= Solution
Take $z$ positive downwards, and let the surface pressure be $p_0$. Static force balance gives the <hydrostatic pressure> equation $dp/dz=\rho g$. Hence
$$
\boxed{p(z)=p_0+\rho gz}.
$$
On a submerged body's surface, pressure traction is $-p\boldsymbol n$, where $\boldsymbol n$ points out of the body. The <divergence theorem> gives the resultant
$$
\boldsymbol F=-\int_{\partial V}p\boldsymbol n\,dS
=-\int_V\nabla p\,dV=-\rho gV\boldsymbol e_z.
$$
Thus \b[the buoyancy is upward with magnitude $\rho gV$], proving <Archimedes' principle>. The constant surface pressure makes no contribution because $\int_{\partial V}\boldsymbol n\,dS=0$.
For a floating iceberg of total volume $V$, the displaced seawater volume is $V-V_I$. Neglecting air buoyancy, equality of weight and buoyancy gives $\rho_IgV=\rho_wg(V-V_I)$. Consequently
$$
\boxed{V=\frac{\rho_w}{\rho_w-\rho_I}V_I},
$$
with $0<\rho_I<\rho_w$, as required for partial flotation.
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