How do I calculate the following surface area? I have encountered this problem during my vector calculus studies which I cannot seem to setup correctly. This is a paraphrasing of the question:
Calculate the surface area of the surface $S$, a cone such that $z^2=x^2+y^2$, bounded above the disc
$x^2+y^2\le4$.


What formula would I use to compute this, or what is the process to getting the correct answer?
What I'd do is to plot the region the exercise is talking about (it's just a cone with the vertex in the bottom at (0,0,0) and a it's top at z = 2) and solve for the triple integral that gives you the volume. That is $\displaystyle\int_{\{(x,y,z) \in \mathbb{R}^3 : x^2+y^2 \leq z^2 \leq 4\}}dxdydz$.

Or if you're used to cylindrical coordinates, try those. They're much easier in this case.
Here, given cone $$z = \sqrt{x^2 + y^2}$$ and as it is given that the required region is bounded above the disc $ \Rightarrow  z \in (0,2] $

You can solve the problem either by relying on the formula of lateral surface area $ S = \pi r l$ where $r =$ Radius, and $ l = $ slanted height.

Or you can use calculus.
Here I will use cylindrical coordinates. 

My area element would be $dA = r d\theta dh \sec \frac{\pi}{4}$ ,Here $r = h$, so
$$ A = \sec \biggr (\frac{\pi}{4} \biggl )\int_0^2 \int^{2\pi}_0 h dr  d \theta $$

Your task is to figure out why there is $\sec \bigr (\frac{\pi}{4} \bigl )$ and $r = h$