Answer
$F=\pi R^2\Delta p$
Work Step by Step
First consider an infinitesimal area on the surface of the sphere.
From the figure, the infinitesimal area is: $dA=R^2\cos\theta d\theta d\phi$
The direction of the above area $dA$ is radially outward.
The horizontal component of the area $dA$ is
$dA_h=dA\sin\theta$
or, $dA_h=R^2\cos\theta\sin\theta d\theta d\phi$
Therefore the horizontal force on the area $dA$ is given by
$dF=\Delta p dA_h$
Therefore, the total horizontal force acting on the sphere is
$F_h=\int \Delta p dA_h$
or, $F_h=\Delta pR^2\int_0^{\frac{\pi}{2}}\cos\theta\sin\theta d\theta \int_0^{2\pi}d\phi$
or, $F_h=2\pi\Delta pR^2\int_0^{\frac{\pi}{2}}\cos\theta\sin\theta d\theta$
or, $F_h=\pi\Delta pR^2\int_0^{\frac{\pi}{2}}\sin2\theta d\theta$
or, $F_h=2\pi\Delta pR^2$
or, $F=2\pi R^2\Delta p$
Therefore, the horizontal force on each hemisphere is
$F=\frac{F_h}{2}$
or, $\boxed{F=\pi R^2\Delta p}$
