PX275 - B4b - example - a cone

taking a cone of radius, $1 m$ , and height, $3 m$ , by revolving the line, $y = \frac{x}{3}$ , around the $x$ -axis

\begin{aligned} A & = \int_{s_{1}}^{s_{2}} 2 π y d s \\ d s & = d x \sqrt{1 + {(\frac{d y}{d x})}^{2}} \\ A & = \int_{x = 0}^{3} 2 π y \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x \\ = \int_{x = 0}^{3} 2 π \frac{x}{3} \sqrt{1 + {(\frac{1}{3})}^{2}} d x \\ = \int_{x = 0}^{3} 2 π \frac{x}{3} \sqrt{\frac{10}{9}} d x \\ = \frac{π}{3} \sqrt{\frac{10}{9}} [x^{2}]_{0}^{3} \\ ∴ A & = π \sqrt{10} \end{aligned}

returning to centroids

centroid in y of the function/curve that is being revolved:

2 π \bar{y} = \frac{2 π \int y d s}{\int d s}

2 π \bar{y} = \frac{A}{S}

where, $A$ is surface area of the rotation, and $S$ is the length of the original curve

pappus's second theorem:

A = 2 π \bar{y} S