I am looking at the following exercise:

The spherical circle of centre $p \in S^2$ and radius $R$ is the set of points of $S^2$ that are a spherical distance $R$ from $p$.

If $0 \leq R \leq \frac{\pi}{2}$ a spherical circle of radius $R$ is a circle of radius $\sin R$.

Show that, if $0 \leq R \leq \frac{\pi}{2}$, the area inside a spherical circle of radius $R$ is $2\pi (1 − \cos R)$.


To find the area inside a spherical circle of radius $R$ do we have to calculate the area inside a circle of radius $\sin R$ ?

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    $\begingroup$ No, it's not necessary to compute the circle of radius $\sin R$. Use the integral formula and you will see that only needs to use the circumference. $\endgroup$ – Tom-Tom Jan 5 '16 at 13:59
  • $\begingroup$ Do you mean the formula $$\mathcal{A}_{\sigma}(R)=\iint_R\|\sigma_u\times\sigma_v\|dudv$$ ? @Tom-Tom $\endgroup$ – Mary Star Jan 5 '16 at 14:02
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    $\begingroup$ I do not know what are the $\sigma$ in you formula. I meant the formula to compute the area of the spherical cap : $$\mathcal{A}=\int_0^R2\pi\sin\theta\;\mathrm d\theta$$ where $2\pi\sin\theta$ is the circumference of the circle of radius $\sin\theta$. $\endgroup$ – Tom-Tom Jan 5 '16 at 14:06
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    $\begingroup$ Yes: it is a spherical disk, see en.wikipedia.org/wiki/Spherical_cap $\endgroup$ – Tom-Tom Jan 5 '16 at 14:08
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    $\begingroup$ The first fundamental form on the sphere is simple $\mathrm ds^2=\mathrm d\theta^2+\sin^2\theta\;\mathrm d\phi^2$. Using this, you will find immediately the integral I have written. $\endgroup$ – Tom-Tom Jan 5 '16 at 14:22

The metric tensor on the sphere is given by the identity $$\mathrm ds^2=\mathrm d\theta^2+\sin^2\theta\,\mathrm d\varphi^2$$ where $\theta\in[0,\pi]$ is the colatitude and $\varphi\in\mathbb R/2\pi\mathbb Z$ is the longitude on the unit sphere $\mathcal S$. A spherical "disk" is usually called spherical cap and refers to $$\mathscr C_R=\left\{(\theta,\varphi)\in\mathcal S^2\, \big| \;0\le\theta\le R\right\}.$$ The area of the spherical cap is given by the integral $$\mathcal A(\mathscr C_R)=\int_{\mathscr C_R}\sqrt{1\times\sin^2\theta-0\times0} \,\mathrm d\theta\,\mathrm d\varphi =\int_0^{2\pi}\mathrm d\varphi\int_0^R \left|\sin\theta\right|\mathrm d\theta.$$ The sinus $\sin\theta$ is always positive and one can integrate to find $$\mathcal A(\mathscr C_R)=2\pi(1-\cos R).$$ Note that if $R=\pi$ (its upper limit), the cap covers the whole sphere and we have $\mathcal A(\mathscr C_R)=4\pi=\mathcal A(\mathcal S^2)$.

It is possible to compute the area of a spherical "ring", a frustum, which is given by $\mathcal A(\mathscr F_{R_1,R_2})=2\pi\left|\cos R_1-\cos R_2\right|$.

  • $\begingroup$ Thank you very much for your answer!! :-) $\endgroup$ – Mary Star Jan 6 '16 at 14:28

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