Finding the volume of the cone using cylindrical polar coordinates?

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The discussion focuses on finding the volume of a cone using cylindrical polar coordinates, specifically with a cone centered on the z-axis, having a base radius of 1 and a height of 1. The limits for the coordinates are φ from 0 to 2π, z from 0 to 1, and ρ from 0 to (1-z). The (1-z) limit arises because, on the cone's surface, ρ and z are related by the equation ρ + z = 1, which distinguishes the cone from a cylinder. A limit of 0 to 1 for ρ would incorrectly define a cylinder, which maintains equal width at both ends. The explanation clarifies the relationship between the coordinates and the geometry of the cone.
sarubobo28
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The cone centre is the z-axis and has base ρ=1 and height z=1,
I'm looking at the lecture notes and it says the limit φ=0 to 2pi, z=0 to 1,
ρ=0 to (1-z).
Could someone tell me where the (1-z) comes from please?
Why is it not 0 to 1?
 
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On the cone surface, ρ and z are related through ρ+z=1. This is the defining equation for that surface. 0 to 1 would define a cylinder, which is equally wide in top and bottom.
 
clamtrox said:
On the cone surface, ρ and z are related through ρ+z=1. This is the defining equation for that surface. 0 to 1 would define a cylinder, which is equally wide in top and bottom.

I see, thank you I get it now :)
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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