Proof that projection of area element $\text{d}A$ over sphere is $\text{d}A/\cos(\theta)$

Question

Let's say I have a ray of area $\text{d}A$ travelling in the $z$ direction. The ray encounters a sphere and is incident on it at a region which lies at an angle $\theta$ with respect to the $z$ axis. The projection of $\text{d}A$ on the sphere is $\text{d}A'$. The following figure illustrates the situation better than my words (sorry for the low quality):

I'd like to show that

$\begin{eqnarray} \text{d}A' & = & \frac{\text{d}A}{\cos(\theta)}. \end{eqnarray}$

This is what I've done so far (with help from a figure I found on another post here in finding $\text{d}A'$; said figure is reproduced below; note that the $\text{d}A$ of the figure is my $\text{d}A'$):

I start with

$\begin{eqnarray} \text{d}A & = & \text{d}x\text{d}y,\\ \text{d}A' & = & r^2\sin(\theta)\text{d}\theta\text{d}\varphi,\\ x & = & r\sin(\theta)\cos(\varphi),\\ y & = & r\sin(\theta)\sin(\varphi), \end{eqnarray}$

whereby

$\begin{eqnarray} \text{d}x & = & r\left(\cos(\theta)\cos(\varphi)\text{d}\theta-\sin(\theta)\sin(\varphi)\text{d}\varphi\vphantom{a^2}\right),\\ \text{d}y & = & r\left(\cos(\theta)\sin(\varphi)\text{d}\theta+\sin(\theta)\cos(\varphi)\text{d}\varphi\vphantom{a^2}\right) \end{eqnarray}$

and thus, ignoring all instances of $\text{d}^2\theta$ and $\text{d}^2\varphi$,

$\begin{eqnarray} \text{d}A & = & r^2\cos(\theta)\sin(\theta)\left(\cos^2(\varphi)-\sin^2(\varphi)\vphantom{a^2}\right)\text{d}\theta\text{d}\varphi. \end{eqnarray}$

This is, obviously, not equal to $\cos(\theta)\text{d}A'$ because of the $-$ sign. As far as I can tell, my derivation of $\text{d}A$ in spherical coordinates is perfectly fine. The only things I think could be wrong are the expression I am using for $\text{d}A'$ and what I am trying to prove in the first place.

Now, if I think about what happens when $\theta\rightarrow\pi/2$, I become confused. On one hand, the projection of $\text{d}A$ onto the sphere must initially grow as $\theta$ increases. On the other hand, at $\theta=\pi/2$ the projection is reduced to a single point at the pole of a sphere, so the area of the projection must equal zero, which does not happen if $\text{d}A'=\text{d}A/\cos(\theta)$. This makes me think that what I am trying to prove is wrong. Is this the case?

Thanks in advance.

Answer 1

I've figured it out thanks to another figure I found on someone's lecture notes:

It turns out we don't need to use surface elements in spherical coordinates or anything like that; all that is required is to assume that $\text{d}A$ (and thus also $\text{d}A'$) is small enough for us to be able to take $\text{d}A'$ as being flat (because the sphere is locally flat).

The angle formed by $\text{d}A$ in the figure (which is my $\text{d}A'$) and the horizontal plane is $\theta$ because the angle formed by $\hat{n}$ and $\hat{n}_A$ is $\theta$. Therefore, simple geometry tells us that the projection of $\text{d}A$ onto the horizontal plane at the bottom is $\text{d}A\cos(\theta)$ (or, in the case of my figure, $\text{d}A'=\text{d}A/\cos(\theta)$).

And no, it doesn't work when $\theta$ is close to $\pi/2$, which bothers me a little, but only a little. ;)