For the most general calculation of the surface area of an arbitrary potato-shaped object, see http://en.wikipedia.org/wiki/Arc_length. Basically it says that to get the length of a function f(x) on the interval [a,b], you use approximations by calculating the length of straight lines between f(x_k) and f(x_(k+1)) for k = 0, ..., n-1 with x_0 = a and x_n = b, and then you let n -> infinity to get better and better approximations. So using the right integrals this method will give you the surface area of anything, as long as you know the shape of the object.
In case of spheres or circles with radius r, surface area S(r) and volume V(r), things are easy and S(r) = V'(r). So since the volume of a sphere is 4/3 pi r^3, the surface area is 4 pi r^2. Half a sphere would thus have a surface area of 2 pi r^2. And a circle has "volume" (surface area) pi r^2, so the "surface area" (circumference) is 2 pi r.
I don't quite remember the integrals needed to find the surface area of a curve, say z=sqrt(1-x^2-y^2), the top half of the unit sphere. I do remember the triple integrals needed to find the volume, but what was needed to find the surface areas?
Also, I don't remember the surface area for the equation p=1, in spherical coordinates, so knowing that would also help, thanks!
Likewise for cylindrical coordinates.