# Area inside Astroid

## Theorem

The area inside an astroid $H$ constructed within a circle of radius $a$ is given by:

$\mathcal A = \dfrac {3 \pi a^2} 8$

## Proof

Let $H$ be embedded in a cartesian coordinate plane with its center at the origin and its cusps positioned on the axes.

By symmetry, it is sufficient to evaluate the area shaded yellow and to multiply it by $4$.

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

Thus:

 $\displaystyle \mathcal A$ $=$ $\displaystyle 4 \int_0^a y \rd x$ $\displaystyle$ $=$ $\displaystyle 4 \int_{x \mathop = 0}^{x \mathop = a} y \frac {\d x} {\d \theta} \rd \theta$ $\displaystyle$ $=$ $\displaystyle 4 \int_{x \mathop = 0}^{x \mathop = a} a \sin^3 \theta \, 3 a \cos^2 \theta \sin \theta \paren {-\sin \theta} \rd \theta$ differentiating $a \cos^3 \theta$ with respect to $\theta$ $\displaystyle$ $=$ $\displaystyle 4 \int_{\theta \mathop = \pi / 2}^{\theta \mathop = 0} a \sin^3 \theta \, 3 a \cos^2 \theta \sin \theta \paren {-\sin \theta} \rd \theta$ $x = 0$ when $\theta = \pi / 2$, $x = a$ when $\theta = 0$ $\displaystyle$ $=$ $\displaystyle 12 a^2 \int_0^{\pi / 2} \sin^4 \theta \cos^2 \theta \rd \theta$ simplifying

Simplifying the integrand:

 $\displaystyle \sin^4 \theta \cos^2 \theta$ $=$ $\displaystyle \frac {\paren {2 \sin \theta \cos \theta}^2} 4 \frac {2 \sin^2 \theta} 2$ $\displaystyle$ $=$ $\displaystyle \frac {\sin^2 2 \theta} 4 \frac {2 \sin^2 \theta} 2$ Double Angle Formula for Sine $\displaystyle$ $=$ $\displaystyle \frac {\sin^2 2 \theta} 4 \frac {1 - \cos 2 \theta} 2$ Square of Sine $\displaystyle$ $=$ $\displaystyle \frac {\sin^2 2 \theta - \sin^2 2 \theta \cos 2 \theta} 8$ $\displaystyle$ $=$ $\displaystyle \frac {1 - \cos 4 \theta} {16} - \frac {\sin^2 2 \theta \cos 2 \theta} 8$ Square of Sine

Thus:

 $\displaystyle \mathcal A$ $=$ $\displaystyle 12 a^2 \int_0^{\pi / 2} \paren {\frac {1 - \cos 4 \theta} {16} - \frac {\sin^2 2 \theta \cos 2 \theta} 8} \rd \theta$ $\displaystyle$ $=$ $\displaystyle \frac 3 4 a^2 \int_0^{\pi / 2} \paren {1 - \cos 4 \theta} \rd \theta - \frac 3 2 a^2 \int_0^{\pi / 2} \sin^2 2 \theta \cos 2 \theta \rd \theta$ Linear Combination of Integrals $\displaystyle$ $=$ $\displaystyle \frac 3 4 a^2 \intlimits {\theta - \frac {\sin 4 \theta} 4} 0 {\pi / 2} - \frac 3 2 a^2 \int_0^{\pi / 2} \sin^2 2 \theta \cos 2 \theta \rd \theta$ Primitive of $\cos a x$ $\displaystyle$ $=$ $\displaystyle \frac 3 4 a^2 \intlimits {\theta - \frac {\sin 4 \theta} 4} 0 {\pi / 2} - \frac 3 2 a^2 \intlimits {\frac {\sin^3 2 \theta} 6} 0 {\pi / 2}$ Primitive of $\sin^n a x \cos a x$ $\displaystyle$ $=$ $\displaystyle \frac {3 \pi a^2} 8 - \frac {3 a^2} {16} \sin 2 \pi - \frac {3 a^2} {12} \sin^3 \pi$ evaluating limits of integration $\displaystyle$ $=$ $\displaystyle \frac {3 \pi a^2} 8$ $\sin 2 \pi = 0$ and $\sin^3 \pi = 0$

$\blacksquare$