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