Volume of 4-ball by Cavalieri's principle?

[yucheng]

Homework Statement

Find the volume of a 4-space ball, with radius R.
(Alternatively you can integrate this via Cavalieri’s principle, by using the fact that the crosssectional volume is given by $\frac{4}{3} \pi r^3$
, the volume of a 3-ball)
See example at end of page 1 of https://www.math.ucla.edu/~bonsoon/math32bh.winter2020/notes/short_notes_-_More_on_Change_of_variables_formula.pdf

Relevant Equations

N/A

I tried integrating the 4-volume of a 4-hemisphere, that is, $$\int^{R}_{0} \frac{4}{3} \pi r^3 dw$$ (along w-axis), since $r$ is proportional to $w$, where $r=\frac{w}{R} R$, $r=w$, thus the integral becomes $$\int^{R}_{0} \frac{4}{3} \pi w^3 dw = \frac{\pi}{3} R^4$$ The volume of a 4-D hypersphere is $\frac{2\pi}{3} R^4$

Clearly, something is not right.

 

[fresh_42]

Are you sure you use the correct formula? I do not see this, neither on what you linked to, nor on Wikipedia:

https://en.wikipedia.org/wiki/N-sphere#Spherical_volume_and_area_elements
https://de.wikipedia.org/wiki/Sphäre_(Mathematik)#Inhalt_und_Volumen

Do you have a description of Cavalieri's principle? It is not explained in your paper.

 

[yucheng]

Are you sure you use the correct formula? I do not see this, neither on what you linked to, nor on Wikipedia:

https://en.wikipedia.org/wiki/N-sphere#Spherical_volume_and_area_elements
https://de.wikipedia.org/wiki/Sphäre_(Mathematik)#Inhalt_und_Volumen

Do you have a description of Cavalieri's principle? It is not explained in your paper.

I don't know the correct formula. I just tried integrating ) and it failed :\

Cavalieri's principle

Just to confirm, a hypersphere has something called a 'radius' right? What I essentially tried was to use the cross sectional volume, integrate across a hemisphere, and see where it goes...

I thought I could do it analogous to integrating the disks in a sphere.

On the top of page 2, it my link in the homework statement states that you can get the volume equation by using Cavalieri's principle. I just took it as a challenge.

 

[yucheng]

@fresh_42 I realized I could not even calculate the volume of the 3-ball with my argument. Looks like it must have been wrong!

 

[fresh_42]

The English Wikipedia tells you how to do it via polar coordinates, the German version has a closed formula, and in order to use the cross section, you have to consider a rotation: https://en.wikipedia.org/wiki/Solid_of_revolution

 

[yucheng]

the German version has a closed formula

Do you always look at German versions?

 

[fresh_42]

Do you always look at German versions?

I look at all versions I can read a bit. English and German are those with the most pages. But I also look at Spanish, French or Italian if I see a chance to find a better version, mostly a better formula. Formulas can be read in any language, hence I do not care the explanations very much and just look at the math.

I made the experience that the English version is often more theoretical and abstract, whereas the German version tends to provide good formulas and specific examples. I also use Wikipedia for translations: technical terms (search the word you know in your native language and then switch to English) as well as for the correct spelling of names (Ångström, de L'Hôpital, Gauß etc.).

 

[Office_Shredder]

I think if you try to repeat this for calculating the area of a circle you will find your mistake. You don't have the right radius of the 3-ball as a function of where you are in the 4-ball.

 

[yucheng]

I think if you try to repeat this for calculating the area of a circle you will find your mistake. You don't have the right radius of the 3-ball as a function of where you are in the 4-ball.

let me try, $r=\sqrt{x^2+y^2}$, $y=\sqrt{r^2-x^2}$,

\begin{align*}
\int \frac{4}{3} \pi y^3 dx &= \frac{4}{3} \pi \int^{r}_{-r} (r^2-x^2)^{3/2} dx\\
&= (\frac{4}{3} \pi)(\frac{3}{8} \pi r^4) \\
&= \frac{1}{2} \pi^2 r^4
\end{align*}

Somehow, I spent a lot of time wondering whether this is the correct integral, when I could have just plugged it into Mathematica to check :-\

Anyway, thanks!