TITLE: Find surface area by calculating surface integrals
QUESTION [0 upvotes]: Fix a radius $r > 0$ and two angles $ϕ_1$ and $ϕ_2$, with $−π/2 < ϕ_1 < ϕ_2 < π/2$ 
Find the surface area of the portion of the sphere of radius r with latitudes between $ϕ_1$ and $ϕ_2$.  
By using Green's theorem we can find the surface area of the sphere
Let $S(u,v) = (sin(u)cos(v),cos(u)cos(v),sin(v))$ be a parametrization for the sphere.
But this uses only one latitude, how do I compute the surface area between the two latitudes?

REPLY [0 votes]: You cannot use Green's theorem here, because the surface is not planar. 
You must start by parametrizing the surface. Since it belongs to the sphere, and that $\phi_2 < \frac{\pi}{2}$, you can write:
$$
x=x, \quad y=y, \quad z=\sqrt{R^2-x^2-y^2},
$$
with $(x,y)$ belonging to the projection of the surface in the $xy$ plane, namely the ring (in polar coordinates) $ \in D=\{(r,\theta)\;|\; r_1 \le r \le r_2, 0\le \theta \le 2\pi \}$. $r_1$ and $r_2$ are given by $\phi_1$ and $\phi_2$, more precisely, since $z=R\cos(\phi)$, the intersection with $x^2+y^2+z^2=R^2$ yields:
$$
x^2+y^2=R^2\sin^2(\phi),
$$
in other words $r_1=R\sin(\phi_1)$ and $r_2=R\sin(\phi_2)$.
You now have everything you need to finish:
$$
A = \iint_D \|r_x \times r_y \| dA = \iint_D \sqrt{\frac{R^2 +2(x^2+y^2) }{R^2+x^2+y^2} }dA,
$$
which is easier to compute with polar coordinates:
$$
A=\int_0^{2\pi}\int_{R\sin(\phi_1)}^{R\sin(\phi_2)}\sqrt{\frac{R^2 +2r^2 }{R^2+r^2} }rdrd\theta
$$
You can take it from there I think, the rest is computing…
Alternative solution. Spherical coordinates is not a bad idea either, the computation looks much easier:
$$
x=R\sin(\phi)\cos(\theta), \quad y=R\sin(\phi)\sin(\theta), \quad z=R\cos(\phi),
$$
with $\phi_1 \le \phi \le \phi_2$ and $0 \le \theta \le 2\pi$. It follows that
$$
A = \int_0^{2\pi}\int_{\phi_1}^{\phi_2}\|r_{\phi}\times r_{\theta} \| d\phi d\theta = \int_0^{2\pi}\int_{\phi_1}^{\phi_2} R^2 \sin(phi) d\phi d\theta \\
= 2\pi R^2(\cos(\phi_1)-\cos(\phi_2)).
$$
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