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As picture above, although it is ugly, but the red ball is a sphere with radial $r$ , i.e $$\{(x,y,z)\in R^3 : x^2+y^2+z^2=r^2\},$$ and the green is a spherical neighbourhood of North pole with radial $0<\delta\leq r$.

I want to calculate the area of green spherical neighbourhood.

Consider the coordinate of half-sphere $$ u:(x,y)\rightarrow (x,y,\sqrt{r^2-x^2-y^2}). $$ Then I have $$ u_x=(1,0,\frac{-x}{\sqrt{r^2-x^2-y^2}})~~~u_y=(0,1,\frac{-y}{\sqrt{r^2-x^2-y^2}}), $$ Hence $$ g_{11}=\frac{r^2-y^2}{r^2-x^2-y^2} ~~~ g_{12}=\frac{xy}{r^2-x^2-y^2} ~~~ g_{22}=\frac{r^2-x^2}{r^2-x^2-y^2} \\ \det(g)=\frac{r^2}{r^2-x^2-y^2} $$ Finally the area is $$ S =\int\int_{x^2+y^2\le \delta ^2} \frac{r^2}{r^2-x^2-y^2} dx dy \\ = \int_0^\delta\int_{x^2+y^2=t^2} \frac{r^2}{r^2-x^2-y^2} dx dy dt \\ = \int_0^\delta \frac{2\pi r^2 t}{r^2-t^2} dt = -2\pi r\ln \frac{\sqrt{r^2-\delta^2}}{r}. $$ But obviously, it is not right. Where is the error?


I think that you forgot to take the square-root of $|\det(g)|$. The area of the spherical cap is given by $$\iint_{x^2+y^2\le \delta^2} \frac{r}{\sqrt{r^2-x^2-y^2}} dx dy =2\pi r^2\int_{\rho=0}^{\delta/r} \frac{\rho}{\sqrt{1-\rho^2}} d\rho =2\pi rh.$$ where $h=r-\sqrt{r^2-\delta^2}$ is the height of the cap.

  • $\begingroup$ I’m too pressed for time to check, but as I recall, the answer is much more satisfying when you specify the radius as an angle, just as you specify the length of a geodesic as an angle. $\endgroup$ – Lubin Dec 8 '16 at 14:09
  • $\begingroup$ Thanks, I calculate it again. I think you are right. $\endgroup$ – lanse7pty Dec 9 '16 at 7:34

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