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Area of a Hypocycloid

Consider the hypocycloid with equation  
\[x^{\frac{2}{3}}+y^{\frac{2}{3}}= a^{\frac{2}{3}}\]

A parametrization of the curve is  
\[(x,y)=(a \: cos^3 \: \theta , a \: sin^3 \: \theta) \]
}
From these parameters,  
\[dx= -3a \: cos^2 \: \theta sin \: \theta \: d \theta, dy=3a \: sin^2 \: \theta cos \: \theta \: d \theta \]

\[\begin{equation} \begin{aligned} A &= \frac{1}{2} \oint_C x \: dy - y \: dx \\ &= \frac{1}{2} \int^{2 \pi}_0 3a^2 (sin^2 \: \theta cos^4 \: \theta + sin^4 \: \theta cos^2 \: \theta )d \theta \\ &= \frac{3a^2}{2} \int^{2 \pi}_0 sin^2 \: \theta cos^2 \: \theta d \theta \\ &= \frac{3a^2}{8} \int^{2 \pi}_0 sin^2 \: \theta d \theta \\ &= \frac{3a^2}{8} \int^{2 \pi}_0 \frac{1}{2} - \frac{cos \: 2 \theta}{2} \: d \theta \\ &= \frac{3a^2}{8} [\frac{\theta}{2} - \frac{ sin \: 2 \theta}{4}]^{2 \pi}_0 \\ &= \frac{3 a^2 \pi}{8} \end{aligned} \end{equation}\]