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Volume of a cone

  • Thread starter daftjaxx1
  • Start date
Can someone help me with this problem?:

We will define a cone in n-dimensions as a figure with a cross - section along its height [tex]X_n [/tex] that has a constant shape, but each of its dimensions is shrunk linearly to 0.

a)let D be a cone in [tex] R^n [/tex] with height h [tex] (ie. [/tex] [tex] X_n [/tex] [tex] \epsilon [/tex] [tex] [0, h]) [/tex] and let the volume of its cross-section at h=0 be [tex] V_o [/tex]. Find the volume of D in terms of [tex]V_o [/tex].

b)Find the volume of the region defined by [tex]|x_1| +...+ |x_n| \le r [/tex] in [tex] R^n [/tex], using a)


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(a) The "volume" of an n-1 figure is proportional to the product of the dimensions. Since the dimensions depend linearly on z (you used h as both height of the entire cone and the variable in that direction- I'm going to call thevariable z) and goes to 0 at z= h, the cross section volume is proportional to (h-z)n-1. Since the volume at z= 0 is V0, we must have [tex]V(0)=V_0(\frac{h-z}{h})^{n-1}[/tex]. The "n-dimensional" volume of a thin "slab" of thickness Δz will be [tex]V_0(\frac{h-z}{h})^{n-1}\Delta z[/tex]. Convert that to an integral.

(b) What does this volume look like? Sketch it for n= 1, 2, 3.
Last edited by a moderator:
thanks for the response. just to clarify, when it says the "volume of its cross section at h=0 is [tex] V_o [/tex], is it really refering to the "area" of the cross section? (eg. if we're talking about 3 dimensions)? its sort of hard to visualize.

i still don't get how to do b). so if n=1, the volume is a line, if n=2, its a triangle, and if n=3, its the cone we're used to, right? i don't know how to work with the given region [tex]|x_1| +...+ |x_n| \le r [/tex]. is "r" just some constant??

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