# Surface area of a solid of revolution, general approach

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.

• 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. – Ninad Munshi Nov 6 '19 at 6:18
• We are only working with the X and Y axes of rotation! – Max Nov 6 '19 at 6:54

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