Hypervolume of hypercone

Hypervolume of hypercone

Postby Angelina » Tue Mar 28, 2017 8:04 am

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
triangle.png (3.37 KiB) Viewed 224 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?





Angelina
 
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Re: Hypervolume of hypercone

Postby Guest » Tue Mar 28, 2017 2:41 pm

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.

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