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I was trying to derive the formula for the surface area of a sphere and thought of deriving it this way. If we have a circle with radius $r$ and we rotate it along its center by $180$ degrees, the circumference of the circle would cover each part of the sphere once, so, the circumference multiplied by the amount it rotated, which was half the circumference, would give us the surface area. This would result in

$$ 2 \pi r \times \pi r ,$$

or equivalently

$$ 2 (\pi r)^2.$$

But this differs from the actual surface area of a sphere which is

$$ 4 \pi r^2.$$

Though I understood the derivation of the second formula using Cavalieri's principle, I couldn't understand what was wrong with my approach. Can anyone explain using high school level mathematics?

Just_A_User
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Rattandeep
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    In your rotation, the distance traveled by a point is variable. Only two points travel a distance equal to a full circumference. The rest travel less. – quasi May 21 '20 at 08:00
  • @quasi As far as I am concerned, that is an answer, – José Carlos Santos May 21 '20 at 08:04
  • @quasi Thank you for the answer. I don't know how I overlooked that. – Rattandeep May 21 '20 at 08:08
  • See https://math.stackexchange.com/q/2478287/265466, of which this seems like a duplicate. – amd May 22 '20 at 07:25
  • I know this is late but I had already seen that question when I posted it. The question at hand used integrals to describe the same thing of which I had no knowledge at the time, therefore, I did not completely understand their question or solution. That's why I asked if somebody could explain it using high school level maths. I'm sorry if I was not supposed to do that. – Rattandeep Mar 16 '21 at 12:50
  • Your attempted formula is for a square, almost. – Тyma Gaidash Oct 24 '21 at 22:26

1 Answers1

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As in my comment . . .

In your rotation, the distance traveled by a point is variable. Only two points travel a distance equal to a half circumference. The rest travel less.

quasi
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