${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS$,where $S$is the portion of the surface

$x=z^3$that lies between the planes $y=13$and $y=42$and where

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 calculated using the Divergence Theorem, where the surfacerole="math" localid="1650777168836" $x=z^3$is the piece of the surface $S$that falls between the planes $y=13$and$y=42$

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 $n$denotes the outwardly unit normal vector

The surface integral is converted to a triple integral using the Divergence Theorem. In $R^3$, the triple integral is calculated over the bounded region $W$. A region $W$can only be found inside closed surfaces. A closed surface encloses a certain area. As a result, the Divergence Theorem does not apply to non-closed surfaces.

The surface $S$is the portion of the surface $x=z^3$that is not closed and lies between the planes $y=13$and $y=42$. As a result, it does not enclose any area.

No, the integral ${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS$, cannot be evaluated by means of Divergence Theorem.