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?
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
$$ A = \int_0^{2\pi}\int_{\phi_1}^{\phi_2}\|r_{\phi}\times r_{\theta} \| d\phi d\theta = \int_0^{2\pi}\int_{\phi_1}^{\phi_2} R^2 \sin(phi) d\phi d\theta \\ = 2\pi R^2(\cos(\phi_1)-\cos(\phi_2)). $$