Let F vector = <x,y,z> and use the Divergence Theorem to calculate the (nonzero) volume of some solid in IR3 by calculating a surface integral. (You can pick the solid).

Let $F\stackrel{\to }{\to }r=$ and use the Divergence Theorem to calculate the (nonzero) volume of some solid in IR3 by calculating a surface integral. (You can pick the solid).
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Cristiano Sears
Step 1
Consider the provided question,
Given, $\stackrel{\to }{F}=⟨x,y,z⟩=\xi +yj+zk$
Let s be a closed surface enclosing some volume V.
Find ${\int }_{s}\stackrel{\to }{F}\cdot Nds$
By gauss divergence theorem,
${\int }_{s}\stackrel{\to }{F}\cdot Nds={\int }_{V}÷\stackrel{\to }{F}dv$
Since, $\stackrel{\to }{F}=\xi +yj+zk$
div $\stackrel{\to }{F}=\frac{\partial }{\partial x}\left(x\right)+\left(\partial y\right)\left(y\right)+\left(\partial z\right)\left(z\right)$
=1+1+1
=3
Step 2
Now, find the (nonzero) volume V of some solid in ${R}^{3}$ by calculating a surface integral.
${\int }_{s}\stackrel{\to }{F}\cdot Nds={\int }_{V}÷\stackrel{\to }{F}dv$
$={\int }_{V}\left(3\right)dv$
$=3{\int }_{V}1dv$
${\int }_{s}\stackrel{\to }{F}\cdot Nds=3V$
$V=1.3{\int }_{s}\stackrel{\to }{F}\cdot Nds$
Hence, nonzero volume, $V=\frac{1}{3}{\int }_{s}\stackrel{\to }{F}\cdot Nds$