Find the area enclosed by the following curve: (x²+y²)³ =

Find the area enclosed by the following curve: $$ (x^2+y^2)^3 = x^4+y^4. $$

Step 1 · Convert to polar coordinates

With $x=r\cos\theta$, $y=r\sin\theta$: $x^2+y^2=r^2$, and $x^4+y^4=r^4(\cos^4\theta+\sin^4\theta)$, so:

$$ r^6 = r^4(\cos^4\theta+\sin^4\theta) \quad\Longrightarrow\quad r^2=\cos^4\theta+\sin^4\theta \quad(r\neq0). $$

Step 2 · Simplify the right-hand side

$$ \cos^4\theta+\sin^4\theta = (\cos^2\theta+\sin^2\theta)^2-2\cos^2\theta\sin^2\theta = 1-\frac{\sin^22\theta}2 = 1-\frac{1-\cos4\theta}4 = \frac{3+\cos4\theta}4. $$

So $r^2=\dfrac{3+\cos4\theta}4$, which is strictly between $\frac12$ and $1$ for every $\theta$ — the curve is a single smooth closed loop enclosing the origin, traced once as $\theta:0\to2\pi$.

Step 3 · Set up and evaluate the area integral

$$ A = \frac12\int_0^{2\pi} r^2\,d\theta = \frac18\int_0^{2\pi}(3+\cos4\theta)\,d\theta. $$

Since $\displaystyle\int_0^{2\pi}\cos4\theta\,d\theta=\left[\frac{\sin4\theta}4\right]_0^{2\pi}=0$:

$$ A = \frac18\left(3\cdot2\pi+0\right) = \frac{6\pi}8 = \frac{3\pi}4. $$

Additional Notes

• Verified numerically: direct quadrature of $\frac12\int_0^{2\pi}\frac{3+\cos4\theta}4\,d\theta$ gives $\approx2.35619$, matching $\frac{3\pi}4\approx2.35619$.
• The curve is a smoothly rounded, four-fold-symmetric shape (symmetric under $\theta\to\theta+\frac\pi2$ and under reflection across either diagonal), with $r$ ranging between $r_{\min}=\frac1{\sqrt2}$ (at $\cos4\theta=-1$) and $r_{\max}=1$ (at $\cos4\theta=1$).