## Derivation of Volume Formula Using Divergence Theorem

Theorem
Let
$S$
be the boundary surface of a volume
$V$
andlket
$\mathbf{n}$
be the outward normal. Then
$V= \int \int_S x \: dy \: dz = \int int_S y \: dx \: dz = \int \int_S z \: dx \: dy = \frac{1}{3} \int \int_S x \: dy \: dz + y \: dx \: dz + z \: dx \: dy$

Proof
From the divergence theorem
$\int \i \int_S (\frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} dx \: dy \: dz = \int \int _S F_1 dy \: dz + F_2 \: dx \: dz + F_3 \: dx \: dy$

If
$\mathbf{\nabla} \cdot \mathbf{F}=1$
then again from the divergence theorem
$\int \int \int_V \mathbf{\nabla} \cdot \mathbf{F} \: dx \: dy \: dz = \int \int \int_V dx \: dy \: dz = \int \int_S F_1 \: dy \: dz + F_2 \; dx \: dz + F_3 \: dy \: dz$

But
$V = \int \int \int_V dx \: dy \: dz$

Hence
$V= \int \int_S F_1 \: dy \: dz + F_2 \; dx \: dz + F_3 \: dy \: dz$

Take
$\mathbf{F} = x \mathbf{i}$
then
$\mathbf{\nabla} \cdot \mathbf{F} =1$
and
$V= \int \int_S x \: dy \: dz$

Take
$\mathbf{F} = y \mathbf{j}$
then
$\mathbf{\nabla} \cdot \mathbf{F} =1$
and
$V= \int \int_S y \: dx \: dz$

Take
$\mathbf{F} = y \mathbf{i}$
then
$\mathbf{\nabla} \cdot \mathbf{F} =1$
and
$V= \int \int_S y \: dx \: dz$

Hence
$V= \frac{1}{3} \int \int_S x \: dy \: dz + y \: dx \: dz + z \: dx \: dy$
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