Change of Coordinates for Surface Area Integral? When using the formula for the surface area integral and using some change of coordinates, i.e. spherical coordinates, does the area element dA change to $r^2sin(\phi)\partial\phi\partial\theta$ ?
Because I noticed in Dr. Paul Lamar's final example here he doesn't use the spherical area element. He simply uses the area element $\partial\phi\partial\theta$.
That's confusing to me. I feel like if he does a spherical change of coordinates, the area element he should be using is the spherical area element. 
 A: If you are referring to Example 2 (the only example with spherical), then notice that he does conclude the surface element changes is $r^2\sin(\phi) d\phi d\theta$, where he calculates that $\|r_{\theta} \times r_{\phi}\| = 4 \sin{\phi}$. This is precisely what you want, since the radius of the sphere was $r = 2$. In that problem, he uses $dA$ to mean $d\phi d\theta$. 
A: My answer supplements Christopher's.  You are referring to Example 2 of Dr. Paul Dawkins's notes.
Even though the partial sphere $S$ lives in Euclidean space $\mathbb R^3,$ notice that it had directly been parameterised in spherical coordinates from the get-go. Since there had not actually been a change of variables/coordinates from Cartesian to spherical, there was no occasion to explicitly change the area differential from, say, $\mathrm dx\mathrm dz,$ to $R^2\sinφ\:\mathrm dφ\mathrm dθ.$
In the formula (second cyan box) for the surface integral of a function on a surface $\mathbf r(u,v),$ note that the area differential $\mathrm dA$ lives in parameter space and simply stands for $\mathrm du\mathrm dv.$ So, it could be $$\mathrm dx\mathrm dz$$ or $$\mathrm dφ\mathrm dθ$$ or $$\mathrm dz\mathrm dxθ,$$ etc. It is not necessarily Cartesian, nor is it automatically substitutable with $r^2\sinφ\:\mathrm dφ\mathrm dθ$ whenever there are spherical coordinates.
