Hypervolume of hypercone

[Angelina]

Hi everyone! I want to calculate the formula of hypervolume (I label it "H") of 4D hypercone, which consists of 3D base which is ball with radius R, apex and lines from apex to points of ball that are not in the same 3D-space as ball. Let's label its height as "h". If 3D volume can be calculated through definite integral:
[tex]V=\int\limits_{a}^{b}S(x)dx[/tex]
I think that hypervolume can be calculated:
[tex]H=\int\limits_{a}^{b}V(x)dx[/tex]
Cross-section which is parallel to base is also ball with radius r(x), and distance between apex and such cross-section is x. Radiuses and height make a triangle:

triangle.png (3.37 KiB) Viewed 4168 times

As full and partial triangles are similar:
[tex]\frac{x}{h}=\frac{r(x)}{R} \Rightarrow r(x)=\frac{Rx}{h}[/tex]
So hypervolume is:
[tex]H=\int\limits_{0}^{h}{V(x)dx}=\int\limits_{0}^{h}\frac{4\pi r^3(x) dx}{3}=\frac{4\pi}{3}\int\limits_{0}^{h}{r^3(x)dx}=\frac{4\pi}{3}\int\limits_{0}^{h}{\frac{R^3 x^3}{h^3}}=\frac{4\pi R^3}{3 h^3}\int\limits_{0}^{h}{x^3 dx}=\frac{4\pi R^3}{3 h^3} \frac{x^4}{4}\begin{array}{|l} h \\ 0 \end{array}=\frac{4\pi R^3 h^4}{3 \cdot 4 h^3}=\frac{\pi R^3 h}{3}[/tex]
Is this formula correct?

[Guest]

Yes. That's correct.

In general the hypervolume of an n-dimensional conic object is
[tex]\frac{h}{n} \times \text{hypervolume of "n-1" dimensional base object}[/tex]

For example when n=2, the conic object is a triangle, and the area is [tex]\frac{h}{2}\times\text{base length}[/tex].
When n=3, the volume is [tex]\frac{h}{3}\times\text{base area}[/tex].
When n=4, the hypervolume is [tex]\frac{h}{4}\times\text{base volume}[/tex].

Hope this helped,

R. Baber.

[Guest]

You have to integer:

V(4D) = 2*[tex]\int\limits_{0}^{h}[/tex]V(h)dh = 2*[tex]\int\limits_{0}^{h}[/tex]1/3*π*r^2*h dh = 2*1/3*π*r^2*h^2*1/2 = π*r^2*h^2/3

You project the cone via bc line or height through the fourth dimension and you continue to the other side. It also depends how you define the hyer cone in 4D. You can stop at c if you like. Be careful if you integered from h to -h it would add the negative volume and you get the wrong result 0!