Volume of a cone in n dimensions with linear shrinking

[daftjaxx1]

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 $$X_n$$ that has a constant shape, but each of its dimensions is shrunk linearly to 0.

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

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

 

[HallsofIvy]

(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 $$V(0)=V_0(\frac{h-z}{h})^{n-1}$$. The "n-dimensional" volume of a thin "slab" of thickness Δz will be $$V_0(\frac{h-z}{h})^{n-1}\Delta z$$. Convert that to an integral.

(b) What does this volume look like? Sketch it for n= 1, 2, 3.

 

[daftjaxx1]

hi,
thanks for the response. just to clarify, when it says the "volume of its cross section at h=0 is $$V_o$$, is it really referring 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 $$|x_1| +...+ |x_n| \le r$$. is "r" just some constant??