${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS$, where $S$is the cone $y=\sqrt{x^2+z^2}$between $y=0$and $y=4$and where$\mathbf{F}(x,y,z)=xyz\mathbf{i}+2^{xyz}\mathbf{j}+\frac{x}{y^2+1}\mathbf{k}.$

Given the vector field $\mathbf{F}(x,y,z)=xyz\mathbf{i}+2^{xyz}\mathbf{j}+\frac{x}{y^2+1}\mathbf{k}.$The objective is to find 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 cone $y=\sqrt{x^2+z^2}$between$y=0$ and$y=4$

Divergence theorem states that

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

$\int {\int }_{s}F(x,y,z).nds=\int \int {\int }_{w}divF(x,y,z)dV.........\left(1\right)$

Where n is the outwards unit normal vector

A cone with equation $y=\sqrt{x^2+z^2}$between $y=0$and $y=4$is not smooth or piece-wise smooth surface because of vertex $(0,0,0)$. At the vector the normal vector is $0$. So this surface is not smooth or piece wise smooth surface

Divergence theorem can only be applied to smooth or piece wise smooth surface

No, the integral${\iint }_{S}F(x,y,z)·ndS$, cannot be evaluated by means of divergence theorem