Consider an infinitesimal area \(dA\) as shown. On this area, due to pressure a force \(\overrightarrow{dF}\).
\[\overrightarrow{dF} = P\overrightarrow{dA}\]
This force has 3 components. By careful observation, we can say that,
\[dF_x = dF \cos \phi \sin \theta\]
\[dF_y = dF \cos \phi \cos \theta\]
\[dF_z = dF \sin \phi\]
\[\hat{n} = \cos \phi \sin \theta \hat{i} + \cos \phi \cos \theta \hat{j} + \sin \phi \hat{k}\]
Some other trivial observations are,
\[{r \over l} = {R \over L} ( = \tan \phi)\]
\[h \buildrel \rm def \over = R - r \cos \theta\]
Now, we have,
\[\overrightarrow{dF} = \rho g h (r d\theta dl) \hat{n}\]
\[= \rho g (R - r \cos \theta) ( r d \theta \csc \phi dr) \hat{n}\]
Integrating, we get,
\[\overrightarrow{F} = \rho g H R^2 \hat{i} + \rho g H R^2 {\pi \over 6} \hat{j} + \rho g R^3 {\pi \over 2} \hat{k}\]
Note that,
\[F_x = F_z = (\text{projected area})(\text{Pressure at mean height})\]
\[F_y = (\text{Weight of liquid displaced by body})\]
Which is generally true.