Let $c>0$ and $A\subseteq R^3$ be a green-region (which is defined by: $A\subseteq R^2$, $A=B_1\cup ...\cup B_m$, and $B^{\circ}_i\cap B^{\circ}_j=\emptyset$) with the outward normal unit vectors $n=(n_1,n_2,n_3)^T$ on $\partial A$. Let $\left\{e_1,e_2,e_3\right\}$ be the standard basis of the $(x,y,z)$-space and
$K=\begin{bmatrix} K_1\\ K_2\\ K_3\end{bmatrix}:=\iint\limits_{\partial A}c~z~n~d\sigma$.
Use Gauss's theorem on the vector fields
$F_i(x,y,z):=cze_i, ~~~~~ (i=1,2,3)$
and prove archimede's principle
$K_1=K_2=0, ~~~~K_3=cv_3(A)$
First I looked up the definition of Gauss's theorem we have in our textbook.
$\iiint\limits_{A}div~k~ d^3x = \iint\limits_{\partial A} k\cdot d\Sigma$ where $k=\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}$ is a vector field.
Since $\partial_xF_1$ and $\partial_yF_2$ are zero then $K_1$ and $K_2$ must be zero as well, right? But if I take $\partial_zF_3$ I get just $c$, which consequently means that it's $\iint\limits_{\partial A}c~n~d\sigma$ and not $\iint\limits_{\partial A} c~z~n~d\sigma$. Did I make some mistake? If not, then, since c is a cosntant I could take it out of the integral and it would just leave me with $K_3=c~v_3(A)$, right?
