Area of Surface of Revolution

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Theorem

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

$\SS$ is embedded in a Cartesian $3$-space
the axis of revolution of $\SS$ is aligned with the $x$-axis
the curve $\CC$ being rotated to generate $\SS$ is the plane curve $y = \map f x$
$\CC$ has endpoints at $x = a$ and $x = b$.


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

$\ds \map \AA \SS = 2 \pi \int_a^b y \sqrt {1 + \paren {\dfrac {\d y} {\d x} }^2} \rd x$


Polar Form

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:

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


Parametric Form

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

$\SS$ is embedded in a Cartesian $3$-space
the axis of revolution of $\SS$ is aligned with the $x$-axis
the curve $\CC$ being rotated to generate $\SS$ is the plane curve described by the parametric equations:
\(\ds \quad \ \ \) \(\ds x\) \(=\) \(\ds \map x t\)
\(\ds y\) \(=\) \(\ds \map y t\)
where $t$ is in the closed interval $\closedint a b$.


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

$\ds \map \AA \SS = 2 \pi \int_a^b y \sqrt {\paren {\dfrac {\d x} {\d t} }^2 + \paren {\dfrac {\d y} {\d t} }^2} \rd t$


Proof


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Sources

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