Area inside Astroid

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.

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

1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 11$: Special Plane Curves: Hypocycloid with Four Cusps: $11.10$
1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.21$: The Cycloid: Problem $6$
2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 9$: Special Plane Curves: Hypocycloid with Four Cusps: $9.10.$