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I have to resolve the integral $$\iint_{\Omega} x \sqrt{x^2+y^2} dxdy$$ with $\Omega=\{(x,y)\in \mathbb{R}^2| x^2+y^2<1, x^2+y^2<2y, x<0 \}$

The geometric interpretation of this set is not to difficult, the first information is a ball of radius $1$ and center $(0,0)$, the second information is a ball of radius 1 and center $(0,1)$. So the set is the intersection of this two balls in the second Cartesian plane dial.

I think that the polar coordinates are inappropriate but I may be wrong: with them we have $$\rho<1 \qquad \rho<2\sin\theta, \qquad \frac{\pi}{2}<\theta<\pi$$ And after that?

Context: This is an exercise of a past exam of calculus II, so theoretically I should have all the possibility to do that.

Mario
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  • If you tried to do the radial integral first you would need three integrals. Instead, do the angular integral first and you will only have one integral. – Ninad Munshi Dec 03 '23 at 16:08
  • @NinadMunshi But In which way can I write the extremes of integration? – Mario Dec 03 '23 at 16:22

1 Answers1

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There are two methods for solving this integral in the simplest way possible.

$\textbf{Method 1:}$ Polar Coordinates

Consider the equation of the translated circle

$$\rho = 2 \sin\theta \implies \theta = \sin^{-1}\frac{\rho}{2}, \pi- \sin^{-1}\frac{\rho}{2}$$

Since we're in the second quadrant only we know to only take the second bound. From there the integral set up and execution is easy

$$I = \int_0^1\int_{\frac{\pi}{2}}^{\pi-\sin^{-1}\frac{\rho}{2}}\rho^3\cos\theta\:d\theta d\rho = \int_0^1\left[\rho^3\sin\theta\right]_{\frac{\pi}{2}}^{\pi-\sin^{-1}\frac{\rho}{2}}d\rho $$

$$= \int_0^1\frac{\rho^4}{2} -\rho^3\:d\rho = \boxed{-\frac{3}{20}}$$

$\textbf{Method 2:}$ Cartesian coordinates

Simply integrate the function over the region in Cartesian coordinates. Even though the $x$ integral will be two integrals, it will be easier to do than thr $y$ integral

$$I = \int_0^\frac{1}{2}\int_{-\sqrt{2y-y^2}}^0 x\sqrt{x^2+y^2}dxdy + \int_\frac{1}{2}^1\int_{-\sqrt{1-y^2}}^0x\sqrt{x^2+y^2}dxdy$$

$$=\int_0^\frac{1}{2}\frac{y^3-(2y)^\frac{3}{2}}{3}dy + \int_\frac{1}{2}^1 \frac{y^3-1}{3}dy = \boxed{-\frac{3}{20}}$$

Ninad Munshi
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