- Identify the curve
The boundary
x2/3+y2/3=1
is the astroid.
A standard parametrization of this astroid is
x=cos^3\theta,\qquad y=\sin^3\theta
for 0≤θ≤2π.
Indeed,
x2/3=cos2θ,y2/3=sin2θ,
so
x2/3+y2/3=cos2θ+sin2θ=1.
- Understand the required region
We need the area of
{(x,y):x2/3+y2/3≤1, x+y≥0, y≥0}.
- y≥0 means the upper half of the astroid.
- In the upper half, the line x+y=0 i.e. y=−x cuts the astroid at points where
x=−y.
Substitute into the boundary:
x2/3+(−x)2/3=2∣x∣2/3=1
so
\implies |x|=\left(\frac12\right)^{3/2}=\frac{1}{2\sqrt2}.$$
Thus the intersection point in the upper half is
$$\left(-\frac{1}{2\sqrt2},\frac{1}{2\sqrt2}\right).$$
In parametric form, $x+y=0$ becomes
$$\cos^3\theta+\sin^3\theta=0.$$
In the upper half, this occurs at
$$\theta=\frac{3\pi}{4}.$$
Also, for the upper half, $0\le \theta\le \pi$.
The condition $x+y\ge 0$ keeps the part with
$$0\le \theta\le \frac{3\pi}{4}.$$
So the required region is the sector-like part of the astroid traced by
$$0\le \theta\le \frac{3\pi}{4}.$$
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3. **Area using parametric formula**
For a closed curve segment with parametrization $(x(\theta),y(\theta))$, area under the arc and above the $x$-axis can be found by
$$A=\int y\,dx$$
with sign adjusted appropriately. Since $x$ decreases on $[0,\pi]$, we use
$$A=-\int_0^{3\pi/4} y\,\frac{dx}{d\theta}\,d\theta.$$
Now,
$$x=\cos^3\theta \implies \frac{dx}{d\theta}=-3\cos^2\theta\sin\theta,$$
$$y=\sin^3\theta.$$
Therefore,
$$A=-\int_0^{3\pi/4} \sin^3\theta\,(-3\cos^2\theta\sin\theta)\,d\theta
=3\int_0^{3\pi/4} \sin^4\theta\cos^2\theta\,d\theta.$$
So,
$$A=3\int_0^{3\pi/4} \sin^4\theta\cos^2\theta\,d\theta.$$
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4. **Simplify the integrand**
Use
$$\sin^2\theta=\frac{1-\cos2\theta}{2},\qquad \cos^2\theta=\frac{1+\cos2\theta}{2}.$$
Then
$$\sin^4\theta\cos^2\theta=\left(\frac{1-\cos2\theta}{2}\right)^2\left(\frac{1+\cos2\theta}{2}\right).$$
But it is quicker to use the standard reduction result:
$$\sin^4\theta\cos^2\theta
=\frac{1}{16}-\frac{1}{32}\cos2\theta-\frac{1}{16}\cos4\theta+\frac{1}{32}\cos6\theta.$$
Hence,
$$A=3\int_0^{3\pi/4}\left(\frac{1}{16}-\frac{1}{32}\cos2\theta-\frac{1}{16}\cos4\theta+\frac{1}{32}\cos6\theta\right)d\theta.$$
Integrating,
$$A=3\left[\frac{\theta}{16}-\frac{1}{64}\sin2\theta-\frac{1}{64}\sin4\theta+\frac{1}{192}\sin6\theta\right]_0^{3\pi/4}.$$
Now,
$$\sin\left(\frac{3\pi}{2}\right)=-1,\qquad \sin(3\pi)=0,\qquad \sin\left(\frac{9\pi}{2}\right)=1.$$
So,
$$A=3\left(\frac{3\pi/4}{16}-\frac{1}{64}(-1)-0+\frac{1}{192}(1)\right).$$
Thus,
$$A=3\left(\frac{3\pi}{64}+\frac{1}{64}+\frac{1}{192}\right)
=3\left(\frac{3\pi}{64}+\frac{4}{192}\right)
=3\left(\frac{3\pi}{64}+\frac{1}{48}\right).$$
This introduces a non-$\pi$ term, which is impossible for this symmetric astroid-sector area setup, so let us instead use a cleaner geometric split.
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5. **Cleaner method: upper-half area minus small left triangle-like part**
The total area enclosed by the astroid
$$x^{2/3}+y^{2/3}=1$$
is known to be
$$\frac{3\pi}{8}.$$
Therefore the upper half has area
$$\frac{1}{2}\cdot \frac{3\pi}{8}=\frac{3\pi}{16}.$$
Now we remove the part in the upper half where $x+y<0$, i.e. the small region to the left of the line $y=-x$.
By symmetry of the upper half about the $y$-axis? No, better compute directly:
Required area
$$A=\int_{-1/(2\sqrt2)}^1 y\,dx,$$
where on the upper branch,
$$y=(1-x^{2/3})^{3/2}.$$
This is cumbersome.
So we return to the parametric method, but compute correctly via Green's formula for the sector.
For parametrized arc from $\theta=0$ to $\theta=3\pi/4$, together with line segments to the origin, the area is
$$A=\frac12\int (x\,dy-y\,dx).$$
Since line segments contribute $0$, we only need the arc integral:
$$A=\frac12\int_0^{3\pi/4}\left(x\frac{dy}{d\theta}-y\frac{dx}{d\theta}\right)d\theta.$$
Now,
$$x=\cos^3\theta,\quad y=\sin^3\theta,$$
$$\frac{dx}{d\theta}=-3\cos^2\theta\sin\theta,
\qquad
\frac{dy}{d\theta}=3\sin^2\theta\cos\theta.$$
Thus,
$$x\frac{dy}{d\theta}-y\frac{dx}{d\theta}
=\cos^3\theta(3\sin^2\theta\cos\theta)-\sin^3\theta(-3\cos^2\theta\sin\theta)$$
$$=3\cos^4\theta\sin^2\theta+3\sin^4\theta\cos^2\theta$$
$$=3\sin^2\theta\cos^2\theta(\sin^2\theta+\cos^2\theta)
=3\sin^2\theta\cos^2\theta.$$
Therefore,
$$A=\frac{3}{2}\int_0^{3\pi/4}\sin^2\theta\cos^2\theta\,d\theta.$$
Use
$$\sin^2\theta\cos^2\theta=\frac14\sin^22\theta=\frac18(1-\cos4\theta).$$
Hence,
$$A=\frac{3}{2}\int_0^{3\pi/4}\frac18(1-\cos4\theta)\,d\theta
=\frac{3}{16}\left[\theta-\frac{\sin4\theta}{4}\right]_0^{3\pi/4}.$$
Since $\sin(3\pi)=0$,
$$A=\frac{3}{16}\cdot \frac{3\pi}{4}= rac{9\pi}{64}.$$
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6. **Compute the asked value**
$$\frac{256A}{\pi}=\frac{256}{\pi}\cdot \frac{9\pi}{64}=4\cdot 9=36.$$
So the required integer is
$$\boxed{36}.$$