Confusion in using double integrals & projection to calculate hemispherical surface area As a preamble, if we have a surface given by $z=z(x,y)$ in $\mathbb{R}^3$, i.e. $z =z(x,y) = \sqrt{a^2-x^2-y^2}$ then $z-  z(x,y)=0$ gives us a constant surface of a scalar field $f$, the $grad$ of which will tell us the normal vector to our surface: $\nabla f = \mathbf{k} -\frac{∂z}{∂x}\mathbf{i}-\frac{∂z}{∂y}\mathbf{j}$, the norm of which is $\sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2}$


A hemisphere of radius $a$ (with the $z$ axis as the axis of symmetry) is described by the equation $x^2+y^2+z^2 (=r^2) = a^2$ and has a surface area given by
$$\iint dS = \iint \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2}dA$$
If you project onto the $xy$ plane.
This formula is obtained by considering an element of surface area $\mathbf{dS}$ and its projected area $\mathbf{dA}$, and relating the two by $dA = \cos(\alpha) dS = \mathbf{\hat{n}}\cdot\mathbf{k}\, dS \rightarrow dS = dA/\mathbf{\hat{n}}\cdot\mathbf{k} =  \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2}dA $ using $\mathbf{n}$, the normal to the surface, from the preamble and $\mathbf{k}$ the unit vector in the $z$ direction.

For the above example we can use $z =\sqrt{a^2-x^2-y^2}$ to find our partial derivatives, and then evaluate the integral using plane polar co-ordinates ($0 \leq \theta \leq 2\pi, 0 \leq r \leq a $) to reach an answer of $2\pi a^2$ as expected.
However, if we use a normal vector to this surface expressed in 3D polar co-ordinates:
$$\mathbf{\hat{n}} = \mathbf{r}/\|\mathbf{r}\| = (x\mathbf{i} + y\mathbf{j} + z\mathbf{r} )/ \|\mathbf{r}\| = \sin\phi \cos\theta \mathbf{i}+\sin\phi \sin\theta \mathbf{j}+\cos\phi \mathbf{k}$$
our logic above the included diagram should still hold: $dS = dA/\mathbf{\hat{n}}\cdot\mathbf{k}$ (here projecting onto the $xy$ i.e. $r,\theta$ plane).
This yields $dS = dA/\cos\phi$ for our 3d polar case, which has got me stuck on the next step of evaluating
$$\iint dS = \iint \sec\phi dA = \iint \sec(\phi) rdrd\theta$$
It is intuitively true that as the angle $\phi$ increases, our surface area element on the hemisphere becomes more and more vertical w.r.t. the $xy$ plane and so this $1/\cos\phi$ makes our corresponding area element larger, but how should I actually evaluate the integral?
$\iint \sec\phi dA = \iint \sec(\phi) rdrd\theta$ is certainly ringing alarm bells, which makes me think that projecting on the $r,\theta$ plane in the first place was a mistake, which is a problem not encountered when working in 3D cartesians.
Is there such a thing as projecting onto the "$\theta,\phi$ plane" and then integrating over those two variables? - if this were to be a valid method I believe it could solve my problem here. Please do ask for further clarification if it is needed.
 A: First of all, they are usually called spherical coordinates, not polar, when you are in $3D$.  The radial variable is usually called $\rho$ instead of $r$.
The element of area on the surface of a sphere is the product of two infinitesimal lengths.  $\rho d\phi$ is the "changing latitude" length, and we usually say $r = \rho \sin \phi$ for the horizontal radius at that "latitude", so that $\rho \sin \phi d\theta$ is the "east-west" length, giving a surface area element $dS = \rho^2\sin\phi d\phi d\theta$.  On this surface, $\rho = a$.  So yes, you can absolutely integrate that way.
A: Trigger warning: in my explanation I'll be using the physics convention when referring to spherical coordinates.
I see your confusion. I think that yeah, as you point out $\iint \sec\phi dA = \iint \sec(\phi) rdrd\theta$ is a mistake - not in the sense that there's anything logically wrong but it doesn't actually provide us any neat way in calculating the surface area since $\theta \neq \phi$ of course so you have 3 degreees of freedom when you can really just do with 2. For this, it would probably be best reverting to cartesian coordinates - or even better parameterising the surface and finding the normal from there. Hope this helps!

