${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}ds$,

where $S$is the surface of the torus

parametrized by $x=(3+\cos v)\cos u,y=(3+\cos v)\sin u,z=\sin v$and where$\mathbf{F}(x,y,z)=(3x-z)\mathbf{i}-x^{2}z\mathbf{j}+x{y}^7\mathbf{k}$.

Consider the vector field below:

The goal is to determine whether the integral${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}ds$can be computed using the Divergence Theorem in the case where the surface is torus parametrized as follows:

$x=(3+\cos v)\cos u,y=(3+\cos v)\sin u,z=\sin v$

Divergence According to the theorem,

"Let $W$ be a bounded region in $R^3$ with a smooth or piecewise-smooth closed oriented surface as its boundary $S$. If on an open region containing $W$, a vector field $F(x,y,z)$ is defined, then

${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS={\iiint }_{W}div\mathbf{F}(x,y,z)dV,".........\left(1\right)$

where $n$is the unit normal vector pointing outwards. "

The smooth surface $S$is torus parametrized by $x=(3+\cos v)\cos u,y=(3+\cos v)\sin u,z=\sin v$. Throughout the region encompassed by $S$, the vector field $\mathbf{F}(x,y,z)=(3x-z)\mathbf{i}-x^{2}z\mathbf{j}+x{y}^7\mathbf{k}$is defined.

Divergence Only smooth or piecewise-smooth surfaces with a vector field $F(x,y,z)$defined on the region encompassed by $S$can be used to prove the theorem.

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