Area of Surface of Revolution from Astroid

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:

Hence:

Thus: