Find the area enclosed by the following curve: (x²+y²)³ =
Find the area enclosed by the following curve: $$ (x^2+y^2)^3 = 4x^2y^2. $$
Explanation of Method
Converting to polar coordinates and using the double-angle identity for $\sin2\theta$ reduces the curve to the classical four-petaled rose $r=|\sin2\theta|$. By symmetry, all four petals have equal area, so it suffices to integrate over a single petal (where $\sin2\theta\geqslant0$) and multiply by $4$.
- Convert to polar coordinates, obtaining $r=|\sin2\theta|$.
- Identify the four petals and their symmetry.
- Integrate one petal and multiply by $4$.
Step 1 · Convert to polar coordinates
With $x=r\cos\theta$, $y=r\sin\theta$: $x^2+y^2=r^2$ and $x^2y^2=r^4\cos^2\theta\sin^2\theta$, so:
$$ r^6 = 4r^4\cos^2\theta\sin^2\theta \quad\Longrightarrow\quad r^2 = 4\cos^2\theta\sin^2\theta=\sin^22\theta \quad\Longrightarrow\quad r=|\sin2\theta|. $$
Step 2 · Identify the petals
The curve $r=|\sin2\theta|$ forms four petals, one in each quadrant, corresponding to the four intervals where $\sin2\theta$ keeps a constant sign: $\left[0,\frac\pi2\right]$, $\left[\frac\pi2,\pi\right]$, $\left[\pi,\frac{3\pi}2\right]$, $\left[\frac{3\pi}2,2\pi\right]$. By symmetry all four petals are congruent.
Step 3 · Set up and evaluate the area integral
For the first petal ($0\leqslant\theta\leqslant\frac\pi2$, where $r=\sin2\theta\geqslant0$):
$$ A_{\text{petal}} = \frac12\int_0^{\pi/2}\sin^22\theta\,d\theta. $$
Total area:
$$ A = 4A_{\text{petal}} = 2\int_0^{\pi/2}\sin^22\theta\,d\theta. $$
Using $\sin^22\theta=\dfrac{1-\cos4\theta}2$:
$$ A = \int_0^{\pi/2}(1-\cos4\theta)\,d\theta = \left[\theta-\frac{\sin4\theta}4\right]_0^{\pi/2} = \frac\pi2-0 = \frac\pi2. $$
Additional Notes
- Each petal has area $\dfrac\pi8$; there are four congruent petals, giving a total of $\dfrac\pi2$.
- Verified numerically: direct quadrature of $2\int_0^{\pi/2}\sin^22\theta\,d\theta$ gives $\approx1.57080$, matching $\frac\pi2\approx1.57080$.
Result
The area is: