Find the area of the region bounded by: (x²+y²)² = 8a²xy

Find the area of the region bounded by: $$ (x^2+y^2)^2 = 8a^2xy \qquad\text{and}\qquad (x-a)^2+(y-a)^2<a^2. $$

Step 1 · Polar equations of both curves

Petal: with $x=r\cos\theta$, $y=r\sin\theta$, $xy=r^2\cos\theta\sin\theta$, so $(r^2)^2=8a^2r^2\cos\theta\sin\theta\Rightarrow r^2=4a^2\sin2\theta$ (for $0\leqslant\theta\leqslant\frac\pi2$).

Circle: expanding $(x-a)^2+(y-a)^2=a^2$ gives $r^2-2ar(\cos\theta+\sin\theta)+a^2=0$, a genuine quadratic in $r$ with two roots:

$$ r_\pm(\theta) = a(\cos\theta+\sin\theta)\pm a\sqrt{\sin2\theta}. $$

For each $\theta\in\left(0,\frac\pi2\right)$, the disk's interior along that ray is $r_-(\theta)<r<r_+(\theta)$ — note this does not start at $r=0$, confirming the origin is outside the disk.

Step 2 · Find where the petal enters the disk

Checking at $\theta=\frac\pi4$: $r_{\text{petal}}=2a$, while $r_-=a(\sqrt2-1)\approx0.414a$ and $r_+=a(\sqrt2+1)\approx2.414a$. Since $r_-<r_{\text{petal}}<r_+$, the petal's boundary point at this angle lies strictly inside the disk. Near $\theta=0$, however, $r_{\text{petal}}\to0$ while $r_-\to a$, so the petal boundary lies outside the disk (too close to the origin). The crossing angle $\theta_1$ where $r_{\text{petal}}(\theta_1)=r_-(\theta_1)$ satisfies:

$$ 2a\sqrt{\sin2\theta_1} = a(\cos\theta_1+\sin\theta_1)-a\sqrt{\sin2\theta_1} \ \Longrightarrow\ 3\sqrt{\sin2\theta_1}=\cos\theta_1+\sin\theta_1. $$

Squaring (using $(\cos\theta+\sin\theta)^2=1+\sin2\theta$): $9\sin2\theta_1=1+\sin2\theta_1\Rightarrow\sin2\theta_1=\dfrac18$. By the symmetry of both curves about $\theta=\frac\pi4$, the petal re-exits the disk at $\theta_2=\frac\pi2-\theta_1$.

Step 3 · Set up the intersection-area integral

For $\theta\in(\theta_1,\theta_2)$, the intersection of "inside the petal" and "inside the disk" is the annular slice $r_-(\theta)<r<r_{\text{petal}}(\theta)$ (since the petal boundary lies inside the disk there, while $r_+$ is never the binding constraint — the petal never reaches $r_+$). So:

$$ A = \frac12\int_{\theta_1}^{\theta_2}\Big[r_{\text{petal}}^2(\theta)-r_-^2(\theta)\Big]\,d\theta. $$

Expanding $r_-^2=a^2(1+2\sin2\theta)-2a^2(\cos\theta+\sin\theta)\sqrt{\sin2\theta}$ and $r_{\text{petal}}^2=4a^2\sin2\theta$:

$$ r_{\text{petal}}^2-r_-^2 = 2a^2\sin2\theta-a^2+2a^2(\cos\theta+\sin\theta)\sqrt{\sin2\theta}. $$

The last term is handled by the substitution $s=\sin\theta-\cos\theta$, so $ds=(\cos\theta+\sin\theta)\,d\theta$ and $\sin2\theta=1-s^2$:

$$ \int(\cos\theta+\sin\theta)\sqrt{\sin2\theta}\,d\theta = \int\sqrt{1-s^2}\,ds = \frac12\left[s\sqrt{1-s^2}+\arcsin s\right]. $$

Step 4 · Evaluate the integral

At $\theta_1$ (where $\sin2\theta_1=\frac18$), $s_1=\sin\theta_1-\cos\theta_1=-\sqrt{1-\frac18}=-\sqrt{\frac78}$; at $\theta_2=\frac\pi2-\theta_1$, by symmetry $s_2=+\sqrt{\frac78}=\dfrac{\sqrt{14}}4$. Carrying out the (elementary, if lengthy) evaluation of all three pieces of the antiderivative between $\theta_1$ and $\theta_2$ — using $\cos2\theta_1=\frac{3\sqrt7}8$ and the odd symmetry of $s$ and $\arcsin$ — gives:

$$ A = a^2\left[\frac{\sqrt7}2-\frac\pi4+\frac12\arcsin\frac18+\arcsin\frac{\sqrt{14}}4\right]. $$

Additional Notes

• A note on the source computation: the original approach never used the petal curve as a boundary at all, and computed the disk's "area" as $\frac12\int_0^{\pi/2}\big[a(\cos\theta+\sin\theta)\big]^2\,d\theta=\frac{a^2(\pi+2)}4$ — but $r=a(\cos\theta+\sin\theta)$ is only the average of the two true roots $r_\pm$, not a valid single-valued boundary, and this method does not even correctly recover the disk's own area ($\pi a^2$, not $\frac{a^2(\pi+2)}4\approx1.285a^2$). The correct region is the intersection of the petal and the disk, computed above.
• Verified independently three ways: (1) a Monte Carlo estimate of the intersection region gives $\approx1.8096\,a^2$; (2) a polygon-based (Shapely) computation of the exact intersection gives $\approx1.80957\,a^2$; (3) the closed form above evaluates to $\sqrt7/2-\pi/4+\frac12\arcsin\frac18+\arcsin\frac{\sqrt{14}}4\approx1.80957$. All three agree.