vector $\overrightarrow {u}$ on the axes $Ox$, $Oy$, $Oz$; in vector form the same formula reads $\iiint_{V}\operatorname {div}\overrightarrow {u} \, dv=\iint_{\sigma }\overrightarrow {u}\cdot \overrightarrow {n} \, d\sigma =\iint_{\sigma }u_{n} \, d\sigma$.

Prove the equality $\iint_{S}\cos (\overrightarrow {n},x) \, d\sigma =0$, where $S$ is a closed surface and $\overrightarrow {n}$ is the outward normal to it.

Find the integral $\displaystyle \iint \limits_{S}(x^{3}\cos \alpha +y^{3}\cos \beta +z^{3}\cos \gamma ) \, ds$, taken over the surface of the sphere $x^{2}+y^{2}+z^{2}=a^{2}$. Here $\alpha$, $\beta$, $\gamma$ are the angles of the outward normal with the axes.

Find the integral $\displaystyle \iint \limits_{S}\big((z^{n}-y^{n})\cos \alpha +(x^{n}-z^{n})\cos \beta +(y^{n}-x^{n})\cos \gamma \big) \, ds$, taken over the upper half of the surface of the sphere $x^{2}+y^{2}+z^{2}=a^{2}$; $\alpha$, $\beta$, $\gamma$ are the angles of the normal outward from the sphere.

Formulate the analog of the Gauss-Ostrogradsky theorem for plane regions and fields.