In order to find the volume of a sphere radiud $R$, one way is to slice it up into a stack of thin, concentric disks, perpendicular to the $z$-axis. a disk at any point $z$ will have radius $r=\sqrt{R^2-z^2}$ and infinitesimal thickness $dz$, so the volume integral is:
$$ V = \int dV = \int_{-R}^{R} \pi r^2 dz = \int_{-R}^{R} \pi (R^2-z^2) \,dz $$
The above integral of course evaluates to $\frac{4}{3}\pi r^3$, the desired answer. Now, when I tried to do the same thing for the surface area, I treated it as a thin, hollow shell, sliced it up into a stack of concentric rings (or simply the outer surface area of the disks). The area integral in that case is:
$$ A = \int dA = \int_{-R}^{R} 2\pi r \,dz = \int_{-R}^{R} 2\pi \sqrt{R^2-z^2} dz $$
Unfortunately, this integral does not evaluate to $4\pi r^2$. I discovered that instead of taking the thickness of the thin rings as $dz$, I have to take the infinitesimal arclength $dL$ at point $z$ on the cross-section $x = \sqrt{R^2-z^2}$
$$ dL = \sqrt{\left(\frac{z}{\sqrt{R^2-z^2}} \right)^2 +1} \,dz = \frac{R}{\sqrt{R^2-z^2}} dz $$
Putting that in place of $dz$, the integral evaluates perfectly. Conversely, if the thickness in the volume integral was $dL$, in won't work out.
This baffles me. Why do the disks have to have different thickness when you're taking the volume, compared to when you're taking the surface area? Why does it matter? When you take the limit as the number of disk slices goes to infinity and the thickness goes to zero, the sum will approach the sphere, won't it?
Is this a case where the sum of those volumes will approach the sphere, but the sum of their surface areas approaches some different shape, or doesn't approach anything if the thickness is $dz$ and reverse if the thickness is $dL$? If so, why?
The same thing applies for any arbitrary surface/solid of evolution. The surface area integral always involves $dL$, but the volume only has $dx$
Try doing what you did to find the circumference of a circle. It's like you approximated the circumference with a bunch of vertical line segments. But no matter how many you make and how small you make them, they still just have a total lenght of $4R$. They're not doing a good job of approximating the curve. To do that, you need to make the ones closer to the top and bottom slanted and therefore longer.
The total length of these segments is just $4R$, not $2\pi R$, and even using more of them at a smaller size would not change this. Try to see the next higher dimension analogy.
Now, if you used the rectangles that these segments define (by adding in horizontal segments) you do get a good approximation of the area, which does get better as the height of these segments gets smaller. The missing area omitted by such rectangles will approach zero as $n\to\infty$. But again, this is not the case with the lengths, which are a constant $4R$ regardless of $n$.