Proof of Archimedes Principle Say I have an arbitrary body $\Omega$ and surface $\partial \Omega$ submerged in a hydrostatic fluid of density $\rho$ under the influence of gravity. How does one show Archimedes Principle? i.e.
$$\int_{\partial \Omega}(P_0 + \rho g h)d\vec{S}=\text{vol}(\Omega)\rho g\vec{e_2}.$$
 A: This is a straightforward application of the divergence theorem. First, split the integral on the left into its vectorial components:
$$
\int_{\partial \Omega}(P_0 - \rho g z)\left(-\mathrm d\mathbf{S}\right)
=
\sum_j \mathbf e_j\int_{\partial \Omega}(-P_0 + \rho g z)\mathbf e_j\cdot \mathrm d\mathbf{S}
.$$
Then, apply the divergence theorem:
\begin{align}
\int_{\partial \Omega}(P_0 - \rho g z)\left(-\mathrm d\mathbf{S}\right)
&=
\sum_j \mathbf e_j\int_{ \Omega}\nabla\cdot\left[(-P_0 + \rho g z)\mathbf e_j\right] \mathrm dV
\\&=
\sum_j \mathbf e_j\int_{ \Omega} \rho g \delta_{zj} \mathrm dV
\\&=
\rho g \mathbf e_z\mathrm{vol}(\Omega)
.\end{align}
A: This answer uses Figures instead calculus as in the excellent Emilio Pisanty's answer. 
($\:h\:$ = depth of immersed horizontal surface from the rest open surface of the fluid)

(1) Firstly : Horizontal hydrostatic pressure force cancels out  
Cut your body horizontally and take any section with infinitesimal height $\:dh_{1}\:$ as in Figure. Then
\begin{align}
\mathbf{F}_{\text{horizontal}}&=\sum_{m=1}^{m=N}\left(- p\right)\Delta\mathbf{s}_{m}=\sum_{m=1}^{m=N}\left(- p\right)\left[\Delta\mathbf{r}_{m}\boldsymbol{\times}\left( dh_{1}\mathbf{k}\right)\right]
\nonumber\\
&=\left(- p\right)\underbrace{\left(\sum_{m=1}^{m=N}\Delta\mathbf{r}_{m}\right)}_{=\mathbf{0}}\boldsymbol{\times}\left( dh_{1}\mathbf{k}\right)=\mathbf{0}
\tag{01}
\end{align} 
Don't worry if the perimeter of your cross section is a closed curve instead of a closed polygon. Then we have differentials $\:d\:$ in place of Deltas $\:\Delta \:$ and integrals instead of sums
\begin{align}
\mathbf{F}_{\text{horizontal}}&=\oint\left(- p\right)d\mathbf{s}=\oint\left(- p\right)\left[d\mathbf{r}\boldsymbol{\times}\left( dh_{1}\mathbf{k}\right)\right]
\nonumber\\
&=\left(- p\right)\underbrace{\left(\oint d\mathbf{r}\right)}_{=\mathbf{0}}\boldsymbol{\times}\left( dh_{1}\mathbf{k}\right)=\mathbf{0}
\tag{02}
\end{align}


(2) Secondly : Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object. 
Of course if this was a plate alone in a fluid then the upward buoyant force exerted by the fluid would be
\begin{equation}
\mathbf{B}_{\text{buoyant}}= p\left(h\right)\mathbf{S}_{A}-p\left(h-dh_{1}\right)\mathbf{S}_{A}
=\rho g\underbrace{dh_{1}S_{A}}_{V_{A}}\mathbf{k}=\left(\rho g V_{A}\right)\mathbf{k}  
\tag{03}
\end{equation}
Now, on the first plate $\:A\:$ of horizontal surface $\:S_{A}\:$ and infinitesimal height $\:dh_{1}\:$ put the next plate $\:B\:$ of the body of horizontal surface $\:S_{B}\:$ and infinitesimal height $\:dh_{2}\:$. Then
\begin{align}
\mathbf{B}_{\text{buoyant}}&= \underbrace{\left[-p\left(h\right)\left(-\mathbf{S}_{A}\right)\right]}_{A\: bottom}+\underbrace{\left[-p\left(h-dh_{1}\right)\left(\mathbf{S}_{A}-\mathbf{S}_{B}\right)\right]}_{step}+\underbrace{\left[-p\left(h-dh_{1}-dh_{2}\right)\mathbf{S}_{B}\right]}_{B\: top}
\nonumber\\
&=\rho g\underbrace{dh_{1}S_{A}}_{V_{A}}\mathbf{k}+\rho g\underbrace{dh_{2}S_{B}}_{V_{B}}\mathbf{k}=\rho g \left(V_{A}+V_{B}\right)\mathbf{k}
\tag{04}
\end{align}
Any body could be cut in horizontal plates of finite surface area and infinitesimal height.



