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

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.

• Hint: think about the inequality in the new coordinates and sketch in the $uv$ plane. What do you see? Commented May 20 at 17:28
• @SeanRoberson ty for the hint. I don't know how to sketch the domain in the new polar plane Commented May 20 at 17:30

Use the polar coordinate transformation:

$$r^4 \leq r^2(cos^2\theta-sin^2\theta)=r^2cos(2\theta)$$ 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 $$\frac{\pi}{8}$$

• You should check your bounds again as they are not correct. What happens, for example, at $\theta =\frac{\pi}{3}$? Commented May 20 at 17:34
• Thanks ! How did you know the domain of theta ? Why didn't you choose 2pi for example Commented May 20 at 17:35
• @Roei Ninad is completely right, I need to check my bounds once again, will edit and fix!
– J.D
Commented May 20 at 17:39
• @Roei I have fixed it now.
– J.D
Commented May 20 at 17:48
• It is probably easier to use symmetries rather than splitting the integral. $\theta\in[0,\frac\pi 4]$ represents the area of half a lobe of the lemniscate, so just multiply $\frac\pi{32}$ by $4$.
– zwim
Commented May 20 at 18:10