andrew.tzeva wrote:
I tried drawing D but I got confused. I represent $x=\sqrt{y}$ as $y=x^2$ and plot the lines $y=2$, $x=1$. It seems to me that D is divided in two sections: $0\leq x\leq1$ , $0\leq y\leq x^2$ and $1\leq x\leq\sqrt{2}$ , $x^2\leq y\leq 2$ , because the curve intersects the vertical line. Is this possible?
Attachment:
Andrew, I apologize. I wasn't careful and I consider $D$ to be the region $D_1$ in the scheme.
\[D=\big\{(x,y)\in\mathbb{R}\;|\; \sqrt{y}\leqslant {x}\leqslant 1, \; 0\leqslant {y}\leqslant 2\big\}\] is not a representation of $D_1\cup D_2$. $x$ must be $\leqslant 1$ and for $D_1\cup D_2$ this not the case.
So, must be a typo in the upper and lower limits of the integrals!