# Talk:Annulus (mathematics)

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Field:  Geometry

As to the annulus in geometry, all definitions of the annulus somehow proceed from the body having the form of a finger ring. In elementary geometry we have three-dimensional annuli (having the form of a finger ring) and two-dimensional annuli (figures limited by two concentric circles). So we can generalize the concept to the n-dimensional Euclidean space, (the union of all (n-1)-dimensional balls whose center lies (ADDENDUM) at a fixed distance from a given point outside of the balls) and further to complex spaces and maybe (I am not sure) even to any metrical space. The condition that the annulus be open is optional.

Might be this needs a correction, at least. For your 3D ring I use the word "torus" http://en.wiktionary.org/wiki/torus , so this perhaps does not come to be "annulus".
ADDENDUM = nothing -- in this case you get a shell = difference of two balls. This is what I was looking for but it is NOT a generalization of your ring in 3D
ADDENDUM = "in a (2D) plane" (tentatively through the fixed point)
ADDENDUM = "in a hyper-plane" (n-1 -dimensional) (tentatively through the fixed point)
Last two possibilities are both generalization of the ring from 3D.
90.180.192.165 (talk) 15:19, 5 December 2012 (UTC)

The definition of the annulus on the complex space is a special case of such a general concept. It might be that that special case has some special importance in the complex analysis. Andres 12:52, 5 Nov 2003 (UTC)

## This result can be obtained via calculus

Why, pray tell, would you ever bother to calculate the area of one of these through integration?

Because one can then use that formula to calculate the volume certain solids obtained through rotation. For example, the volume of the region defined by ${\displaystyle y=(x+2)^{2}+2}$ and ${\displaystyle y=6}$, rotated about the y-axis can be calculated as an integral with respect to y by taking the areas of infinitely thin horizontally-oriented annuli.
Aside from that, the way to prove the area of any curved shape is through integration. siafu 18:15, 6 March 2006 (UTC)

## Way too many diambiguations

So I was looking for annulus, and I went through the ring link, then it went to circular, then finally I got to circle. Isn't their some easier way to get where I want.

128.12.39.120 00:25, 18 May 2007 (UTC)