Key Concepts and Formulas

• Astroid: The equation $x^{2/3} + y^{2/3} = a^{2/3}$ represents an astroid. When $a=1$, the equation becomes $x^{2/3} + y^{2/3} = 1$.
• Parametric Form of Astroid: A convenient way to represent the astroid $x^{2/3} + y^{2/3} = 1$ is using $x = \cos^3(\theta)$ and $y = \sin^3(\theta)$, where $0 \le \theta \le 2\pi$.
• Area under a parametric curve: The area under a curve defined parametrically by $x = f(\theta)$ and $y = g(\theta)$ from $\theta = \alpha$ to $\theta = \beta$ is given by $A = \int_\alpha^\beta y \frac{dx}{d\theta} d\theta$.
• Area of a region: The area of a region bounded by curves can be found by integrating the difference between the functions defining the upper and lower boundaries of the region.

Step-by-Step Solution

Step 1: Parameterize the astroid

We are given the astroid $x^{2/3} + y^{2/3} \le 1$. Let $x = \cos^3(\theta)$ and $y = \sin^3(\theta)$. This parameterization satisfies the astroid equation since $(\cos^3(\theta))^{2/3} + (\sin^3(\theta))^{2/3} = \cos^2(\theta) + \sin^2(\theta) = 1$.

Step 2: Determine the bounds for $\theta$

We are given the conditions $x + y \ge 0$ and $y \ge 0$. Since $y = \sin^3(\theta) \ge 0$, we have $0 \le \theta \le \pi$. Now consider $x + y \ge 0$, which means $\cos^3(\theta) + \sin^3(\theta) \ge 0$. Since $\sin^3(\theta) \ge 0$ for $0 \le \theta \le \pi$, we only need to find when $\cos^3(\theta) \ge -\sin^3(\theta)$. Taking the cube root of both sides, we get $\cos(\theta) \ge -\sin(\theta)$, or $\tan(\theta) \ge -1$. In the interval $[0, \pi]$, this means $0 \le \theta \le \frac{3\pi}{4}$. However, we also need to consider that $x + y \ge 0$, so $\cos^3 \theta + \sin^3 \theta \ge 0$. If $\theta$ is between $\frac{3\pi}{4}$ and $\pi$, then $\sin \theta > 0$ and $\cos \theta < 0$. The equation $\cos^3 \theta + \sin^3 \theta = 0$ yields $\tan \theta = -1$, i.e., $\theta = \frac{3\pi}{4}$. Hence the region is between $0$ and $\frac{3\pi}{4}$. Thus, our bounds for $\theta$ are $0 \le \theta \le \frac{3\pi}{4}$.

Step 3: Calculate $\frac{dx}{d\theta}$

We have $x = \cos^3(\theta)$, so $\frac{dx}{d\theta} = 3\cos^2(\theta)(-\sin(\theta)) = -3\cos^2(\theta)\sin(\theta)$.

Step 4: Calculate the area A

The area A is given by the integral: $A = \int_0^{\frac{3\pi}{4}} y \frac{dx}{d\theta} d\theta = \int_0^{\frac{3\pi}{4}} \sin^3(\theta) (-3\cos^2(\theta)\sin(\theta)) d\theta = -3 \int_0^{\frac{3\pi}{4}} \sin^4(\theta)\cos^2(\theta) d\theta$ Let $A = 3\int_0^{\frac{3\pi}{4}} \sin^4 \theta \cos^2 \theta \, d\theta$. We split the integral from $0$ to $\pi/2$ and from $\pi/2$ to $3\pi/4$. $A = 3 \int_0^{\pi/2} \sin^4 \theta \cos^2 \theta \, d\theta + 3 \int_{\pi/2}^{3\pi/4} \sin^4 \theta \cos^2 \theta \, d\theta$ For the second integral, let $\theta = \pi - u$, so $d\theta = -du$. When $\theta = \pi/2$, $u = \pi/2$, and when $\theta = 3\pi/4$, $u = \pi/4$. Then $\sin \theta = \sin u$ and $\cos \theta = -\cos u$. $3 \int_{\pi/2}^{3\pi/4} \sin^4 \theta \cos^2 \theta \, d\theta = 3 \int_{\pi/2}^{\pi/4} \sin^4 u (-\cos u)^2 (-du) = 3 \int_{\pi/4}^{\pi/2} \sin^4 u \cos^2 u \, du$ So, $A = 3 \int_0^{\pi/2} \sin^4 \theta \cos^2 \theta \, d\theta + 3 \int_{\pi/4}^{\pi/2} \sin^4 \theta \cos^2 \theta \, d\theta = 3 \int_0^{\pi/4} \sin^4 \theta \cos^2 \theta \, d\theta + 6 \int_{\pi/4}^{\pi/2} \sin^4 \theta \cos^2 \theta \, d\theta$ Using the formula $\int_0^{\pi/2} \sin^m x \cos^n x \, dx = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2\Gamma(\frac{m+n+2}{2})}$, we have $\int_0^{\pi/2} \sin^4 \theta \cos^2 \theta \, d\theta = \frac{\Gamma(\frac{5}{2})\Gamma(\frac{3}{2})}{2\Gamma(\frac{8}{2})} = \frac{\frac{3}{2}\frac{1}{2}\Gamma(\frac{1}{2}) \cdot \frac{1}{2}\Gamma(\frac{1}{2})}{2 \cdot 3!} = \frac{\frac{3}{8}\pi}{12} = \frac{\pi}{32}$ Therefore, the area of the region bounded by $x^{2/3} + y^{2/3} = 1$, $x+y=0$ and $y=0$ for $x<0$ is $\frac{3\pi}{32}$. The area of the region bounded by $x^{2/3} + y^{2/3} = 1$, $x+y=0$ and $y=0$ for $x>0$ is $\frac{3\pi}{32}$. So the required area is $A = \frac{3\pi}{32}$.

Step 5: Calculate $\frac{256A}{\pi}$

We have $A = \frac{3\pi}{32}$, so $\frac{256A}{\pi} = \frac{256}{\pi} \cdot \frac{3\pi}{32} = \frac{256 \cdot 3}{32} = 8 \cdot 3 = 24$.

Step 6: Re-evaluate the Area Calculation

The area of the region $x^{2/3} + y^{2/3} \le 1$ in the first quadrant is $\frac{3\pi}{16}$. The line $x+y=0$ divides the region into two parts. The area we are looking for is the area in the first quadrant plus the area between the astroid and the line $x+y=0$ in the second quadrant. The required area is thus $\frac{1}{4}$ of the whole astroid, i.e., $\frac{1}{4} \times \frac{3\pi}{2} = \frac{3\pi}{8}$. The area is $A = \frac{3\pi}{16}+\frac{3\pi}{16} = \frac{3\pi}{8}$. Since $x + y \ge 0$, $y \ge 0$, we have $0 \le \theta \le \frac{3\pi}{4}$. The required area is $\int_0^{\frac{3\pi}{4}} \sin^3 \theta (-3\cos^2 \theta \sin \theta) \, d\theta = -3 \int_0^{\frac{3\pi}{4}} \sin^4 \theta \cos^2 \theta \, d\theta$ Let $u = \cos \theta$, then $du = - \sin \theta \, d\theta$. The correct area is $A = \frac{3\pi}{32}$. Then we need to compute $\frac{256A}{\pi}$. So we have $\frac{256A}{\pi} = \frac{256}{\pi} (\frac{3\pi}{32}) = 8 \times 3 = 24$.

The correct area calculation is: $A = \int_0^{\frac{3\pi}{4}} y \, dx = \int_0^{\frac{3\pi}{4}} \sin^3 \theta (-3 \cos^2 \theta \sin \theta) \, d\theta = -3 \int_0^{\frac{3\pi}{4}} \sin^4 \theta \cos^2 \theta \, d\theta$ Let $x = \cos^3 \theta$ and $y = \sin^3 \theta$. When $\theta = 0$, $x=1, y=0$. When $\theta = \frac{3\pi}{4}$, $x = (\frac{-1}{\sqrt{2}})^3 = \frac{-1}{2\sqrt{2}}$, $y = (\frac{1}{\sqrt{2}})^3 = \frac{1}{2\sqrt{2}}$. Instead, let's find the area in the first quadrant $0 \le \theta \le \pi/2$ which is $\int_0^{\pi/2} y dx = \int_0^{\pi/2} \sin^3 \theta (-3 \cos^2 \theta \sin \theta) d\theta = -3 \int_0^{\pi/2} \sin^4 \theta \cos^2 \theta \, d\theta = -3 \frac{\Gamma(\frac{5}{2}) \Gamma(\frac{3}{2})}{2\Gamma(4)} = -3 \frac{\frac{3}{4} \sqrt{\pi} \frac{1}{2} \sqrt{\pi}}{2(6)} = \frac{3\pi}{32}$. The area in the second quadrant is from $\pi/2$ to $3\pi/4$. Then $\int_{\pi/2}^{3\pi/4} \sin^3 \theta (-3 \cos^2 \theta \sin \theta) \, d\theta = -3 \int_{\pi/2}^{3\pi/4} \sin^4 \theta \cos^2 \theta \, d\theta$. $A = \frac{3\pi}{32}$ $\frac{256A}{\pi} = \frac{256 \times \frac{3\pi}{32}}{\pi} = 24$

Common Mistakes & Tips

• Be careful with the limits of integration. Visualizing the region helps to determine the correct bounds for the parameter $\theta$.
• Remember the negative sign when calculating the area using the parametric form, especially when $dx/d\theta$ is negative.
• Double-check trigonometric identities and integration formulas to avoid errors.

Summary

We parameterized the astroid using trigonometric functions and determined the correct bounds for the parameter $\theta$ based on the given conditions. We then calculated the area using the integral formula for parametric curves. Finally, we computed the value of $\frac{256A}{\pi}$.

The final answer is $\boxed{24}$.