Surface Element in Spherical Coordinates In spherical polars, 
$$x=r\cos(\phi)\sin(\theta)$$
$$y=r\sin(\phi)\sin(\theta)$$
$$z=r\cos(\theta)$$
I want to work out an integral over the surface of a sphere - ie $r$ constant. I'm able to derive through scale factors, ie $\delta(s)^2=h_1^2\delta(\theta)^2+h_2^2\delta(\phi)^2$ (note $\delta(r)=0$), that:
$$h_1=r\sin(\theta),h_2=r$$
$$dA=h_1h_2=r^2\sin(\theta)$$
I'm just wondering is there an "easier" way to do this (eg. Jacobian determinant when I'm varying all 3 variables). I know you can supposedly visualize a change of area on the surface of the sphere, but I'm not particularly good at doing that sadly.

Because of the striking fact that
(a)  The area of [a slice of the spherical surface between two parallel planes (within the poles)] is proportional to its width.
. . . here's a rarely (if ever) mentioned way to integrate over a spherical surface. We assume the radius = 1.
(b) Note that every point on the sphere is uniquely determined by its z-coordinate and its counterclockwise angle phi, $0 \leq\phi\leq 2\pi$, from the half-plane y = 0,
x >= 0.
From (a) and (b) it follows that an element of area on the unit sphere centered at the origin in 3-space is just dphi dz.
Then the integral of a function f(phi,z) over the spherical surface is just
$$\int_{-1 \leq z \leq 1, 0 \leq \phi \leq 2\pi} f(\phi,z) d\phi dz$$.
