Area of Surface of Revolution/Polar Form

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\SS$ be a surface of revolution such that:

$\SS$ is embedded in a cylindrical coordinate space $\polar {r, \theta, z}$
the axis of revolution of $\SS$ is aligned with the polar axis
the curve $\CC$ being rotated to generate $\SS$ is the plane curve:
$r = \map r \theta$
where $\theta$ is in the closed interval $\closedint a b$.


Then the area of $\SS$ is given by:

$\ds \map \AA \SS = 2 \pi \int_a^b r \sin \theta \sqrt {r^2 + \paren {\dfrac {\d r} {\d \theta} }^2} \rd \theta$




Proof




Sources