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The following is a simple proof of divergence theorem :

$$ \begin{align} & \phantom{={}} \iiint (\nabla\cdot F) \, dV \\ & = \iiint \frac{(∂F_x)}{∂x} \, \text{dx dy dz} + \iiint \frac{(∂F_y)}{∂y} \, \text{dy dx dz} + \iiint \frac{(∂F_z)}{∂z} \text{dz dx dy} \\ & =\iint F_x\, \text{dy dz} + \iint F_y \, \text{dx dz} + \iint F_z \text{dx dy} \\ & =\iint F_x dS_x +\iint F_y \, dS_y +\iint F_z \, dS_z \end{align} $$ (since the projected area onto (for eg) $x$-$y$ plane is the $z$-component of area vector)

$$ =\iint (F_x+F_y+F_z)\cdot( dS_x+ dS_y+ dS_z) $$

$$ =\iint F\cdot dS$$

Is this proof enough? Then why we use the long proof given in most textbooks?

Michael Hardy
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faheemahmed400
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    Because what you've written doesn't mean anything, sadly. – Ted Shifrin Oct 12 '13 at 14:58
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    There is a problem in the very first step. To express a triple integral as a single iterated integral with, say, the order of integration $dx dy dz$, you must assume that the region of integration has the form $f(y,z) \leq x \leq g(y,z)$ for $y,z$ in some domain in the plane. This is rarely possible. – Paul Siegel Oct 12 '13 at 14:59
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    @PaulSiegel: Even if you assume the region is $x$-, $y$-, and $z$-"simple," this isn't a proof without a whole lot more. – Ted Shifrin Oct 12 '13 at 15:00
  • That said, any suitably nice region can be approximated arbitrarily well by the union of cubes, and some version of your calculation works on a cube. This is the idea behind many of those long proofs. – Paul Siegel Oct 12 '13 at 15:01
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    It looks more or less correct to me on a simple region. (Though the last step should be written out correctly using a parametrization of the boundary.) – Paul Siegel Oct 12 '13 at 15:02

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