Surface Area of a Sphere Without Calculus I'm having trouble with finding the surface area of a sphere, without using any calculus.
What I thought, was that the surface area of a sphere is fundamentally an infinite number of rings, decreasing in size as you go up or down the actual circumference, stacked up on top each other. By adding the infinite circumferences of the rings up, you should be able to obtain the actual value of the surface area, as by the formula, $4\pi r^2$. The more circumferences you add up in your sum, the closer your value will be to $4\pi r^2$:
If I were to decrease the radii of each of the circumferences by a constant, I would end up calculating the surface area of a cone. Because of this, I used the equation of a circle, $x^2 + y^2 = r^2$. Suppose we divide a sphere of diameter $10$ into five layers with a line of symmetry cutting through the centre horizontally; each layer would have a constant height of 2 units. I used the centres of each of the layers as the y values; the y value for the centre layer would be $0$, giving an $x$ value of $5$. The $y$ value for the second layer would be 2 (half of the centre layer's height added to half of the second layer's height), which would give the $x$ axis value of root $21$. Finally, the $y$ axis value for the third layer would be 4, which would give an $x$ axis value of 3. To get the total surface area produced by these five layers, I multiplied all of the x axis values (5 of them including the bottom hemisphere) by 2π to get the circumference, and then by 2 (the height) to get the surface area. After adding all of this up, I got $80.66\pi$ as the final value. This is a lot less than the actual value of $100\pi$ calculated by the formula $4\pi r^2$.
This was only the first step as I thought that my value of the surface area would get closer as I measured the surface area using more and more layers. Here is what I got:
005 Layers: 80.66π 
015 Layers: 78.96π
025 Layers: 78.73π
035 Layers: 78.66π
055 Layers: 78.60π
095 Layers: 78.57π
155 Layers: 78.55π
495 Layers: 78.54π
What I had predicted was not the case. A huge chunk, 20π (a fifth of the actual value) is missing from my perceived sphere. Why is this? Why is the value of the surface area of a sphere getting farther and farther away as I increase the number of layers? 
 A: 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. 
A: I find your description a bit difficult to follow, but it looks like you're neglecting the fact that the "layers" are are not slices of cylinders, but slices of cones, and therefore are wider than their projection on their axis.
For example, your radius-5 circle contains points at approximately $(1,4.9)$, $(2,4.6)$, $(3,4)$.
For the band covering $x$-values between $1$ and $3$ you would estimate its length as $4.6\times 2\pi$. However, its width should not be $2$, but the distance between $(1,4.9)$ and $(3,4)$ -- that is, $\sqrt{(3-1)^2+(4-4.9)^2} \approx 2.2$.
A: In fact if you have one "layer" for the whole sphere you should get the right result.
As you add more layers, your cylinders get smaller on average, which is why the value here is decreasing.
The problem you have is that the smaller the cylindrical layers get, the greater the angle the surface of the sphere makes with the side of the cylinder, and the greater the error you get from estimating the area of the slice of the spherical surface by the area of the cylinder (consider a thin cylinder at one of the poles, for example). If you correct for this, the calculation should work out.
Also what you are doing here is essentially computing an integral longhand, by taking the limit as the cylinders get smaller.
A: It's better to find the voulme, and then suppose it's a pyramid with the apex at the centre and the base as a surface.  The reason for this approach is that the surface may never reduce.  If you were trying to find the perimeter of the circle by using ever smaller squares, it never leaves 4, but the area of the circle, and hence the true circumference, is correctly found.
