In analyzing the hydrostatic force on an incline plane, we need to set an appropriate axis. For now, we assume that the fluid surface is open to the atmosphere. Let the plane in which the surface lies intersect the free surface at 0 and make an angle  with this surface. We define the x-y coordinate system where 0 is the origin, y is directed along the surface with x directed out of the page. The area on which the force acts is arbitrary. The illustration below sets the stage for our analysis.

Note that in drawing the x-y­ coordinate axis, we have panned the x-axis from shooting out of the page to lying on it so that we are looking at the area from inside the fluid.  Our object is to determine the direction, location and magnitude of the result force acting on one side of the area due to the liquid in contact with the area.

At any given depth, h, the force acting on dA is  and is perpendicular to the surface. We find this resultant force by summing these differential forces over the entire surface.

where . Assuming constant  and ,

We observe here that the integral appearing in the above equation is the first moment of the area with respect to the x axis, which is given by the equation

where  is the y coordinate of the centroid measured from the x axis which passes through 0. This allows us to write

or equivalently,

where  is the vertical distance from the fluid surface to the centroid of the area. The angle  has dropped out of the equation telling us that the magnitude of the force is independent of the angle, instead it depends on the specific weight of the fluid, total area, and depth of the centroid of the area below the surface. As for the direction, we argue that since all the differential forces we summed to obtain  are perpendicular to the surface, the resultant  must also be perpendicular to the surface.

It may be suggested that the resultant force should pass through the centroid of the area. This, however, is not the case. Further calculations tell us otherwise.

We will take moments about the x axis by summing the y-coordinate of each of the differential force dF multiplied by its respective dF and equating it to the y-coordinate,  of the resultant force  multiplied by .

Since , we have

Now, the integral in the numerator is the second moment of the area (moment of inertia), , with respect to an axis formed by the intersection of the plane containing the surface and the free surface. Thus, we have

Using the parallel axis theorem, we will reexpress  as

where  is the second moment of the area with respect to an axis passing through its centroid and parallel to the x-axis. Thus,

Notice that the resultant force does not pass through the centroid but is always below it, since .

Using similar calculations, we can find the x coordinate, , for the resultant force by summing moments about the y axis.

where  is the product of inertia with respect to the x and y axes. We now employ the parallel axis theorem for the product of inertia given by

to write

where  is the product of inertia with respect to an orthogonal coordinate system passing through the centroid of the area and formed by the translation of the x-y coordinate system.

The result force  acts at the point given by the coordinates  and we label this point as the centre of pressure. To facilitate the use of the formulas, we usually use the results of moment of inertia of either the x or y axis passing through the centroid for various geometric shapes.

NOTE: For an explanation of the first and second moment of area, I suggest you view the videos, on which I talked about it briefly.