This is the first time I write in this forum so hope I get everything right! I'm trying to resolve this double integral but I'm having trouble with it :/. First thing I thought about was a polar coordinate substitution but then I don't know where do $\rho$ and $\theta$ vary. Could anybody give me some help? Thank you all! $$\iint_\omega 2x+x^2+y^2$$ where $\omega = \{(x,y)| x>0,y>0,(x^2+y^2)^\frac 32 \le 2xy\}$
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$\begingroup$$x > 0$ and $y>0$, so we are in the first quadrant: $\theta$ varies between $0$ and $\pi/2$.
Since the region is given by $x,y > 0$, $(x^2 + y^2)^{3/2} \le 2xy$, substituting $x = \rho \cos \theta$ and $y = \rho \sin \theta$, we get $0 < \rho \le \sin 2 \theta$.
Therefore,
$$\int_0^{\pi/2}\int_0^{\sin 2 \theta} (2\rho \cos \theta + \rho^2)\rho d\rho d\theta$$
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