Ok. I'm going to take a stab at this.
Essentially what we will do is calculate the weight of an infinitesimally thin column of water, then find the component of that force normal to the surface of the sphere, then integrate. It will be a little difficult to see without a diagram but here goes.
The
volume of a sphere in cylindrical coordinates is
\int_{0}^{2\pi}{\int_0^R{\int_{-\sqrt{R^2-\rho^2}}^{+\sqrt{R^2-\rho^2}}{\rho dz d\rho d\phi}}}
If we drop the rho and phi integration and perform just the z integration we get
2\rho \sqrt{R^2-\rho^2} d\rho d\phi
This is the volume of a column that lies at angle phi and runs from z=-\sqrt{R^2-\rho^2} to z=+\sqrt{R^2-\rho^2}
(phi doesn't appear in the expression because of the symmetry)
Multiplying by 3W/4\pi R^3, the weight-density of the water, we get
\frac{3W}{2\pi R^3}\rho \sqrt{R^2-\rho^2} d\rho d\phi
This is the weight of that column of water. If we performed the integration over rho and phi of this quantity as is, we would get W, the weight of the water in the sphere. The force of this infinitesimal column weight is in the vertical direction. What we want to do is find the component normal to the surface of the sphere and integrate
that over rho and phi. I'll let you take it from here. I got \frac{3}{8}W. Do you know if that is the answer?
