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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.

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  • $\begingroup$ Everything looks good to go. Just know how to translate these formulas if necessary to other axes of rotation parallel to either the $x$ or $y$ axes. $\endgroup$ – Ninad Munshi Nov 6 '19 at 6:18
  • $\begingroup$ We are only working with the X and Y axes of rotation! $\endgroup$ – Max Nov 6 '19 at 6:54
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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

$$\mathbf{R}=\frac{\int \mathbf{r}ds}{\int ds}=\frac{1}{L} \int \mathbf{r}ds$$

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