If there's a constant force F in the y-direction acting over some surface S, the pressure is going to vary over the surface.
Would it be right to say that the total pressure on the surface is the sum of dP's at each point on the surface? (since pressure is a scalar quantity, so i'm assuming it's algebraically additive)
Would this 'total pressure' have any physical significance?
You can define "total pressure" like that if you like, but it doesn't have any particular physical significance. The main significant quantity is the total force due to pressure on an object, but that is not equal to this total pressure you have described, but instead equal to a vector sum over the surface: $$\vec F=\oint Pd\vec A$$
Where $d\vec A$ is an infinitesimal vector of magnitude $dA$ (the infinitesimal area) in the direction normal to the surface.
For a constant pressure, you can show that this force is equal to the pressure times the cross-sectional area over which it is applied.
You can only add Forces. On an infinitesimal surface element $dS$ the force $dF = p dS$ with pressure $p$ acts on it. If you add These Forces up, you get a total force on the surface
$F = \int_{S} p dS$
or in vectorial form
$\vec{F} = \int_S p \vec{n} dS$
because the pressure acts every time in direction of the unit normal $\vec{n}$ of the surface.
The average pressure you obtain by calculating $\frac{|\vec{F}|}{S_{tot}}$ with $S_{tot} = \int_S dS$. This is the effective pressure and measures how many normal force acting on a Body per surface element in average.
Therefore, a weighted Addition (it is weighted with surface elements) holds for the pressures.
In engineering we call this distributed load and when we calculate static equilibrium, eg. if at the start point and end point there are pin and roller, objective would be to calculate reactions at those supports. In this case continuous load can be replaced with one concentrated force acting at the centroid of the area, calculated using formula provided in one of the previous answers by @Chris, we commonly for this kind of load or pressure use $q$ in kN/m or kN/m$^2$. $$\vec F=\int qd A$$
If internal forces, moments and later stress and deflection of this shell or curved beam are considered than continuous load must be used as such.
here are some sources:distributed forces and centroids
