Find the surface area cut from the cone $2x^2+2y^2=5z^2,\;z>0$, by the cylinder $x^2+y^2=2y$. In this PDF I found looking for a solution to another problem I had earlier I see that the author claims that $$\iint dA = \iint\sec(\gamma)dydx$$
Could someone explain where this formula comes from?

Bonus points if you can explain how the author could somehow evaluate the first problem on that PDF using this formula without having to find the intersection of the two surfaces and then parametrize the region.
Here's the problem I'm asking about

Find the surface area cut from the cone $2x^2+2y^2=5z^2,\;z>0$, by the cylinder $x^2+y^2=2y$.

Given Solution

 A: First, The projection ("shadow" if you prefer heuristically viewing this) of the differential area $dA$ onto the $x−y$ plane is given by $dxdy/\cos\gamma$, where $\gamma$ is the angle that the $z$-axis makes with the normal to the differential area $dA$.
Now, suppose a surface is described by $\phi(x,y,z)=0$.  A normal vector $\vec N$ to the surface is given by $\vec N=\nabla \phi$.  
Then, the cosine of the angle , say $\gamma$, that $\vec N$ makes with the $z$ axis is given by
$$\cos \gamma =\hat z\cdot\left( \frac{\nabla \phi}{|\nabla \phi|}\right)=\frac{\frac{\partial \phi}{\partial z}}{\sqrt{\left(\frac{\partial \phi}{\partial x}\right)^2+\left(\frac{\partial \phi}{\partial y}\right)^2+\left(\frac{\partial \phi}{\partial z}\right)^2}} \tag 1$$
where $ \frac{\nabla \phi}{|\nabla \phi|}$ is a unit normal to the surface.  Note that there are two unit normal vectors to a surface.  In order to ensure that the integral $\int dA$ is positive, one can take the absolute value of $\cos \gamma$ as given in $(1)$.  Then, we have $\int dA=\int |\sec \gamma|\,dx\,dy$.
