To calculate the surface area of a solid of revolution I have the following formulae for "4 cases" I've observed
1-When the curve rotates around the X-axis
i) When the limit of integration is in the X-axis
$S = \int 2 \pi f(x) \sqrt{1+(f'(x))^2}dx\ $ |Where $f(x)=function$
ii) When the limit of integration is in the Y-axis
$S = \int 2 \pi y \sqrt{1+(f'(y))^2}dy$ |Where $y = a\ number\ on\ the\ Y-axis$
2-When the curve rotates around the Y-axis
i) When the limit of integration is in the X-axis
$S = \int 2 \pi x \sqrt{1+(f'(x))^2}dx $ |Where $x = a\ number\ on\ the\ X-axis$
ii) When the limit of integration is in the Y-axis
$S = \int 2 \pi f(y) \sqrt{1+(f'(y))^2}dy$ |Where $f(y)=function$
I was wondering if my assessment of the math was correct since I still get confused when working on exercises.
The general approach you seek expressed quite succinctly in Pappus's $(1^{st})$ Centroid Theorem.
This theorem states that the surface area $A$ of a surface of revolution generated by rotating a plane curve $C$ about an axis external to $C$ and on the same plane is equal to the product of the arc length $s$ of $C$ and the distance $d$ traveled by its geometric centroid. Simply put, $S=2\pi RL$, where $R$ is the normal distance of the centroid to the axis of revolution and $L$ is curve length. The centroid of a curve is given by
$$\mathbf{R}=\frac{\int \mathbf{r}ds}{\int ds}=\frac{1}{L} \int \mathbf{r}ds$$