Find the area bounded by the astroid: x^(2/3) + y^(2/3) =

Find the area bounded by the astroid: $x^{2/3} + y^{2/3} = a^{2/3}$.

We will use the parametric representation of the astroid together with Wallis integrals.

Step 1: Understand the astroid

The astroid is a star-shaped curve with four cusps located at $(a, 0)$, $(0, a)$, $(-a, 0)$, and $(0, -a)$. The curve is symmetric about both axes and about both diagonals.

Step 2: Use parametric representation

The astroid can be parameterized as:

\[x = a\cos^3 t, \quad y = a\sin^3 t, \quad t \in [0, 2\pi]\]

Verification:

\[x^{2/3} + y^{2/3} = (a\cos^3 t)^{2/3} + (a\sin^3 t)^{2/3}\]

\[= a^{2/3}(\cos^3 t)^{2/3} + a^{2/3}(\sin^3 t)^{2/3}\]

\[= a^{2/3}\cos^2 t + a^{2/3}\sin^2 t = a^{2/3}(\cos^2 t + \sin^2 t) = a^{2/3} \quad \checkmark\]

Step 3: Use symmetry

Due to symmetry, we can calculate the area in the first quadrant and multiply by 4. For the first quadrant: $t \in [0, \frac{\pi}{2}]$

Step 4: Apply the parametric area formula

For a parametric curve $x = x(t)$, $y = y(t)$, the area is:

\[A = \int y \, dx\]

With $x = a\cos^3 t$ and $y = a\sin^3 t$:

\[\frac{dx}{dt} = a \cdot 3\cos^2 t \cdot (-\sin t) = -3a\cos^2 t \sin t\]

Therefore:

\[dx = -3a\cos^2 t \sin t \, dt\]

Step 5: Set up the integral for the first quadrant

As $t$ goes from $0$ to $\frac{\pi}{2}$, $x$ goes from $a$ to $0$ (decreasing).

\[A_{\text{first quadrant}} = \int_0^{\pi/2} y \, dx = \int_0^{\pi/2} a\sin^3 t \cdot (-3a\cos^2 t \sin t) \, dt\]

\[= -3a^2 \int_0^{\pi/2} \sin^4 t \cos^2 t \, dt\]

Since $x$ decreases as $t$ increases, we need the absolute value:

\[A_{\text{first quadrant}} = 3a^2 \int_0^{\pi/2} \sin^4 t \cos^2 t \, dt\]

Step 6: Evaluate the integral

Using the identity:

\[\sin^4 t \cos^2 t = \sin^4 t (1 - \sin^2 t) = \sin^4 t - \sin^6 t\]

We need to evaluate:

\[I = \int_0^{\pi/2} (\sin^4 t - \sin^6 t) \, dt\]

Using Wallis integrals:

\[\int_0^{\pi/2} \sin^{2n} t \, dt = \frac{(2n-1)!!}{(2n)!!} \cdot \frac{\pi}{2}\]

For $\sin^4 t$ (where $2n = 4$, so $n = 2$):

\[\int_0^{\pi/2} \sin^4 t \, dt = \frac{3!!}{4!!} \cdot \frac{\pi}{2} = \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} = \frac{3}{8} \cdot \frac{\pi}{2} = \frac{3\pi}{16}\]

For $\sin^6 t$ (where $2n = 6$, so $n = 3$):

\[\int_0^{\pi/2} \sin^6 t \, dt = \frac{5!!}{6!!} \cdot \frac{\pi}{2} = \frac{5 \cdot 3 \cdot 1}{6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2} = \frac{15}{48} \cdot \frac{\pi}{2} = \frac{5}{16} \cdot \frac{\pi}{2} = \frac{5\pi}{32}\]

Therefore:

\[I = \frac{3\pi}{16} - \frac{5\pi}{32} = \frac{6\pi - 5\pi}{32} = \frac{\pi}{32}\]

Step 7: Calculate the area

\[A_{\text{first quadrant}} = 3a^2 \cdot \frac{\pi}{32} = \frac{3\pi a^2}{32}\]

Total area (all four quadrants):

\[A = 4 \cdot \frac{3\pi a^2}{32} = \frac{12\pi a^2}{32} = \frac{3\pi a^2}{8}\]

\[\boxed{A = \frac{3\pi a^2}{8} \text{ square units}}\]

Step 8: Numerical value

\[A = \frac{3\pi a^2}{8} \approx 1.178 a^2 \text{ square units}\]

Comparison with circle

A circle of radius $a$ has area $\pi a^2$. The astroid has area $\frac{3\pi a^2}{8} = \frac{3}{8}\pi a^2 \approx 0.375\pi a^2$. So the astroid encloses about 37.5% of the area of its circumscribed circle.

Summary

The area enclosed by the astroid $x^{2/3} + y^{2/3} = a^{2/3}$ is:

\[\boxed{A = \frac{3\pi a^2}{8} \approx 1.178a^2 \text{ square units}}\]

Key steps:

1. Use parametric form: $x = a\cos^3 t$, $y = a\sin^3 t$
2. Apply symmetry: calculate first quadrant and multiply by 4
3. Use parametric area formula: $A = \int y \, dx$
4. Compute: $dx = -3a\cos^2 t \sin t \, dt$
5. Evaluate: $\int_0^{\pi/2} \sin^4 t \cos^2 t \, dt = \frac{\pi}{32}$
6. Result: $A = 4 \cdot 3a^2 \cdot \frac{\pi}{32} = \frac{3\pi a^2}{8}$

Key insight:

The astroid is a beautiful curve with cusps at the coordinate axes. Its area is exactly $\frac{3}{8}$ of the area of its circumscribed circle of radius $a$.