areacalculus
I am trying to find the surface area of a sphere of radius r by adding the circumferences of circular rings whose radii range from 0 to r. Because the radius ranges from 0 to r, the surface area equals
$$SA = 2\int_{0}^{r}2\pi rdr$$
$$SA = 2\pi r^2 $$
What is wrong about this approach? I know it doesn't take into account the slope but as the height of each circular ring gets infinitesimally small if the rings were stacked on top of each other wouldn't they accurately approximate a sphere?
You might want to look for a treatise by Archimedes on "the method", in which he does this computation using Cavalieri's principle (more or less) and a rather clever mechanical argument. There's a good case to be made that in doing so, he was inventing some of the key ideas of calculus, and the proofs he offers don't really meet the modern standard of rigor, but nonetheless, it's a remarkable achievement.
This is the same as for the circumference of a circle. If you approximate it by a polygon, the perimeter will tend to the correct value. But if you approximate it by horizontal segments, the total length of the segments is always twice the diameter, and $\pi=2$ !
Best Answer
You might compare your approach to the surface of a sphere with the approach taken in Why is area of a surface of revolution integral $2\pi y~ds$? not '$dx$'? or Areas versus volumes of revolution: why does the area require approximation by a cone?
In each of those two questions, someone tried to measure the area of a surface of revolution (which your sphere is) essentially by fitting a "layer cake" of thin disks inside the surface and then measuring the curved lateral sides of the disks while ignoring the exposed flat areas of the disks. You are measuring the exposed flat areas (in the upper part of the sphere, this is the part of the top of each disk that is not covered by the disk above it) while ignoring the curved lateral sides.
The problem is that either the curved surface of the disk or the exposed flat area is less than the surface area of a frustum that (on the upper half of the sphere) connects the upper edge of one disk to the upper edge of the disk below. If you were measuring the area of a cone, that frustum would be the exact area; when measuring a sphere, a "slice" of the sphere looks a lot more like a conical frustum than like either the side of a cylinder or an annular ring, and its area is relatively much closer to that of the frustum than to the areas of either of the other two objects (where "relatively close" means the ratio between the two areas is close to $1$).
In this answer to one of the other questions I detailed how the calculations differ when computing the area of a cone, where it is easy to see how this works (because we have other ways of computing the area of a cone without calculus, and because the relative error is the same no matter what part of the cone you look at). For a sphere, the argument is a bit more subtle, and the error in your calculation varies over the surface of the sphere: at the "pole" of the sphere (near the axis) your approximation is quite good, and could be used to find the area of a small spherical cap with pretty good accuracy; but near the "equator" of the sphere (where the radius of your ring is nearly $r$) you are missing almost all the area of each slice of the sphere no matter how thin you slice it.