Area of a circle on sphere On a (flat) Euclidean plane, the area of a circle with a radius $r$ can be described by the function $A(r) = \pi r^2.$
But how can one describe the area of the same circle on a spherical manifold? Assuming that the radius of the sphere is an Euclidean distance of $d,$ how would $A(x)$ look?
I'm assuming this can be found using calculus and/or trigonometric functions, but I'm not exactly sure how to do it.
 A: If $S$ is a sphere of radius $R$ in Euclidean three-space, $p$ is a point of $S$, and $S'$ is a sphere of radius $r$, $0 < r \leq 2R$, centered at $p$, then the area of the portion of $S$ inside $S'$ is $\pi r^{2}$. (!) 
(A detailed argument is given in my answer to Why do disks on planes grow more quickly with radius than disks on spheres?. I learned this nifty fact way back from Vector Calculus by Marsden and Tromba.)
If instead the "radius" $0 < r \leq \pi R$ of a disk on a sphere of radius $R$ is measured intrinsically (as an arc length of a great circle along the sphere), the area is
$$
2\pi R^{2}\left(1 - \cos\frac{r}{R}\right),
$$
as can be shown by straightforward calculation.
A: It depends on how you define $r$. If $r$ is the length of the arc on the sphere, then your area is still $\pi r^2$. If $r$ is the radius in the plane, you need to calculate the length of the arc given by a point on the circle, and the intersection between the sphere and the line that goes through the center of the sphere and the center of the circle. For the radius of the sphere $d$, the arc length id $d\theta$, where $\sin(\theta)=r/d$. The area of the "circle" is then $\pi d^2 \theta^2$ 
My mistake. Here is the solution:
Suppose you call $r$ the length of the arc along the sphere, and $x$ the radius in the plane. At a position $l$ along the arc, $x=d\sin(\theta)$, where $\theta=l/d$. A small strip on the sphere of width $dl$ has area $2\pi x dl$. Then 
$$A=2\pi d\int_0^l dl \sin(l/d)=2\pi d^2\int_0^{l/d} dy \sin(y)=2\pi d^2(1-\cos(l/d))$$
For $l=\pi d$, $\cos(\pi)=-1$, so $A=4\pi d^2$
For $d\rightarrow \infty$, we should recover the plane geometry. $\cos(l/d)\approx1-l^2/d^2/2$, so $A=\pi l^2$  
