I'm given a surface $S = \{(r, \theta, z): a \leq z \leq b, r = g(z)\}$ in cylindrical coordinates. Now I have to derive the area of surface of rotation which is $\int_{a}^{b}2\pi g(z)\, \sqrt{\,1 + (g`(z))^{\vphantom{\large A}2}\,}\, \,{\rm d}z$.
The way I'm told to do this is by turning the cylindrical coordinates into Cartesian coordinates and then using the normal surface area formula but I don't know how to do it. I know that the bounds for $z$ will remain the same but I don't know how to get the bounds for $x$ and $y$ and how to change cylindrical coordinates to Cartesian. Also I don't know how to get the equation that I have to get the derivatives of to make the surface area equation. Any help will be appreciated.
Here is one way to do it, using Cartesian coordinates as you suggest:
Notice that by symmetry, we may compute the area of 1/4 of it, and multiply by 4. So let's just compute 4 times the area from $\theta=0$ to $\theta=\pi/2$. Then, since $r^2 = x^2+y^2$, and also $r=g(z)$, we have $g(z)^2=x^2+y^2$. We may then solve for $y$ and view this part of the surface as the graph of $$ y = f(x,z) = \sqrt{g(z)^2-x^2} $$ Now you may use the standard techniques in Cartesian coordinates. The area differential is then $$ dA = \sqrt{1+f_x^2+f_z^2} \, dx \, dz = \frac{\sqrt{1+(g'(z))^2}}{\sqrt{1-\left(\frac{x}{g(z)}\right)^2}} \, dx \, dz $$ Projecting to the $x,z$-plane, you see that the domain is $D = \{(x,z) ~|~ a \leq z \leq b ~ \mathrm{and} ~ 0 \leq x \leq g(z)\}$.
Now do the standard double-integral, remembering to multiply by 4: $$ \begin {align*} A &= 4 \iint\limits_D dA \\ &= 4 \int_a^b \int_0^{g(z)} \frac{\sqrt{1+(g'(z))^2}}{\sqrt{1-\left(\frac{x}{g(z)}\right)^2}} \, dx \, dz \\ \end {align*} $$
At this point, do a substitution $u=\frac{x}{g(z)}$, and the integral becomes
$$ \begin {align*} A &= 4 \int_a^b \int_0^1 g(z) \sqrt{1+(g'(z))^2} \, \frac{1}{\sqrt{1-u^2}} \, du \, dz \\ &= 4 \int_a^b g(z) \sqrt{1+(g'(z))^2} \, \left[ \sin^{-1}(x) \right]_0^1 \, dz \\ &= 2 \pi \int_a^b g(z) \sqrt{1+(g'(z))^2} \, dz \end{align*} $$