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I wonder if it is possible to derive surface area and volume of a sphere seperately using techniques involving pappus' theorem.

I did some calculation and found out the ratio of surface area and volume. Here is my work,

  1. My key observation is finding out that the centroid of a semidisk of raidus $r$ is also the centroid of a semicircle of raidus $2r/3$ when the centers coincide. (By "centers" I mean the centers of full circle and disc induced by half of them.)

  2. I sliced the semidisk identical pieces of triangles(infinitely many) and by locating each triangle's centroid I form a semicircle of radius $2r/3$ which must have the same centroid with semidisk.

  3. Then let's say our centroid is located $h$ distance above the center. We still don't know it.

  4. Using the theorem, the circular path taken by the centroid of the semidisk times the area of the centroid should give the volume of the sphere of radius $r$. And similarly, the circular path taken by the same centroid of the semicircle times the arc length of semicircle should give the surface area of a sphere of radius $2r/3$.

  5. We know that when the radius increases with a proportion, corresponding surface area will also increase with the square of that proportion. Thus, we need to multiply the surface area of sphere of radius $2r/3$ by the factor $9/4$ to get the surface area of the sphere of radius $r$.

Here is the calculations,

$$V = 2\pi h * \pi r^2/2=\pi^2r^2h$$ $$S = 9/4*2\pi h * \pi (2r/3) = 3\pi^2rh$$

Since we don't know $h$, I simply divide them to cancel it out and get, $$V/S = r/3.$$

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The centroid of semi-circle should be $\dfrac{4r}{3\pi}$ from the origin.

So the volume is

$$V=2\pi \times \frac{4r}{3\pi}\times \frac{\pi r^2}{2}=\frac{4}{3}\pi r^3$$

The centroid of semi-circular arc should be $\dfrac{2r}{\pi}$ from the origin.

So the surface area is

$$S=2\pi \times \frac{2r}{\pi}\times \pi r=4\pi r^2$$

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  • $\begingroup$ Yeah but how to get those centroids mathematically without using calculus or hanging them or something? $\endgroup$ – Ahmet Bilal Jan 19 '17 at 5:58
  • $\begingroup$ I didn't claim that the centroids popped out from stone but just point out your mistakes. The key point is how to apply the Pappus' theorem. Lazy guys love wikipedia so that no needs to do everything from scratch. $\endgroup$ – Ng Chung Tak Jan 19 '17 at 11:09
  • $\begingroup$ Where are the mistakes? You just put the value of $h$ and simplified the expressions. $\endgroup$ – Ahmet Bilal Jan 19 '17 at 11:14
  • $\begingroup$ You say $\dfrac{2r}{3}$ $\endgroup$ – Ng Chung Tak Jan 19 '17 at 11:17
  • $\begingroup$ Ok. I said the centroid of a semicircle of radius 2r/3 is the same point that is the centroid of a semidisk of radius r when their centers coincide. Put another way, if they have same radius, disk's centroid should be at a distance above center 2/3 times of the circle's. My english suffers a lot so sorry for that. $\endgroup$ – Ahmet Bilal Jan 19 '17 at 11:34

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