Integral of $x^2+y^2$ on the domain $(x^2 + y^2)^2 \le x^2-y^2$

Question

I need to calculate the double integral $\iint_D (x^2 + y^2)$

$D = \{ (x, y) \in \mathbb{R}^2 \mid (x^2 + y^2)^2 \leq x^2 - y^2 \}$

The answer is $\frac{\pi}{8}$

I've tried using the substitution rule for double integral with polar transformation, and with $u = (x^2 + y^2)$ , $v = (x^2 - y^2)$ but I couldn't figure out the domain after the substitution.

Answer 1

Use the polar coordinate transformation:

The given inequality reads:

$$r^4 \leq r^2(cos^2\theta-sin^2\theta)=r^2cos(2\theta)$$ So:

$$r^2\leq cos(2\theta)$$ so: $0\leq r\leq \sqrt {cos(2\theta)}$

As we have $0\leq r^2\leq cos(2\theta)$, we must restrict $\theta$, so $0\leq \theta\leq \frac{\pi}{4}, \frac{3\pi}{4}\leq \theta \leq \frac{5\pi}{4}, \frac{7\pi}{4}\leq \theta\leq 2\pi$.

Then, one of the integrals becomes:

$$\int^\frac{\pi}{4}_0 \int^{\sqrt{cos(2\theta)}}_0r^3drd\theta=\int^\frac{\pi}{4}_0 [\frac{r^4}{4}]^{\sqrt{cos(2\theta)}}_0d\theta=\int^\frac{\pi}{4}_0 \frac{cos(2\theta)^2}{4}d\theta= \int ^\frac{\pi}{4}_0\frac{cos(4\theta)+1}{8} d\theta= [\frac{sin(4\theta)}{32}+\frac{1}{8}\theta]^\frac{\pi}{4}_0$$= $\frac{\pi}{32}$. You repeat this process with the other integrals :

$$\int^{\frac{5\pi}{4}}_{\frac{3\pi}{4}}\int^{\sqrt{cos(2\theta)}}_0r^3drd\theta$$ $$\int^{2\pi}_\frac{7\pi}{4}\int^{\sqrt{cos(2\theta)}}_0r^3drd\theta$$

Then summing those three integrals, you get $\frac{\pi}{8}$