Talk:Annulus (mathematics)

Sketching a general definition of annulus in 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.

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)


 * 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)

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 $$y=(x+2)^2 + 2$$ and $$y=6$$, 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)

Conceptual problems with this article.
It is obvious that if the area of a disk with radius r is πr², then it is a simple definition that the area of a ring between two circles with radii R and r is π(R²-r²). What I don't understand is how you can use calculus to prove this by integrating 2πrdr. To be clear, I understand is that the definite integral of 2πrdr is π(R²-r²), but what I don't follow is that taking the limit of r+e (as e approaches zero) of the rectangular area r(r+e) is identical to the area between the curves - that area bounded by four curved lines. That is, I fail to see how it follows from the calculus. It probably does, since the resulting formulae are identical, but asserting it doesn't make it true. Just as bad is the nonsense about the longest "interval" inside the ring. If you mean "line segment" then you should SAY line segment. Just to be clear, where is it defined that an interval is a straight line? An interval can be many things, and could certainly be an arc. Finally, asserting that the area can be obtained with the Pythagorean theorem from this line segment seems like another case where you can show it by determining that d is √(R²-r²) but constructing d and then asserting that magically the area of a ring is equal to twice the area of the triangle doesn't follow. Neither assertion (the integration and the construction) are useful, imho, because they aren't obvious until you compare them with the simple difference between the two disks' areas. What is needed is to show, by proof, that it follows that d*r is the ring's area (rather than being algebraically equal to it). Same thing with the integral: you need to show that the rectangle r*dr's area approaches an annulus sector. Seems to me that that entire paragraph should either go into enough detail to establish the proofs, or be removed (along with the figure). I guess what I'm saying is if I were to assert that the area of the ring is (π/2)(r(R-r)) then neither of the methods (integration or the triangle) could demonstrate (given whats in the article) that I was wrong. Sure if I integrate 2πrdr I get π(R²-r²) or if I construct a tangent from one circle it will intersect a concentric circle at R and have a length of 2d, and an area of ½rd, but so what? I am arguing that the paragraph doesn't advance the argument (the reason why the area is π(R²-r²)).216.96.77.203 (talk) 20:41, 5 April 2015 (UTC)

Segment/section of annulus
One should also define a sector ($$\phi_0 \leq \phi \leq \phi_1$$) and a segment ($$y \geq y_0$$) of the annulus. Confer to similar definitions for circles: Circular sector, Circular segment. HerrHartmuth (talk) 12:17, 17 January 2020 (UTC)