53. Show that if ${\iint }_S\mathbf{F}(x,y,z)-\mathbf{n}dS=0$, it does not follow that $\mathbf{F}$ is conservative.

The given is the function $\int {\int }_{s}f(x,y,z)·nds=0$

The objective is to show that the function does not follow that $F$ is conservative

The goal is to demonstrate that if,

${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS=0$

It is not the case $\mathbf{F}$is conservative.

A vector field If and only if the following conditions are met, it is conservative.

Consider the vector field below:

$\mathbf{F}(x,y,z)=⟨y+z,2x+xz,2xy+y⟩$

This vector field is for

$F_1(x,y,z)=y+z,F_2(x,y,z)=2x+xz,and{F}_3(x,y,z)=2xy+y$

Take note of this:

role="math" $$\begin{gathered} \frac{\partial F_3}{\partial y}=\frac{\partial }{\partial y}(2xy+y) \\ =2x+1 \end{gathered}$$

role="math" $$\begin{gathered} \frac{\partial F_2}{\partial z}=\frac{\partial }{\partial z}(2x+xz) \\ =x \end{gathered}$$

As a result, in this example,

$\frac{\partial F_3}{\partial y}\ne \frac{\partial F_2}{\partial z}$

Consequently, this vector field $\mathbf{F}(x,y,z)=⟨y+z,2x+xz,2xy+y⟩$is a vector field that is not conservative.

It is sufficient to establish that the integral is necessary ${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS=0$for the vector field that isn't conservative$\mathbf{F}(x,y,z)=⟨y+z,2x+xz,2xy+y⟩$.

Proof:

To establish the above, use the Divergence Theorem.

According to the Divergence Theorem,

"Let $W$be a bounded region in ${\mathrm{ℝ}}^3$with a smooth or piecewise-smooth closed oriented surface as its boundary $S$. If a vector field exists,

$\mathbf{F}(x,y,z)$is defined on an open region containing $W$, then,

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

Here $\mathbf{n}$is the normal vector pointing outwards. "

First, determine the vector field's divergence $\mathbf{F}(x,y,z)=⟨y+z,2x+xz,2xy+y⟩$.

A vector field's divergence is defined as follows:

$=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}.$

Then there's the vector field's divergence $\mathbf{F}(x,y,z)=⟨y+z,2x+xz,2xy+y⟩$will be,

Then there's the vector field's divergence $F(x,y,z)=<y+z,2x+xz,2xy+y>willbe,divF=\nabla ·F$

$$\begin{gathered} =\frac{\partial }{\partial x}(y+z)+\frac{\partial }{\partial y}(2x+xz)+\frac{\partial }{\partial z}(2xy+y) \\ =0+0+0 \\ =0 \end{gathered}$$

Now evaluate the integral using the Divergence Theorem (1)${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS$as follows:

${\iint }_S\mathbf{F}·\mathbf{n}dS={\iiint }_{R}div\mathbf{F}dV$

$$\begin{gathered} ={\iiint }_{R}0dV \\ =0 \end{gathered}$$

As a result, in the case of non-conservative vector fields, $\mathbf{F}(x,y,z)=⟨y+z,2x+xz,2xy+y⟩$,

${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS=0$

As a result, it has been established that, if ${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS=0$, it does not follow that $\mathrm{F}$ is conservative.