I'm trying to do the area of the domain while using this formula $$A=\int \int _D\:dxdy$$ for the following: $$\left(1\right)\:D:\:x+y=1,\:y=2x+1,\:x=2y+1$$ and $$\left(2\right)\:D:\:y^2=4x+4,\:y^2=4-2x$$
For the first one I tried first isolating the y thus resulting the following:
$y=1-x$ , $y=2x+1$, $y=\frac{x-1}{2}$ so I will have the following drawing:
but I don't know how the polar coordinates should look like in this scenario, here I am stuck.
As for the second one I did the drawing and that's about it for now, I will edit when I get more ideas how to proceed.
edit for 1st one by splitting the integral:
$f1\left(x\right)=f2\left(x\right)\:<=>\:1-x=2x+1\:\rightarrow x=0$ $f1\left(x\right)=f3\left(x\right)\:<=>\:1-x=\frac{x-1}{2}\:\rightarrow x=1$ $f2\left(x\right)=f3\left(x\right)\:<=>\:2x+1=\frac{x-1}{2}\:\rightarrow x=-1$ $\rightarrow \:A=\int _{-1}^0\:f2\left(x\right)-f3\left(x\right)dx\:+\:\int _0^1\:f1\left(x\right)-f3\left(x\right)dx$