Magdeburg Hemispheres The Magdeburg Hemisphere experiment was the experiment that showed the effect of pressure differences on a vacuumed sphere. 
We know that the Force caused by pressure is $\Delta p A$ and so you can calculate the force by using the area of the base of one of the hemispheres of the the vacuumed ball. 
It is intuitive to me that the area we use is the area created by the diameter of the sphere, a line through the center of the sphere, but is there a clearer reason that we use this area? I'm not sure how to explain why this "important" area. 
 A: Integrating the force over the surface of the half-sphere yields
$$F=\int_\text{C}\Delta p\hat{\mathbf{n}}dS=\int_0^{\pi/2}\int_0^{2\pi}\Delta p\left(
\begin{array}{c}
 \sin (\theta ) \cos (\phi ) \\
 \sin (\theta ) \sin (\phi ) \\
 \cos (\theta ) \\
\end{array}
\right)R^2\sin(\theta)\,d\phi d\theta=\left(
\begin{array}{c}
 0 \\
 0 \\
 \pi R^2\Delta p \\
\end{array}
\right)$$
where $C$ is the surface of one hemisphere, assuming that the base is sitting on the $xy$-plane and that the radius is $R$. Thus $|F|=A\Delta p$, as you originally wrote.
Mathematica proof:
Integrate[\[CapitalDelta]p Sin[\[Theta]] r^2 {Cos[\[Phi]] \
Sin[\[Theta]], Sin[\[Theta]] Sin[\[Phi]], Cos[\[Theta]]}, {\[Theta], 
  0, \[Pi]/2}, {\[Phi], 0, 2 \[Pi]}]


{0, 0, [Pi] r^2 [CapitalDelta]p}

General Case
The general case can be argued in two ways. The traditional way is by geometry, which is what dmckee used in his answer. Here is a second way:
Consider an arbitrary surface $C$ (not necessarily a hemisphere), with a flat base $B$ with area $A$. The total surface of the object is, of course, $D=B\cup C$. The total force acting on the object is 
$$\mathbf{F}=\int_D\Delta p\hat{\mathbf{n}}\,dS=\mathbf{0}.$$
The reason it's zero is because otherwise, the object would spontaneously accelerate without any source of propulsion, which contradicts reality.
We can then break up the integral to give
$$\mathbf{0}=\int_B\Delta p\hat{\mathbf{n}}\,dS+\int_C\Delta p\hat{\mathbf{n}}\,dS=A\Delta p\hat{\mathbf{n}}_B+\int_C\Delta p\hat{\mathbf{n}}\,dS
\\
\Rightarrow \int_C\Delta p\hat{\mathbf{n}}\,dS=-A\Delta p\hat{\mathbf{n}}_B$$
where $\hat{\mathbf{n}}_B$ is a vector normal to the base $B$. Substituting $A=\pi R^2$ and $\hat{\mathbf{n}}_B=-\hat{\mathbf{z}}$ gives the hemisphere result I wrote above as a special case (but again, it doesn't have to be a hemisphere, it can be anything).
Note: implicit in the above derivation was that $\Delta p$ was constant. If $\Delta p$ is not constant, then the object can spontaneously accelerate (that's how balloons work).
A: At all points on the sphere the pressure points normal to the surface (because that it what pressure does...). 
That said this system has a preferred direction: the line between the hoops where we hook the harness on for the horse to pull. And we have built the device such that the joint between the two halves lies at the equator perpendicular to that axis.
Now, the system therefore has axial symmetry, we are are only interested in the component of force parallel to the symmetry axis (because the components transverse to the axis cancel out when we add up the bits).
Measuring angle $\theta$ from the + end of the symmetry axis (either one, just pick your favorite), the axial part of the force from the pressure on a infintesimal area element $\mathrm{d}A$ (or very small element $\Delta A$ if you don't have calculus) is $P \sin (\theta ) \mathrm{d}A$ or the projection of that area onto a place normal to the axis.
Therefore the total force is that exerted on a circle of the same diameter.
This can be shown rigorously in a few lines of calculus which is what DumpsterDoofus set up (I've really just stated the same integral in a lot of words).
