I would appreciate if anyone can verify if I did the boundary right, I am trying to practice only this thing, I don't want to evaluate the integrals, just to learn how to do the boundary right.
$\left(1\right)$ $$\int \:\int _D\:\sqrt{x^2-y^2}dxdy\:$$ $$\text{where }D \text{ is bounded by the sides of the triangle } OAB\: \text{ with }O\left(0,0\right),\:\:A\left(1,-1\right),\:B\left(1,1\right)$$
Here I found out that $y=-x \text{ and }y=\:-1$ from doing OA and AB hence my integral becomes bounded by: $$\int _0^1\:\left(\int _{-x}^{-1}\:\:\sqrt{x^2-y^2}dy\:\right)dx$$
$\left(2\right)$ $$\int \:\int _D\:\:xy \: dxdy$$ $$\text{where } D \text{ is bounded by } y=x^2,\:y=2x+3$$
For this one I did the following: $$\int _{-1}^{0}\:\:\left(\int _{x^2}^{2x+3}xy \: dy\right)\:dx$$ because i took $-1\le x\le 0$
$(3)$ $$\int \:\int _D \arcsin\sqrt{x+y}\:dxdy$$ $$\text{ where } D\text{ is bounded by } x+y=0,\:x+y=1,\:x=0,\:x=1$$ For this one I took: $0\le x\le 1$ and $y=-x\:,\:y=\:1-x$ so my integral will have the following boundary: $$ \int _0^1\:\left(\int _{-x}^{1-x} \arcsin\sqrt{x+y}\:dy\right)dx $$
I hope I didn't do any typos, I am a bit tired and also still not used to the typing format yet.
$$(1)\quad \int\limits_{0}^{1}\int\limits_{-x}^{x}$$ $$(2)\quad \int\limits_{-1}^{3}\int\limits_{x^2}^{2x+3}$$