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I have been thinking about the equation for the area of a circle recently. When you calculate the area of a circle, is that not like calculating a sliver of volume across the diameter of a sphere, and if you slice that in half by dividing by 2, does that not give you the radius three dimensionally?

That was probably worded incorrectly, what I am saying is, would this equation of taking the a circle's area of a circle's area theoretically work?

$$V= \pi \left(\frac {\pi r^2} {2} \right)^2$$

When I solve for the volume of a sphere with the radius of 4, $V = 268.08 $

When I solve using my equation for spherical volume with the radius of 4, $V = 1984.4$

Can someone explain to me why this does not work or give a correction to my beautifully spectacular extra special formula?

Marconius
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Youseflapod
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  • Why do you say that dividing a "sliver of volume" across the diameter of a sphere by $2$, gives you the "radius three-dimensionally"? It doesn't seem to be the case. – uniquesolution Nov 17 '15 at 14:52
  • In the formula $V = \pi(\pi r^{2}/2)^{2} = \pi^{3} r^{4}/4$, doubling $r$ causes $V$ to increase by a factor of $2^{4} = 16$. Volume doesn't behave that way, to say the least. :) – Andrew D. Hwang Nov 17 '15 at 15:00

3 Answers3

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You can compute the volume of a sphere using the area of a circle, but you need a different trick. Take a sphere of radius r with centre the origin, and slice it into circles along a diameter. Well, not quite circles. The slice from $x$ to $x+dx$ with $|x|<r$ is a cylinder of volume $\pi[\rho(x)]^2dx$, where by the Pythagorean theorem $[\rho(x)]^2=r^2-x^2$. The sphere's volume is therefore $$\int^{r}_{-r}\pi[r^2-x^2]dx=\pi[r^2 x-\tfrac{x^3}{3}]^{r}_{-r}=\tfrac{4\pi}{3}r^3.$$ You can use this method to obtain a recurrence relation expressing the $d$-volume of a radius-$r$ $d$-ball (say $V_d r^d$) in terms of a radius-$r$ $(d-1)$-ball, viz. $$V_d r^d = \int^{r}_{-r} V_{d-1}[r^2-x^2]^{\tfrac{d-1}{2}}dx.$$ Hence $$\frac{V_d}{V_{d-1}}=2\int^1_0 [1-x^2]^{\tfrac{d-1}{2}}dx=\text{B}(\frac{1}{2},\,\frac{d+1}{2}).$$ Interestingly, comparing $V_d$ to $V_{d-2}$ gives an especially neat result: $$\frac{V_d}{V_{d-2}}=\frac{\sqrt{\pi} \Gamma(\frac{d}{2})\sqrt{\pi}\Gamma(\frac{d+1}{2})}{\Gamma(\frac{d+1}{2})\Gamma(\frac{d+2}{2})}=\frac{2\pi }{d}.$$ This explains why the series $$V_1 = 2, V_2 = \pi, V_3 = \frac{4\pi}{3}, V_4 = \frac{\pi^2}{2}, V_5 = \frac{8\pi^2}{15}$$ has integer powers of $\pi$, gaining a power at even $d$. Considering the even-$d$ case allows one to conjecture the correct general result, which is $$V_d=\frac{\pi^{d/2}}{\Gamma(\frac{d}{2}+1)}$$ (as induction easily verifies).

J.G.
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let us suppose $r$ changes with constant rate $r$ so we want change in volume of sphere so $d(\frac{4/\pi r^3}{3})/dr$ so note we get $4\pi r^2$ on differentiating but according to you its $\pi^3 r^4/4$ which on differentiating gives $\pi^3.r^3$ hope you get the drastic difference between two volumes.

Archis Welankar
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Hint: Imagine trying to find the volume of a cylinder by this method. Say it has height $h$ and radius $r$. The "three dimensional radius of the cylinder" would be a rectangle with width $r$ and height $h$, and therefore would have area $rh$. So, by your logic, the volume of the cylinder would be $\pi(rh)^2=\pi r^2h^2$. However, the real volume is $\pi r^2h$ ! What happened?

Akiva Weinberger
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