Denote by the volume of the n-ball of radius r. Then
because this is just a line segment twice as long as the radius, i.e.
For all we have:
Volume is proportional to nth power of radius
We shall first show by induction that the volume of an n-ball is proportional to the nth power of its radius. We have already noted that this is true in one dimension. Suppose it is true for n dimensions, i.e.:
Then:
Now we have established that for all n≥1, the volume of an n-ball is proportional to the nth power of its radius; that is, if we denote the volume of the unit n-ball by we have:
First few steps
In the case of we get
which is the area of the unit circle, as we expect. The next derivation, the volume of the unit sphere, is much easier:
General case
Let us try to generalize this derivation for a ball of any dimension:
Here is a graph of the integrand to make it easier to visualize what is going on:
As you can see, the hyperballs becomes squishier and squishier as the dimension increases.
By a change of variables we have:
The integral on the far right is known as the beta function:
Since we can easily verify by induction that for all n ≥ 1:
General form and surface area
The "surface area" of the n-ball, i.e. the (n-1)-dimensional volume measure of the (n-1)-sphere, can easily be found by differentiating the volume of the n-ball with respect to the radius. So, if we denote the volume of the n-ball of radius r by
then its "surface area" is
Further generalizations
This method of integration should carry over to balls in other Lp spaces where Not quite as easy as it looks, this problem has enormous significance for information theory and coding theory. The accurate and rigorous calculation of the volume of hyperballs in many dimensions is also highly relevant to the computation of subatomic forces in the calculation of branching ratios for the simulation of nuclear explosions.
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![{\displaystyle V^{(n)}[r]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d9985b6af5f02c7ba11316eac4407086b361cf1a)
![{\displaystyle V^{(1)}[r]=2r}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d71f9ca407510b9644585aa75fe9e2f90e3f7c59)
![{\displaystyle V^{(n+1)}[r]=\int _{-r}^{r}V^{(n)}\left[{\sqrt {r^{2}-x^{2}}}\,\right]dx.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8ca2275d066f65e2ec8aec91845526e628bf9866)
![{\displaystyle V^{(n)}[r]=r^{n}V^{(n)}[1].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3cca61a3193e807d404d273890e0e5835489b1b7)
![{\displaystyle V^{(n+1)}[r]=\int _{-r}^{r}V^{(n)}\left[{\sqrt {r^{2}-x^{2}}}\,\right]dx,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/73f6174eac450e13718c53a5a69a1f2a136ebba5)
![{\displaystyle V^{(n+1)}[r]=r\int _{-1}^{1}V^{(n)}\left[{\sqrt {r^{2}-(rx)^{2}}}\,\right]dx,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0674120c4f702c1c16c3208a1fa3ec208e52138c)
![{\displaystyle V^{(n+1)}[r]=r\int _{-1}^{1}V^{(n)}\left[r{\sqrt {(1-x^{2})}}\,\right]dx,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6264a0bfa44570b295a07a4b300871b2ab6e068e)
![{\displaystyle V^{(n+1)}[r]=r\int _{-1}^{1}r^{n}V^{(n)}\left[{\sqrt {(1-x^{2})}}\,\right]dx=r^{n+1}V^{(n+1)}[1].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/21cd15d1c4ce46d9bb983e6d83d24786d9d7b9e9)
![{\displaystyle V^{(n)}[r]=r^{n}V^{(n)},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e9d963aed4bff76dbfc13fd549249455164f11d7)
![{\displaystyle V^{(n)}[r]={\frac {\pi ^{n/2}r^{n}}{\Gamma \left({\frac {n}{2}}+1\right)}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4acfc413a3e529dbabd6de330c9f7130511fbe9a)
![{\displaystyle S^{(n-1)}[r]={\frac {\partial }{\partial r}}V^{(n)}[r]={\frac {\pi ^{n/2}nr^{n-1}}{\Gamma \left({\frac {n}{2}}+1\right)}}={\frac {2\pi ^{n/2}r^{n-1}}{\Gamma \left({\frac {n}{2}}\right)}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ee2e5b01f1e44e7059b3d0f37cd16d71cad16b4)
