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.

circle perimeter "approximated" by vertical segments

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 $n$.

  • $\begingroup$ Ok, I see what you're getting at. What would happen if these were rings, since the radius still changes? what does the sum approach? I know it does approach something (actually $\pi^2 R^2$), but what's a good way to visualize that area? $\endgroup$ – Dylan Apr 12 '14 at 5:30
  • $\begingroup$ If they were ring-ribbons, then their total surface area would change as $n$ grew. But not by enough. The surface area of such a ribbon defined by a slanted segment is significantly larger. $\endgroup$ – alex.jordan Apr 12 '14 at 5:36
  • $\begingroup$ What about when approximating the volume using those slanted disks? Why does the sum grow too large? Is there a way to modify the formula using that slanted thickness and make it converge? $\endgroup$ – Dylan Apr 12 '14 at 5:40
  • $\begingroup$ For the volume, it would be fine to not use the slanted edges. This is what you have done when you stacked disks of thickness dz. And it would not grow too large to use the slant-edged disks for volume either; the limit of the sum of volumes of either type of disk will be equal. With either method for volume, you can see that any open ball of volume within the sphere will eventually be mostly covered by disks. But you can't say the same thing (2D version) about surface area, because you are not approximating from within the surface. Careful which meaning of "within" I am using. $\endgroup$ – alex.jordan Apr 12 '14 at 7:28
  • $\begingroup$ By "within" the surface, I don't mean the interior of the sphere. I mean within the set consisting of all points that make up the surface. $\endgroup$ – alex.jordan Apr 12 '14 at 7:28

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