I believe that the point that
@kuruman is trying to make is not so much about integrals as it is about vectors.
Way back up in your original post, you spoke about getting the "total force" on a spherical object. You were going to calculate this total force in a simple way: average pressure times total surface area.
In effect, you were going to calculate "total force" by adding up all of the smaller forces on all of the little bits of surface area. The problems is that forces are vectors.
You cannot add up a bunch of non-parallel vectors by adding up their magnitudes.
If you wanted total force on a
flat surface under a
constant pressure then ##\vec{F} = P\vec{A}## would be correct. That would be the distributive law at work. ##\sum_i ( p_i \times a_i ) = p_\text{constant} \times \sum_i a_i##
But it turns out that the individual area elements here are not scalars. They are vectors. Directed areas. The magnitude of a directed area is just the surface area of the area element. The direction is that of a vector normal to a particular side of the surface. The multiplication of a directed area by a pressure is not technically the multiplication of two scalars yielding a scalar. Technically, it is the multiplication of a tensor (the Cauchy stress tensor) by the directed area vector yielding a force vector.
Do not panic. You'll likely get out of first year physics without ever learning about the
Cauchy stress tensor. Probably without learning about tensors at all. However, I do not want to leave you hanging. Nothing here is beyond the grasp of a willing student...
In component form, the Cauchy stress tensor is a 3 x 3 matrix with direction (x, y and z) on one axis and force components (x, y and z) on the other axis. Each of the nine terms gives a component of stress (force per unit area): "how much force in this direction exists across a unit of directed area in that direction".
In a fluid only the three terms on the main diagonal will be non-zero. Fluids cannot maintain
shear forces. Those main diagonal terms will all be equal to one another and will be equal to the pressure within the fluid. So all you have is an identity matrix scaled up by a factor of pressure.
When you matrix-multiply an area element by this tensor, it is just like multiplication of a vector by a scalar. You get a force in the direction normal to the surface of the area element.
Just as one would expect for the force resulting from a particular pressure on a particular area element.
If we bring this back to the problem at hand, we are trying to compute "total force" on an object. We assume that pressure is approximately constant and are trying to assert that:$$F = \sum_i \vec{a_i} \times P_\text{constant} = P_\text{constant} \times \sum_i \vec{a_i} = P_\text{constant} \times 4\pi r^2$$The difficulty is the same as for average speed versus average velocity:$$0 = |\sum_i \vec{a_i}| \neq \sum_i |\vec{a_i}| = 4 \pi r^2$$The magnitude of the sum is not the same as the sum of the magnitudes.
[One can easily see that the sum of directed areas for a sphere is zero. For every directed area on the one side of a sphere, there is an equal and opposite directed area diametrically opposed. The sum has to be zero. With some vigorous handwaving, one can assert that the same holds for any two-dimensional surface that encloses a volume.]
