At a radial distance of r, the surface is restricted to a thin band with circumference 2πr. If the surface is sloping upward at this point, the band consumes more area. The surface area is multiplied by the length of the upward vector, which has components 1,f′(r). (Since the curve is spinning around the y axis, x and r are synonymous.) Thus we have the following formula.
Area = 2π ∫r×sqrt(1 + f′(r)2)
I know, I didn't prove that rigorously. It's a typical sort of δ ε proof that consumes many pages. For small δ, f is nearly linear, and the inside of the annulus at r isn't that far away from the outside of the annulus at r+δ, and the tilted annulus approaches the surface, and the areas are almost the same, and so on and so on.
Let's check this with the sphere. If q is the square root of 1-x2, then f′ is -x/q. Square this and add 1, then take the square root, giving 1/q. Multiply by r and we have the same integral as before.
Next rotate f about the x axis. Again we find thin tilted bands as the intervals shrink to zero. Each band has radius f(x) and tilt factor sqrt(1+f′(x)2). This gives the following formula.
Area = 2π ∫f(x) × sqrt(1+f′(x)2)
Plug in the equation of the sphere, f(x) = sqrt(1-x2). This time the integrand is simply 1. Integrate from -1 to 1 to get 4π as expected.
Let a cone have radius r at its base, and height h. Orient the cone horizontally, with its apex at the origin. The cone is now a surface of revolution about the x axis, with f(x) = rx/h. Now f′(x) = r/h. Apply the earlier formula and get the following for the area of the cone, including its base.
Area = πr2 + πr×sqrt(h2+r2)
Consider another example, which is rather unusual. Let f(x) = 1/x, as x runs from 1 to infinity. The volume of this infinitely tapered funnel is the integral of 1/x2 from 1 to infinity, which is 1. What is its surface area?
Square the derivative to get 1/x4, then add 1 and take the square root, giving sqrt(1+x4)/x2. Multiply by f, giving a denominator of x3. The numerator is larger than x2, so the integrand is larger than 1/x. The integral does not converge, and the surface area is infinite. We have found a shape with finite volume and infinite surface area. You can fill it up with paint, but you can't paint it.
There are other formulas for surfaces of revolution that are based on the centroid.