I am asked to estimate the force due to hydrostatic pressure exerted to a hemispherical container as shown below, due to the fluid which fills the container to $h$. 
I first parameterized the surface using spherical coordinates:
$$r(\theta, \phi) = <R \sin(\theta)cos(\phi), R\sin(\theta)\sin(\phi), R\cos(\theta)>$$
The differential pressure is given by:
$$dF = -\vec{P} \cdot \vec{n} \ dA$$
Where we can define the pressure based on $P(\theta) = \rho g(h-R\cos(\theta))$ and $\vec{n} \ dA = <R^2 \sin^2(\phi)\cos(\theta), R^2 \sin^2(\phi)\sin(\theta), R^2 \sin(\phi) \cos(\phi)>$. Taking the magnitude of the force as a differential area $dA$ I can find the force to be
$$dF = \rho g (h-R\cos(\phi))R^2 \sin(\phi) d\phi d\theta$$
Which then needs to be integrated over the region $\pi/2 < \phi<\phi_0$ where $\cos(\phi) = h/R$ and $0 < \theta < 2\pi$. An answer provided frames the force to be:
$$F = 2\pi \rho R^2 \int_{\phi_0}^{\pi/2} (h-R\cos(\phi)) \sin(\phi) \cos(\phi)d\phi$$
Frankly, I am not sure where the $\cos(\phi)$ term is coming from. It is like they are only taking the force in the z direction, but I do not understand how that leads to the hydrostatic pressure force exerted on the surface. Could someone help explain this / explain where my solution has gone wrong?
Best Answer
Due to the cylindrical symmetry of the setup, only the forces in the z direction will affect the final integration. The other components will cancel. So, to avoid integrating over a vector $\mathrm{d}\mathbf{F}$, we only keep the z component in the integration.