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

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What is a volume? first define it then try to calculate. However this is what is usually used in elementary proofs. – user59671 May 11 '13 at 15:08
That proof via Cavalieri's principle is what I saw in a high-school geometry course. – Michael Hardy May 11 '13 at 15:10
I mean the explanation for the formula V=4/3 pi*r^3 – user61681 May 11 '13 at 15:10
@user61681 : That the volume of a sphere is $V=\frac43\pi r^3$, where $r$ is the radius, is what is proved by the argument involving Cavalieri's principle, to which CutieKrait linked. – Michael Hardy May 11 '13 at 15:11
@user61681, scroll the wiki page down to this part. – vadim123 May 11 '13 at 15:22

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

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A brute proof:

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

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

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Thanks for your nice answer. – user61681 May 12 '13 at 17:47

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