Proof of Volume of sphere

Question

I want a simple proof for the formula of volume of sphere. Does the proof/explanation without integration possible?

Answer 1

We have that the area of lateral surface $C$ of the trucated cone with basis circles of radii $r$ and $R$ is $$ A(C)= 2 \pi l \frac{r+R}{2} $$ with $l$ the meridian. If the cone is inscripted in the unit sphere $S^2$ we have that $A(C)=2 \pi d h$ with $h$ the hight of truncated cone and $d$ the distance from origin to $C$. If we cut $S^2$ with parallel planes with distance $h=\frac{1}{n}$ we can put $S^2$ in a lot of truncated. So is it easy to show that the sum of lateral surfaces $C_k$ of thes cones is $$ \sum_k A(C_k)=\frac{2\pi}{n} \sum_{k=1}^{2n} d_n. $$ Now if $n \rightarrow + \infty$ we obtain the area of sphere. Using this principle we can calculate the volume os sphere. For all details you can look at http://www.proofwiki.org/wiki/Volume_of_Sphere/Proof_by_Archimedes.

Answer 2

A brute proof:

One can "transform" sphere to some cone/pyramid: $$ V_{sphere} = V_{cone/pyramid} = \frac{1}{3}HS = \frac{1}{3}R\cdot 4\pi R^2 = \frac{4}{3}\pi R^3. $$ (You need to know here that sphere surface area is $4\pi R^2$.)

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The same way one can "prove" that circle area is $\pi R^2$. Just "transform" circle to triangle: $$ S_{circle} = S_{triangle} = \frac{1}{2}HL=\frac{1}{2}R\cdot 2\pi R = \pi R^2. $$ (You need to know here that circle's circumference length is $2\pi R$.)