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Let $P\in{S^2}\subset\mathbb{R^3}$, a point on the unit sphere. Let there be a spherical circle around $P$ of radius $r<\pi$. Show that the circumference of the spherical circle is $2\pi{\sin{r}}$. And show that the spherical disc bounded by this spherical circle has an area of $2\pi{(1-\cos{r})}$.

For the first part I am aware that what needs to be understood is that a spherical circle lies on a plane in $\mathbb{R^3}$. Then we can consider the circle of intersection, and if we can find this radius, we can find the circumference of the intersection circle (i.e. the spherical circle).

Edit: I now have constructed a triangle, with indices of $P$, the origin of $S^2$ and some point on the spherical circle. The two radial lines have side length 1 with an angle of $r$. from this we can use the cosine rule to find the opposite side length and hopefully find the length of the radius of the circle in $\mathbb{R^3}$.

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You have a good start. In the unit circle you can in fact figure out many of the things you need. This way you get the radius $\sin(r)$ for the circle in $\mathbb{R}$ to calculate the circumference.

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The area can be found in some formularies. Alternatively the general formula for surfaces of revolution could be used. The function describing the unit-circle is $f(x) = \sqrt{1-x^2}$ and you want the area of the surface of revolution around the $x$-axis from $\cos(r)$ to $1$. $$ A=\int_{\cos(r)}^{1} f(x) \sqrt{1+f'(x)^2} dx $$ Solving this integral results in the stated area.

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  • $\begingroup$ Brilliant, this has helped me so much. I'm assuming you centred $\mathbb{R^3}$ at the point at which the line OP intersects the plane I was considering. $\endgroup$ – Sam.S May 2 at 21:45
  • $\begingroup$ Yes, I think you figured that out yourself too. $\endgroup$ – Strichcoder May 2 at 21:47

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