>>11
Not really... It depends on the coordinates/metric you're using.
The full differential geometry answer is that dA = sqrt(det(metric))
polar coordinates: metric = ds^2 + s^2 dphi^2
dA = s^2 ds dphi
in higher dimensions...
spherical coordinates: metric = dr^2 + r^2 dtheta^2 + (rsin(theta))^2 dphi^2
dV = r^2 sin(theta) dr dtheta dphi
If you're doing electrostatics, you're probably concerned with an integral over a 2D surface in 3D Euclidean space; that is a far more complicated issue.
If you're brute forcing Gauss' law, you'll probably integrate over a spherical shell of radius R in spherical coordinates... except you're not because of
>> \A=((R^2)*y)/2
polar coordinates? surface of a cylinder at fixed z? ???
>> \R=h*Tan[x]
if x = elevation angle, then you're almost certainly integrating over a surface of constant z = h
Continuing my reverse engineering, it seems you have a charged particle at the origin, and you're integrating over the surface z=h in a region bounded by a circle of radius htan(a)
(cylindrical coordinates)
integral(K/r^2 rhat * dA) =
integral(|K/r^2| (rhat * zhat) |dA|) =
integral(|K/r^2| h/r |s ds dphi|) =
K h 2pi integral(s/sqrt(h^2 + s^2)^3 ds)
If you insist on doing this in spherical coordinates, I can't tell you the answer off the top of my head, because dA will be a horrible function of theta.
tl;dr
Be clear about what you're asking about.
Integrating over a 2D surface in a 2D world is
very different from integrating over a 2D surface in a 3D world.