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Step-by-Step Solution
Step 1: Understand the Region
The region described by
x^{\tfrac{2}{3}} + y^{\tfrac{2}{3}} \leq 1 ,
x + y \ge 0 , and
y \ge 0
is the portion of the standard astroid (given by x^{\tfrac{2}{3}} + y^{\tfrac{2}{3}} = 1 ) in the first quadrant and partly extending into the region where x could be negative but x + y \ge 0 and y \ge 0 . We aim to find its area A and then compute \dfrac{256A}{\pi} .
Step 2: Set Up the Integral
To handle the area part, we break the integral into two regions with respect to x :
The region from x = -\bigl(\tfrac{1}{2}\bigr)^{\tfrac{3}{2}} to 0 , so that x + y \ge 0 and y \ge 0 holds.
The region from x = 0 to x = 1 , the standard portion in the first quadrant.
Therefore, to find the total area, we consider:
A = \int_{-\left(\frac{1}{2}\right)^{\frac{3}{2}}}^{\,0}
\left[ \Bigl(1 - x^{\tfrac{2}{3}}\Bigr)^{\tfrac{3}{2}} + x \right] \,dx
+ \int_{0}^{1}
\Bigl(1 - x^{\tfrac{2}{3}}\Bigr)^{\tfrac{3}{2}} \,dx.
Step 3: Split the Integral and Make Substitution
Observe that
\int_{-\left(\frac{1}{2}\right)^{\frac{3}{2}}}^{\,0}
\Bigl(1 - x^{\tfrac{2}{3}}\Bigr)^{\tfrac{3}{2}} \,dx
\quad\text{and}\quad
\int_{0}^{1}
\Bigl(1 - x^{\tfrac{2}{3}}\Bigr)^{\tfrac{3}{2}} \,dx
can be handled by the substitution
x = \sin^3 \theta.
Then
dx = 3\,\sin^2\theta\,\cos\theta\, d\theta.
In addition, for the linear term
\int_{-\left(\frac{1}{2}\right)^{\frac{3}{2}}}^{\,0} x\,dx,
we can integrate directly.
Step 4: Apply the Bounds for Substitution
With x = \sin^3 \theta ,
When x = -\bigl(\tfrac{1}{2}\bigr)^{\tfrac{3}{2}} , \sin^3 \theta = -\frac{1}{2} , giving \theta = -\tfrac{\pi}{4}.
When x = 0 , \sin \theta = 0, so \theta = 0.
When x = 1 , \sin^3 \theta = 1, so \theta = \tfrac{\pi}{2}.
Step 5: Evaluate the Integrals
After performing the substitution and splitting into the relevant intervals, the integral involving
\bigl(1 - x^{\tfrac{2}{3}}\bigr)^{\tfrac{3}{2}}
becomes an integral of the form
\int \sin^2 \theta \cos^4 \theta \, d\theta,
which can be calculated using standard trigonometric identities and methods. Meanwhile, the integral of x from
-\left(\tfrac{1}{2}\right)^{\tfrac{3}{2}}
to 0 is straightforward:
\int_{-\left(\frac{1}{2}\right)^{\frac{3}{2}}}^{\,0} x \,dx
= \left[\frac{x^2}{2}\right]_{-\left(\frac{1}{2}\right)^{\frac{3}{2}}}^{\,0}
= 0 - \frac{1}{2}\Bigl(-\bigl(\tfrac{1}{2}\bigr)^{\tfrac{3}{2}}\Bigr)^2
= \frac{1}{16}.
Combining all terms carefully (including signs and limits) yields:
A
= \int_{-\left(\frac{1}{2}\right)^{\frac{3}{2}}}^{\,0}
\Bigl(1 - x^{\tfrac{2}{3}}\Bigr)^{\tfrac{3}{2}} \,dx
+ \int_{0}^{1}
\Bigl(1 - x^{\tfrac{2}{3}}\Bigr)^{\tfrac{3}{2}} \,dx
+ \int_{-\left(\frac{1}{2}\right)^{\frac{3}{2}}}^{\,0} x \,dx.
After detailed trigonometric integration and simplification, the result is
A = \frac{36\pi}{256}.
Step 6: Final Calculation
We have
A = \frac{36\pi}{256}.
Hence,
\frac{256A}{\pi}
= \frac{256 \times \frac{36\pi}{256}}{\pi}
= 36.
Step 7: Conclusion
Therefore, the value of
\dfrac{256A}{\pi}
is
36.
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