A 4-D sphere: what is the area??

Question

I have noticed a pattern with circles and applied it to a $4$D sphere with inaccurate results. the formula for a circle is

$$\pi\times r^2\text{ or }\pi\times \left( \frac{d}{2} \right)^2\text{ or }\pi\times \frac{d^2}{4}$$

and the area of a sphere is:

$$\frac{4}{3}\times\pi \times r^3\text{ or }\frac{4}{3}\times\pi \times\left(\frac{d}{2}\right)^3\text{ or }\frac{4}{3}\times\pi \times\frac{d^3}{8}$$

So naturallty I assumed that For a $4$D hypersphere is

$$\frac{16}{9}\times \pi \times r^4\text{or}\frac{16}{9}\times \pi \times \left(\frac{d}{2}\right)^2\text{or }\frac{16}{9}\times \pi \times \frac{d^4}{16}$$

I explained my thinking to my friend, but he said it was wrong.

I did the same thing to the surface area; if curcumfrence is

$$\pi d$$

and the sphere is

$$4\pi d$$

then $4$D should be

$$16\pi d$$

This was also thought to be wrong. what is going on here!!??

Thanks in advance

Do you know how to use an integral? I have a full explanation prepared but it involves integration. — Franklin Pezzuti Dyer, Jun 08 '17 at 15:37
The main reason your analogies are incorrect is that you only have two numbers in this "pattern" - not nearly enough to begin predicting what the next one will be. — Franklin Pezzuti Dyer, Jun 08 '17 at 15:40
The area of an $n$-dimensional unit sphere is computed in formula $(4)$ of this answer. For $n=4$, it gives $2\pi^2$. Actually, what you are calling area, would more commonly be called volume, and that is handled by formula $(5)$. For $n=4$ it gives $\frac{\pi^2}2$. — robjohn, Jun 08 '17 at 15:41
I find it interesting that $$V_n = \frac{S_{n-1}}{n}$$and$$S_n = 2 \pi V_{n-1}$$where $V_n$ is the volume of n-ball ($n+1$-dimensional sphere) and $S_n$ its surface. Thus, $$V_n = \frac{2 \pi}{n} V_{n-2}$$ — Nominal Animal, Jun 08 '17 at 15:48
The surface are of sphere isn't $4\pid$, it is $4\pir^2$. Am I missing something? — Jaideep Khare, Jun 08 '17 at 15:50

score 2 · Answer 1 · edited Jun 08 '17 at 15:48

For the sphere and circle there are wonderful explanations. Think of it this way - when the volume of a sphere increases, it is like you are adding infinitely many infinitely thin (flat) layers (or shells) to the outside of the sphere, each with an area of $4\pi r^2$.

Using these analogies, you should be able to find the "surface volume" and "hypervolume" of a hypersphere (4 dimensional sphere). Let us denote $V_n(r)$ the "interior space" (area, volume, hypervolume) of a sphere in $n$ dimensions, and $A_n(r)$ the "exterior space" (circumference, surface area, surface volume). Since the length of a chord of a circe that is some distance $d$ from the center is given by $2\sqrt{r^2-d^2}$, the area of a circle can be derived by the integral of the lengths of these chords from $d=-r$ to $r$: $$2\int_{-r}^r \sqrt{r^2-d^2} \,\mathrm dd$$ Which gives us $\pi r^2$. You can use a similar formula for the volume of a sphere, since the area of each circular cross section of the sphere a distance $d$ from the center is given by $\pi(r^2-d^2)$, the volume of the sphere is $$\int_{-r}^r \pi(r^2-d^2) \,\mathrm dd$$ also giving us the correct formula, $\frac{4}{3}\pi r^3$. Now you can find the analogous formula for the volume of a hypersphere: $$\int_{-r}^r \frac{4}{3}\pi(r^2-d^2)^\frac{3}{2} \, \mathrm dd$$ and the "surface volume" is the derivative of this: $$\frac{4}{3}\pi(r^2-d^2)^\frac{3}{2}$$

This answer is from my previous answer to this question. Perhaps you would find it helpful as well.

A 4-D sphere: what is the area??

1 Answers1