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Consider the unit sphere $S^2$ and a disc centred at $u\in S^2$. Also, let $A = \{x\in \mathbb{R}^3:x\cdot u = \cos r\}$ (where $r$ is the angle between $u$ and $v$), which is the plane whose nearest point to the origin is $(\cos r) u$. If $B = \{x \in \mathbb{R}^3: x\cdot u = 1\}$, which is the plane whose nearest point to the origin is $u$. Then the distance between A and B is $1 − \cos r$. Then the spherical disc $D(u,r)$ is equal to the portion of $S^2$ which lies between the parallel planes A and B, and so its area is $2\pi\Delta = 2π(1 − \cos r)$.

Why is the area of a spherical circle equal to $2\pi\Delta$? I don't seem to see how exactly this formula is justified.

sequence
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  • Downvote because question is terribly phrased and in sore need of cleanup and clarification. Are you using $\cdot$ as the dot product, since when is area a 3D quantity, where are you getting $r$ from (unless it's the radius, in which case why are you taking the $\cos$ of a linear quantity), etc, etc? – Pockets May 15 '16 at 02:25
  • Sorry, I don't get you. 3D objects do have surface area. $r$ is the angle between vectors $u$ and $v$. – sequence May 15 '16 at 02:28
  • The words "surface area" appear nowhere in your question; neither does that definition of $r$. Please edit your post so that it actually contains all information relevant to your question, in as clear and precise a manner as possible. – Pockets May 15 '16 at 02:37
  • I used the convention that $r$ in $\cos r$ is an angle, which should be clear without further description. Same applies to the area - this term cannot apply to anything else. This is non-Euclidean geometry. The disc lies in the sphere $S^2$, and the sphere itself is a surface. – sequence May 15 '16 at 05:16
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    @Pockets, you are really being picky here. The formulation is not brilliant, but clearly not as bad as you complain. – Alex M. May 15 '16 at 10:50
  • Related (with answers): http://math.stackexchange.com/questions/153472/calculating-the-surface-area-of-sphere-above-a-plane and http://math.stackexchange.com/questions/203953/volume-of-a-region-on-the-sphere; related unanswered questions: http://math.stackexchange.com/questions/971321/interesting-problem-of-finding-surface-area-of-part-of-a-sphere and http://math.stackexchange.com/questions/1640377/area-of-spherical-zone – David K May 15 '16 at 13:12
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    I suspect that in the original context in which this question arose, it is clearer from the outset that $S^2$ is embedded in $\mathbb R^3$ with the usual Euclidean metric, and that while $r$ is so named because it is in some sense a "radius" of the "disc", the "disc" is a spherical cap $\mathbb R^3$ and $r$ is the length of a geodesic path from the center of the "disc" to its boundary; that is, we use an arc on the unit sphere as the "radius" of the "disc", so the length of the "radius" is equal to the angle of that arc. – David K May 15 '16 at 13:37
  • @DavidK, exactly. – sequence May 15 '16 at 17:02
  • @DavidK: the questions you gave links to are somewhat different. My question is more "geometrical", and directly refers to non-Eucledian geometry. I think some people downvoted this question simply because they don't understand the subject matter. – sequence May 15 '16 at 17:10
  • @sequence You are correct, the linked questions are different from this one. That is why I neither downvoted this question nor flagged it as a "duplicate." But this question does appeal to the embedding of the sphere in $\mathbb R^3$, in which the "disc" is a spherical cap and has the same area as the spherical cap, which in turn can be measured using the answers to the other questions. So it's clear why we would expect the area to be $2\pi\Delta$, though not necessarily clear how one proves that using non-Euclidean methods. – David K May 15 '16 at 17:55
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    In fact I have now upvoted the question, since (despite possible gaps in the presentation of the problem) I would like to see how this problem is solved within spherical geometry without appealing to a three-dimensional model. – David K May 15 '16 at 18:09

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