Given the domain $$ D := \left\{ (x,y) \in {\Bbb R}^2 : x^2 + y^2 \leq 4 , y\geq 0 ,y^2\geq 4(1+x) \right\} $$ calculate the following double integral $$\iint_{D}\frac{{\rm d}x \, {\rm d} y}{y-2}$$
I do not understand. I would appreciate some help on this problem that has been leaving me out of options. What I've tried so far was to use the polar coordinates because we have a condition that could imply you use them, so I have that the radius must be $r \in (0,2)$ but I cannot find the condition for the angle $\theta$ that i need to use. I appreciate any kind of help.
The region $D=\{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \leqq 4, y \geqq 0, y^2 \geqq 4(1+x)\}$ is the intersection of inside of a circle $x^2+y^2 \leqq 2^2$ with origin $(0,0)$ and radius $2$ and left (relative to Ox) to the parabola is $y^2 \geqq 4(1+x).$
So when $0 \leqq y \leqq 2$, the other variable (when inside $D$) is between two functions on $y.$ The upper function is the piece of the circle as a function of $y$ that is in the $Oy^{-}$ and the lower function is the piece of the parabola as a function on y. That is $-\sqrt{4-y^2} \leqq x \leqq \frac{y^2-4}{4} .$ This is because from $x^2+y^2=4$ we get two functions $x=\pm \sqrt{4-y^2}$ of $y$ and the one with positive sign coresponds to the half circle sliced along $Oy$ on the right of zero, and the one with negative sign coresponds to the once on the left of zero. So we are interested in the one with negative sign, so the piece of function we are taking is $x=x(y)=-\sqrt{4-y^2}.$ For each point in $D,$ it's $x$ coordinate is greater of equal than the on the curve $-\sqrt{4-y^2}$ so $x\geqq -\sqrt{4-y^2}.$ Analogously for the other inequality. So the only problem is that the function $\frac{1}{y-2}$ has a simple pole at $y=2.$ In theory you should contract the region just a little bit off $D_{\varepsilon}$ where this $D_{\varepsilon}$ is as $D$ but the circle and parabola are throught $2-\varepsilon.$ Then use Fubini's theorem in this new domain and take limit. After formally using Fubini on the whole domain there are 2 integrals the first of which cancels the pole and only the second one stands. \begin{align*} \iint_D \frac{1}{y-2}dxdy=\int_0^2 \left(\int_{-\sqrt{4-y^2}}^{\frac{y^2-4}{4}}\frac{1}{y-2}dx \right)dy&=\int_0^2 \left\{\frac{1}{4} \frac{(y-2)(y+2)}{y-2}+\frac{\sqrt{4-y^2}}{y-2}\right\}dy \\ &=\frac{1}{4}\int_0^2 (y+2)dy + \lim_{\varepsilon \to 0^{+}}\int_0^{2-\varepsilon} \frac{\sqrt{4-y^2}}{y-2}dy. \end{align*}