The applet below applies a symplectic (area-preserving) map to a disk of radius *R*: taking each point with polar coordinates
(*r*, θ) and mapping it to (√(*R*^{2} – *r*^{2}), θ).

Choose an image file to upload from your computer, or enter the full URL of an image to use from the internet.

You can drag anywhere on the source image to move the location of the disk, or drag on the perimeter of the disk to resize it.

How can we check that the function:

f_{R}(r, θ) = (√(R^{2}–r^{2}), θ)

preserves areas? In polar coordinates, area is measured by:

dA=rdrdθ

So we have:

r_{2}= √(R^{2}–r_{1}^{2})

θ_{2}= θ_{1}

dA_{2}=r_{2}dr_{2}dθ_{2}

= √(R^{2}–r_{1}^{2}) d(√(R^{2}–r_{1}^{2})) dθ_{1}

= √(R^{2}–r_{1}^{2}) (–2r_{1}) / (2 √(R^{2}–r_{1}^{2})) dr_{1}dθ_{1}

= –r_{1}dr_{1}dθ_{1}= –dA_{1}

The minus sign tells us that the function *f*_{R} produces a “mirror-reversed” image, but that the absolute value of the
area of any region is unchanged.

One interesting application of this area-preserving property is to show that the volume of the unit-radius ball in 2*n* dimensions is:

V(B^{2n}) = π^{n}/n!

That is to say, the volume is 1/*n*! times the volume of the Cartesian product of *n* unit disks. This is
reminiscent of the fact that a hypercube can be dissected into *n*! simplexes of equal volume. If the hypercube
has vertices (±1, ±1, ..., ±1) we can take one simplex to have the *n*+1 vertices:

v_{0}= (1, 1, 1, ..., 1)

v_{1}= (–1, 1, 1, ..., 1)

v_{2}= (–1, –1, 1, .., 1)

...

v_{n}= (–1, –1, –1, ..., –1)

and then generate the vertices of all *n*! simplexes by permuting these coordinates. The simplexes themselves
are the regions that satisfy:

–1 ≤x_{1}≤x_{2}≤ ... ≤x_{n}≤ 1

for all *n*! permutations of the coordinates *x*_{i}.

If we take a point that lies within the unit ball *B*^{2n} to be given by *polar coordinates* (*r*_{i}, θ_{i}) in *n* planes, we can define the following points
in the unit disk:

d_{1}=f_{1}(r_{1}, θ_{1}), so |d_{1}|^{2}= 1 –r_{1}^{2}

d_{2}=f_{|d1|}(r_{2}, θ_{2}), so |d_{2}|^{2}= 1 –r_{1}^{2}–r_{2}^{2}

d_{3}=f_{|d2|}(r_{3}, θ_{3}), so |d_{3}|^{2}= 1 –r_{1}^{2}–r_{2}^{2}–r_{2}^{3}

...

Clearly the |*d*_{i}| will be in descending order, so we are mapping the unit ball into a region with
just 1/*n*! of the total volume of the product of the unit disks (since every one of the *n*! possible orders for these
magnitudes will occupy an equal volume).
Because *f*_{R} preserves areas, the unit ball must have the same volume as the region into which it is mapped.
So we have shown that:

V(B^{2n}) = π^{n}/n!