Based on Pressley's Elementary Differential Geometry (Definition 5.3 and Proposition 5.2) the area of a unit-sphere ($R=1$) must be $$4\pi=?A_\sigma(R)=\int \int_R ||\sigma_u\times \sigma_v||dudv=\int \int_\text{Sphere} ||\sigma_\theta\times \sigma_\phi||d\theta d\phi=\int_0^{\pi} \int_0^{2\pi} [(\sigma_\theta.\sigma_\theta)(\sigma_\phi.\sigma_\phi)-(\sigma_\theta.\sigma_\phi)^2]^{\frac{1}{2}}d\theta d\phi=\int_0^{\pi} \int_0^{2\pi} [(1)\times (\text{cos}^2\ \theta)-4\text{cos}^2\ \theta\ \text{sin}^2\ \phi\ \text{cos}^2\ \phi]^{\frac{1}{2}}d\theta d\phi=\int_0^{\pi} \int_0^{2\pi} |cos \theta||cos 2\phi| d\theta d\phi,$$

(since $\sigma=(cos \theta \cos \phi, cos \theta \sin \phi, sin \theta)$), but I can't go any further (I can't evaluate the integral because of absolute values) to see if I will reach the result $4\pi$. Please help!

PS - If a surface patch has to be considered open we can exclude 'end-points' of $\theta$ and $\phi$ which won't make difference in calculating the mentioned integral.


1 Answer 1


Ok, Let's go :

$$\sigma _\theta=\begin{pmatrix}-\sin\theta\cos\varphi\\-\sin\theta\sin\varphi\\ \cos\theta\end{pmatrix}\quad \text{and}\quad \sigma _\varphi=\begin{pmatrix}-\cos\theta\sin\varphi\\\cos\theta\cos\varphi\\0\end{pmatrix}.$$

Therefore $$\sigma _\theta\times \sigma _\varphi=\begin{pmatrix}-\cos^2\theta\sin\varphi\\-\cos^2\theta\sin\varphi\\ \cos\theta\sin\theta\end{pmatrix} $$

and thus $$\|\sigma _\theta\times \sigma _\varphi\|=\sqrt{\cos^4\theta+\cos^4\theta\sin^2\varphi+\cos^2\theta\sin^2\theta}=\sqrt{\cos^2 \theta}=|\cos \theta|.$$

We finally get $$\text{Area}(\mathbb S^1)=\int_0^{2\pi}\int_0^\pi|\cos\theta|\mathrm d\theta \mathrm d\varphi=2\pi\int_0^\pi|\cos \theta|\mathrm d\theta=4\pi\int_0^{\pi/2}\cos\theta \mathrm d\theta=4\pi.$$

  • $\begingroup$ the result of my integral is 8 which doesn't give the desired result i.e. $4\pi$ but the method should be correct, why is the contradiction? Thank you $\endgroup$
    – user231343
    Sep 27, 2015 at 13:03
  • $\begingroup$ Yes, I corrected it. It's wrong because $\|\sigma _\theta\times \sigma _\phi\|=|\cos \theta|$ and not what you said. $\endgroup$
    – Surb
    Sep 27, 2015 at 13:09
  • $\begingroup$ Ok, You have the complete solution now :-) $\endgroup$
    – Surb
    Sep 27, 2015 at 13:18
  • $\begingroup$ Thanks a lot. (my mistake was in σθ.σϕ=0) $\endgroup$
    – user231343
    Sep 27, 2015 at 13:31

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