Find the area of the region bounded by: (x²+y²)² =
Find the area of the region bounded by: $$ (x^2+y^2)^2 = 2a^2(x^2-y^2), \qquad x^2+y^2>a^2. $$
Explanation of Method
The first equation is a lemniscate, $r^2=2a^2\cos2\theta$, with two loops along the $x$-axis. The condition $x^2+y^2>a^2$ (i.e. $r>a$) selects only the outer part of each loop — the part farther from the origin than the circle $r=a$. The region of interest is therefore an annular sliver: bounded on the outside by the lemniscate and on the inside by the circle, for the angular range where the lemniscate lies beyond the circle. Crucially, computing this area requires subtracting the inner circular part, not simply integrating $\tfrac12r_{\text{lemniscate}}^2$ over that angular range.
- Convert the lemniscate to polar form and find the angular range where it exists.
- Find the angular sub-range where the lemniscate lies outside the circle $r=a$.
- Integrate $\frac12\big[r_{\text{lemniscate}}^2-a^2\big]$ (not $r_{\text{lemniscate}}^2$ alone) over that sub-range, then double for both loops.
Step 1 · Convert the lemniscate to polar form
With $x=r\cos\theta$, $y=r\sin\theta$: $x^2-y^2=r^2\cos2\theta$, so:
$$ r^4 = 2a^2r^2\cos2\theta \quad\Longrightarrow\quad r^2=2a^2\cos2\theta \quad(r\neq0). $$
This requires $\cos2\theta\geqslant0$, i.e. $\theta\in\left[-\frac\pi4,\frac\pi4\right]\cup\left[\frac{3\pi}4,\frac{5\pi}4\right]$ (a right-hand loop and a left-hand loop).
Step 2 · Find where the lemniscate lies outside the circle $r=a$
Setting $r^2=2a^2\cos2\theta>a^2$ gives $\cos2\theta>\frac12$, i.e. $-\frac\pi6<\theta<\frac\pi6$ for the right-hand loop (and symmetrically $\frac{5\pi}6<\theta<\frac{7\pi}6$ for the left-hand loop). Outside this narrower angular window, the entire lemniscate loop at that angle lies within the disk $r\leqslant a$, contributing nothing to the region $r>a$.
Step 3 · Set up the area integral correctly
For $\theta\in\left(-\frac\pi6,\frac\pi6\right)$, the region satisfying both "inside the lemniscate loop" and "$r>a$" consists of the radii from $a$ (the circle) out to $2a^2\cos2\theta$ (the lemniscate) — not from $0$ to the lemniscate. Doubling for the symmetric left-hand loop:
$$ A = 2\cdot\frac12\int_{-\pi/6}^{\pi/6}\Big[r_{\text{lemniscate}}^2(\theta)-a^2\Big]\,d\theta = \int_{-\pi/6}^{\pi/6}\big[2a^2\cos2\theta-a^2\big]\,d\theta. $$
Step 4 · Evaluate the integral
$$ A = \Big[a^2\sin2\theta-a^2\theta\Big]_{-\pi/6}^{\pi/6} = 2\left[a^2\sin\frac\pi3-a^2\cdot\frac\pi6\right] = 2a^2\left[\frac{\sqrt3}2-\frac\pi6\right] = a^2\sqrt3-\frac{a^2\pi}3. $$
Additional Notes
- A note on the source computation: a naive approach integrates $\frac12\int_{-\pi/6}^{\pi/6}r_{\text{lemniscate}}^2\,d\theta$ (doubled) without subtracting $a^2$, which gives $a^2\sqrt3$ — this is actually the area of the full angular sector from the origin out to the lemniscate over $|\theta|<\frac\pi6$, not the area of the region satisfying $r>a$. Since part of that sector (specifically, $0\leqslant r\leqslant a$) lies inside the circle and must be excluded, the correct area is smaller: $a^2\sqrt3-\frac{a^2\pi}3\approx0.6849\,a^2$, not $a^2\sqrt3\approx1.7321\,a^2$.
- Verified numerically: direct quadrature of $\int_{-\pi/6}^{\pi/6}(2a^2\cos2\theta-a^2)\,d\theta$ gives $\approx0.68485\,a^2$, matching $a^2\left(\sqrt3-\frac\pi3\right)\approx0.68485\,a^2$ to five decimal places.
Result
The area is: