Area of Surface of Revolution from Astroid

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Theorem

Let $H$ be the astroid constructed within a circle of radius $a$.

The surface of revolution formed by rotating $H$ around the $x$-axis:


AstroidSurfaceOfRevolution.png


evaluates to:

$\SS = \dfrac {12 \pi a^2} 5$


Proof

By symmetry, it is sufficient to calculate the surface of revolution of $H$ for $0 \le x \le a$.

From Area of Surface of Revolution, this surface of revolution is given by:

$\ds \SS = 2 \int_0^{\pi / 2} 2 \pi y \, \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2} \rd \theta$


From Equation of Astroid:

$\begin{cases}

x & = a \cos^3 \theta \\ y & = a \sin^3 \theta \end{cases}$

so:

\(\ds \frac {\d x} {\d \theta}\) \(=\) \(\ds -3 a \cos^2 \theta \sin \theta\)
\(\ds \frac {\d y} {\d \theta}\) \(=\) \(\ds 3 a \sin^2 \theta \cos \theta\)


Hence:

\(\ds \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2}\) \(=\) \(\ds \sqrt {9 a^2 \paren {\sin^4 \theta \cos^2 \theta + \cos^4 \theta \sin^2 \theta} }\)
\(\ds \) \(=\) \(\ds 3 a \sqrt {\sin^2 \theta \cos^2 \theta \paren {\sin^2 \theta + \cos^2 \theta} }\)
\(\ds \) \(=\) \(\ds 3 a \sqrt {\sin^2 \theta \cos^2 \theta}\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds 3 a \sin \theta \cos \theta\)


Thus:

\(\ds \SS\) \(=\) \(\ds 2 \int_0^{\pi / 2} 2 \pi a \sin^3 \theta \, 3 a \sin \theta \cos \theta \rd \theta\)
\(\ds \) \(=\) \(\ds 12 \pi a^2 \int_0^{\pi / 2} \sin^4 \theta \cos \theta \rd \theta\)
\(\ds \) \(=\) \(\ds 12 \pi a^2 \intlimits {\frac {\sin^5 \theta} 5} 0 {\pi / 2}\) Primitive of $\sin^n a x \cos a x$
\(\ds \) \(=\) \(\ds \frac {12 \pi a^2} 5 \paren {\sin^5 \theta \frac \pi 2 - \sin^5 \theta 0}\) evaluating limits of integration
\(\ds \) \(=\) \(\ds \frac {12 \pi a^2} 5 \paren {1 - 0}\)
\(\ds \) \(=\) \(\ds \frac {12 \pi a^2} 5\)

$\blacksquare$


Sources

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