Calculate double integral with specific region

Question

I need to calculate

$$\iint\frac{dx\,dy}{\sqrt{x^2 + y^2}}$$ $$x^2 + y^2 < 2y$$

I tried to solve this via polar coordinates

$$0 \leq r < 2\sin(\phi)$$ $$0 \leq \phi \leq 2\pi$$

So, our integral becomes

$$\int_0^{2\pi} d\phi \int_0^{2\sin(\phi)} \, dr $$

which seems to be wrong because of $$r < 2\sin(\phi)$$

You probably only need to integrate from $-\pi/2$ to $\pi/2,$ since there is no portion of your region in the left two quadrants. — Thomas Andrews, May 23 '21 at 21:15
@ThomasAndrews that would be true for $r = 2 \cos\theta$. Here there is no portion of the region in bottom two quadrants. — Math Lover, May 23 '21 at 21:32

score 2 · Accepted Answer · edited May 23 '21 at 23:13

$$ \iint\frac{dx\,dy}{\sqrt{x^2 + y^2}}$$

Region is $x^2+y^2 \leq 2y$.

So if you are using $x = r \cos \theta, y = r \sin\theta$,

$$r \leq 2 \sin\theta, 0 \leq \theta \leq \pi.$$

This is a circle centered at $(0, 1)$ with radius $1$. As you are measuring radius from the origin, $r = 2 \sin\theta$ on the circle and note that the complete circle forms for $0 \leq \theta \leq \pi$.

So integral becomes $ \ \displaystyle \int_0^\pi \int_0^{2 \sin\theta}\,dr \ d\theta $

Calculate double integral with specific region

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