${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS$, where $S$is the surface of the tetrahedron bounded by the three coordinate planes and the plane $3x+2y+6z=12$and where

$\mathbf{F}(x,y,z)=z\sin y\mathbf{i}-x\cos z\mathbf{j}+z\cos y\mathbf{k}.$

The given is a plane $3x+2y+6z=12$and the three coordinate planes is

$\mathbf{F}(x,y,z)=z\sin y\mathbf{i}-x\cos z\mathbf{j}+z\cos y\mathbf{k}.$ The objective is to find whether the integral of${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS$ can be elevated by means of Divergence theorem

Consider the following vector field :

$\mathbf{F}(x,y,z)=z\sin y\mathbf{i}-x\cos z\mathbf{j}+z\cos y\mathbf{k}.$

The objective is to decide whether or not the integral ${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS$, can be evaluated by means of Divergence Theorem, where the surface $S$is the surface of the tetrahedron bounded by the three coordinate planes and the plane $3x+2y+6z=12$

Divergence Theorem states that,

"Let $W$be a bounded region in ${\mathrm{ℝ}}^3$whose boundary $S$is a smooth or piecewise-smooth closed oriented surface. If a vector field $\mathbf{F}(x,y,z)$is defined on an open region containing W, then, ${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS={\iiint }_{W}div\mathbf{F}(x,y,z)dV=\dots \dots \text{(1)}$

where $\mathit{n}$is the outwards unit normal vector."

The surface $S$is the surface of the tetrahedron bounded by the three coordinate planes and the plane $3x+2y+6z=12$is a piecewise-smooth surface. Also, the vector field

$\mathbf{F}(x,y,z)=z\sin y\mathbf{i}-x\cos z\mathbf{j}+z\cos y\mathbf{k}$is defined throughout the region enclosed by $\mathrm{S}$.

Divergence Theorem can be applied only for smooth or piecewise-smooth surface where vector field $\mathbf{F}(x,y,z)$is defined on region enclosed by $S$.

Yes, the integral ${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS$, can be evaluated by means of Divergence Theorem.