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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.

$$\int_{\partial \Omega}(P_0 + \rho g h)d\vec{S}=\text{vol}(\Omega)\rho g\vec{e_2}.$$

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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)

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(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}


enter image description here

(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.


3D image of first Figure

3D image of second Figure

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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}

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  • $\begingroup$ Except that there is no water inside $\partial\Omega$! Ok, I know you know but that's the one single important thing to say as Archimedes theorem has a limited domain of validity because of the part of the argument you did not produce… $\endgroup$ – user154997 Aug 7 '17 at 16:00
  • $\begingroup$ @Spirine In case it wasn't clear, your style edit is inappropriate. Do not submit it again. (If you don't understand the notation, then it is your responsibility to think about it more carefully. The post is correct as is.) $\endgroup$ – Emilio Pisanty Aug 7 '17 at 16:07

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