Answer
$F=-W k$
Work Step by Step
Consider $\iint_S fc \cdot n dS=\iiint_Ediv (fc) dV$
Re-arrange as: $\iint_S fc \cdot n dS=\iiint_E f( \nabla \cdot c) +(\nabla f) \cdot c dV$
This implies that $\iint_S fc \cdot n dS=\iiint_E f(0) +(\nabla f) \cdot c dV$
and $\iint_S fn \cdot c dS=\iiint_E (\nabla f) \cdot c dV$
Thus, $\iint_S f \cdot n dS=\iiint_E (\nabla f) dV$
Here, we have $\nabla P=\dfrac{\partial p}{\partial x}i+\dfrac{\partial p}{\partial y}j+\dfrac{\partial p}{\partial z}k=\rho g k$
$F=-\iiint_E (\nabla P) dV$
or, $F=-(\rho g k) \iiint_E dV$
where, $\rho$ denotes the density of the liquid and $\iiint_E dV$ shows volume of the solid.
The weight of the liquid, $W=\rho g v$
Hence, our result is $F=-W k$
