Answer
$F=-W k$
Work Step by Step
The Divergence Theorem states that $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $
Here, $S$ is a closed surface and $E$ is the region inside that surface.
Here we have $\iint_S fc \cdot n dS=\iiint_Ediv (fc) dV$
or, $\iint_S fc \cdot n dS=\iiint_E f( \nabla \cdot c) +(\nabla f) \cdot c dV$
or, $\iint_S fc \cdot n dS=\iiint_E f(0) +(\nabla f) \cdot c dV$
or, $\iint_S fn \cdot c dS=\iiint_E (\nabla f) \cdot c dV$
and $\iint_S f \cdot n dS=\iiint_E (\nabla f) dV$
$\nabla P=\dfrac{\partial p}{\partial x}i+\dfrac{\partial p}{\partial y}j+\dfrac{\partial p}{\partial z}k=\rho g k$
Thus $F=-\iiint_E (\nabla P) dV=-(\rho g k) \iiint_E dV$
Here, $\rho$ is the density of the liquid and $\iiint_E dV$ is the volume of the solid.
We know that the weight of the liquid is given by $W=\rho g v$
Hence, we have $F=-W k$
