If the area of the region {(x, y) : x^(2/3) + y^(2/3) ≤ 1, x + y ≥ 0, y ≥ 0} is A, then 256A/π is

Q: Wait, why do we use $x = \sin^3 \theta$ instead of $x = \cos^3 \theta$?

It is purely a matter of convenience! Both substitutions work perfectly because of the identity $\sin^2 \theta + \cos^2 \theta = 1$. Using $x = \sin^3 \theta$ maps $x=0$ to $\theta=0$ and $x=1$ to $\theta=\frac{\pi}{2}$, which aligns beautifully with the standard limits for the Wallis formula.

Q: How do we know the line $y = -x$ bisects the second quadrant area?

The astroid is symmetric about both the x-axis and the y-axis, and also about the lines $y = x$ and $y = -x$. Because the astroid is symmetric about the line $y = -x$, the portion of the astroid in the second quadrant is perfectly divided into two equal areas by this line. This symmetry is a powerful tool that saves us from complex geometric calculations.

Q: Why can't we just use energy conservation or some other physics principle?

While physics principles are beautiful, this is a pure mathematical problem involving the area of a region bounded by a curve. The most efficient path here is to use calculus—specifically, definite integration—to find the area. Think of it as calculating the 'work' done by the curve to enclose the space!

Question

If the area of the region $\{(x, y) : x^{\frac{2}{3}} + y^{\frac{2}{3}} \le 1, x + y \ge 0, y \ge 0\}$ is $A$, then $\frac{256 A}{\pi}$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The curve defined by the equation $x^{2/3} + y^{2/3} = 1$ is known as the **Astroid**. It is a hypocycloid with four cusps located at $(\pm 1, 0)$ and $(0, \pm 1)$.

We are tasked with finding the area of the region defined by the constraints:
1. $x^{2/3} + y^{2/3} \le 1$ (Inside the astroid)
2. $y \ge 0$ (Upper half-plane)
3. $x + y \ge 0$ (Region above the line $y = -x$)

### The Power of Symmetry

Visualizing these constraints reveals that the region covers the entire first quadrant portion of the astroid and exactly half of the second quadrant portion.

Because the astroid is symmetric about the line $y = -x$, the area in the second quadrant is bisected by this line. Consequently, the total area $A$ is the sum of the area of the first quadrant and half the area of the second quadrant.

Since the area of the first quadrant equals the area of the second quadrant, the total area $A$ is given by:
$$A = \frac{3}{2} 	imes (	ext{Area of the first quadrant})$$

### The Calculus Journey

To find the area of the first quadrant, we set up the integral:
$$A_{Q1} = \int_{0}^{1} y \, dx = \int_{0}^{1} (1 - x^{2/3})^{3/2} \, dx$$

To solve this, we use the trigonometric substitution $x = \sin^3 	heta$, which implies $dx = 3 \sin^2 	heta \cos 	heta \, d	heta$. The limits of integration change from $[0, 1]$ to $[0, \pi/2]$.

Substituting these into the integral, we obtain:
$$A_{Q1} = \int_{0}^{\pi/2} (1 - \sin^2 	heta)^{3/2} \cdot 3 \sin^2 	heta \cos 	heta \, d	heta = 3 \int_{0}^{\pi/2} \sin^2 	heta \cos^4 	heta \, d	heta$$

### The Wallis Formula Victory

We apply the **Wallis Formula** for integrals of the form $\int_{0}^{\pi/2} \sin^m 	heta \cos^n 	heta \, d	heta$:
$$\int_{0}^{\pi/2} \sin^2 	heta \cos^4 	heta \, d	heta = \frac{(2-1)!!(4-1)!!}{(2+4)!!} \cdot \frac{\pi}{2} = \frac{1 \cdot 3 \cdot 1}{6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2} = \frac{3}{48} \cdot \frac{\pi}{2} = \frac{\pi}{32}$$

Multiplying by the constant $3$ outside the integral, the area of the first quadrant is:
$$A_{Q1} = 3 \cdot \frac{\pi}{32} = \frac{3\pi}{32}$$

### Final Calculation

Using our symmetry relation, the total area $A$ is:
$$A = \frac{3}{2} \cdot \frac{3\pi}{32} = \frac{9\pi}{64}$$

The problem asks for the value of $\frac{256 A}{\pi}$:
$$\frac{256}{\pi} \cdot \left( \frac{9\pi}{64} ight) = 4 \cdot 9 = 36$$

The final answer is **36**.

Animated Solution for Mathematics - Definite Integration: If the area of the region ${(x, y) : x^{\frac{2}{3}} + y^{\frac{2}{3}} \leq 1, x + y \geq 0, y \geq 0}$ is $A$ , then $\frac{256 A}{π}$ is

Enter Numerical Value:

Visualized Solution

The Astroid Curve

Applying Linear Constraints

Identifying the Required Region

Exploiting Symmetry

Setting up the Integral

Trigonometric Substitution

Simplifying the Integral

Applying Wallis Formula

Calculating Total Area $A$

Final Result

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Power of Symmetry

The Calculus Journey

The Wallis Formula Victory

Final Calculation

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the area of the region {(x,y):x32​+y32​≤1,x+y≥0,y≥0} is A, then π256A​ is