Surface Area

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We practice setting up integrals for the surface area of surfaces of revolution.

Online Texts

Examples

Suppose the graph $y = 2\sqrt {x}$ between $x = 0$ and $x = 1$ is revolved around the $x$ -axis. Compute the area of the resulting surface.

The distance from the $x$ -axis to the point $(x,2\sqrt {x})$ is $\answer {2 \sqrt {x}}$ .
For this curve, the arc length element satisfies $ds = \sqrt {1 + \left ( \frac {dy}{dx} \right )^2} dx = \answer {\sqrt {1 + \frac {1}{x}}} dx$ $\begin {aligned} A & = \int _{\answer {0}}^{\answer {1}} \answer {2\sqrt {x}} ds \\ & = \int _{\answer {0}}^{\answer {1}} \answer {2 \sqrt {x+1}} dx \\ &= \answer {\frac {4}{3} \left ( 2\sqrt {2} - 1 \right )}. \end {aligned}$