Can we prove the formula for surface of revolution? This is math. We like to prove things. However, proofs are rigorous processes (for a good reason) and are more than just "that idea looks like it could make sense".
I've seen proofs for many different things, including in calculus, but I have yet to come across a proof for the formula of the surface of revolution
$$
A=2\pi\int_{a}^{b}y\,
\sqrt{\,1 + \left(\frac{{\rm d}y}{{\rm d}x}\right)^{\vphantom{\large A}2}\,}\,
\,{\rm d}x
$$
It obviously "makes sense", when you think about how you're unwrapping thin rectangular strips from the circles and adding them all up. However, that doesn't seem to be a very rigorous and indisputable proof! Right, it's worked so far, but how are we so sure that it always will?
Wikipedia says nothing for both volume and surface of solids of revolution. I've been around on Google, and there's still no evidence of a proof (although it might just be buried under all that math help for Calculus students...)
 A: If you accept the definition of area of a general surface $S$ parametrized by $u$ and $v$:
$$
A = \int_S \bigg| \frac{\partial\vec x}{\partial u} \times \frac{\partial\vec x}{\partial v} \bigg|\,du\,dv,
$$
(this one makes "a lot" of intuitive sense if you think of the area of a parallelogram), you can recover your formula as a special case.
Let's move to polar coordinates $(r,\theta,z)$. Now for a surface of revolution, the surface is given simply by $r(z)$, and $r$ does not depend on $\theta$. 
The simplest way to substitute is to take $u=\theta$ and $v=z$. 
Now $\vec x= (r\cos\theta, r\sin\theta, z)$, so (remember that $r$ is a function of $z$!):
$$
\frac{\partial\vec x}{\partial \theta} = (-r\sin\theta, r\cos\theta, 0),
$$
and:
$$
\frac{\partial\vec x}{\partial z} = \bigg(\frac{dr}{dz}\cos\theta,\frac{dr}{dz}\sin\theta,1\bigg).
$$
We get ($\cos^2+\sin^2=1$):
$$
\frac{\partial\vec x}{\partial \theta} \times \frac{\partial\vec x}{\partial z}  = \bigg( r\cos\theta, r\sin\theta, -r\frac{dr}{dz} \bigg),
$$
so:
$$
\bigg|\frac{\partial\vec x}{\partial \theta} \times \frac{\partial\vec x}{\partial z} \bigg| = \sqrt{r^2 + \bigg(r\frac{dr}{dz}\bigg)^2} = r\sqrt{1+\bigg(\frac{dr}{dz}\bigg)^2}.
$$
We can finally put it in the integral:
$$
A=\int_a^b\int_0^{2\pi} r\sqrt{1+\bigg(\frac{dr}{dz}\bigg)^2}\,d\theta\,dz.
$$
The integrand does not depend on $\theta$, so it becomes:
$$
A=2\pi\int_a^b r\sqrt{1+\bigg(\frac{dr}{dz}\bigg)^2}\,dz.
$$
If you call $r\to y$ and $z\to x$, you get exactly the integral you are asking.
A: We can approximate a circle by regular polygons. 
In particular, instead of a revolution surface, we take a "polygonal surface" whose section is a regular polygon. As an example, the vase in this picture:

Now, the vase is made of strips which are actually unrollable. Their width is the side of the polygon, given by (see image below)
$$
s = 2a\,\tan \bigg( \frac{\pi}{n} \bigg),
$$
where $n$ is the number of sides of the polygon (for a hexagon, 6).
 
The height of the strip is the length of the light brown lines on the vase above, which is simply the length of the curve given by:
$$
h=\int_{z_1}^{z_2} \sqrt{1 + \bigg( \frac{da}{dz} \bigg)^2}\,dz,
$$
where the apothem now depends on $z$ (which is what you called $x$). 
(I hope you are familiar with the arc length formula.)
The area of a strip is therefore:
$$
\int_{0}^h s\,dh = \int_{z_1}^{z_2} 2a(z)\,\tan \bigg( \frac{\pi}{n} \bigg)\sqrt{1 + \bigg( \frac{da}{dz} \bigg)^2}\,dz.
$$
Since we have $n$ strips, the area of the total surface is then:
$$
A = 2n\tan \bigg( \frac{\pi}{n} \bigg) \int_{z_1}^{z_2} a(z)\,\sqrt{1 + \bigg( \frac{da}{dz} \bigg)^2}\,dz.
$$
Now take the revolution surface $S$. We approximate it with 2 polygonal surfaces of $n$ strips, one inscribed ($I_n$) and one circumscribed ($C_n$). For each $n$, we can write that (can you see why?):
$$
I_n \le S \le C_n.
$$
Using the formula above, for the circumscribed surface the apothem is simply the radius, so:
$$
C_n = 2n\tan \bigg( \frac{\pi}{n} \bigg) \int_{z_1}^{z_2} r(z)\,\sqrt{1 + \bigg( \frac{dr}{dz} \bigg)^2}\,dz.
$$
For the inscribed one, it is easy to see (look again at the second picture) that the apothem is:
$$
a = r\cos\bigg(\frac{\pi}{n}\bigg),
$$
so:
$$
I_n = C_n\cos\bigg(\frac{\pi}{n}\bigg).
$$
Now we have that:
$$
\lim_{n\to\infty} C_n = \lim_{n\to\infty} 2n \tan \bigg( \frac{\pi}{n}\bigg) \int\dots = 2\pi \int\dots
$$
and $I_n$ has the same limit, as $\cos(\pi/n)\to 1$. Since $I_n\le S\le C_n$, it must be then that:
$$
S = 2\pi \int_{z_1}^{z_2} r(z)\,\sqrt{1 + \bigg( \frac{dr}{dz} \bigg)^2}\,dz.
$$
A: What you are looking for is called Pappus's centroid theorem. Centroid, because we calculate the distance that is traveled by the geometric center, or centroid, of the curve, in order to calculate the solid of revolution.
https://en.wikipedia.org/wiki/Pappus%27s_centroid_theorem
