Area inside Astroid

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Theorem

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

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


Proof

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


AstroidArea.png


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

By Equation of Astroid:

$\begin{cases}

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

Thus:

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


Simplifying the integrand:

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


Thus:

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

$\blacksquare$


Sources

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