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 you start with a diameter of 10, and you add up the circumferences of rings with diameters of [1, 2, 3, 4, 5, 6, 7, 8, and 9] x 2 (for the top half of the sphere) + 10 (the actual circumference) you get 100π as the result.

But, if you use the formula for the surface area of a sphere, $4\pi r^2$, you also get 100π as a result!

Although I'm taking the sum of only 19 circumferences of the circle, I am still getting identical values for both ways! Why is this? Shouldn't the sum of the method involving the circumferences of the rings be less than the formula value, and get more and more accurate as i increase the number of rings in my sum?

This doesn't only work for a sphere with a diameter of 10, but for any other sphere as well!

  • $\begingroup$ You shouldn't sum the circumferences. It's easy to see that every ring has a circumference larger than zero, so infinitely many circumferences (of which infinitely many are larger than a certain value $a>0$) would approach infinity. It's actually the surface of a few cylinders you add up, so every circumference of every ring should be multiplied by its height. $\endgroup$ – Kaj Apr 4 '14 at 13:17
  • $\begingroup$ The surface area of a spherical segment, corresponding to one of the "rings" of surface area you're using, only depends on the "width" of the belt and the radius of the sphere, or $ \ 2 \pi \ R \ h \ $ (see, for instance, mathworld.wolfram.com/SphericalSegment.html) . $\endgroup$ – colormegone Apr 4 '14 at 13:20
  • $\begingroup$ That makes a lot of sense, Thanks! $\endgroup$ – StopReadingThisUsername Apr 5 '14 at 7:38

Your method usualy won't yield the correct result because the assumption that a sphere is a stack of rings which grow smaller in size (which is correct) is by itself not enough. There are many 3d figures that have this property like a cone.


The right way to slice the sphere in this case is into rings which are not of constant width but rather whose projections to say the $z$-axis have constant width. Archimedes's great insight was that these rings all have the same area. Those that are closer to the poles are wider and those that are closer to the equator are "longer" so the two dimensions compensate each other. In modern terminology this has to do with the moment map in symplectic geometry, but you don't need that to calculate the area.


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