Answer
$B$ is the domain of integration.
Work Step by Step
We have $\mathop \smallint \limits_{x = - \sqrt 2 /2}^0 \mathop \smallint \limits_{y = - x}^{\sqrt {1 - {x^2}} } f\left( {x,y} \right){\rm{d}}y{\rm{d}}x$.
From the iterated integral, we obtain:
1. The limits of outer integral: $ - \frac{{\sqrt 2 }}{2} \le x \le 0$
2. The limits of inner integral: $ - x \le y \le \sqrt {1 - {x^2}} $
So, this is a vertically simple region. The domain of integration is ${\cal D} = \left\{ {\left( {x,y} \right)| - \frac{{\sqrt 2 }}{2} \le x \le 0, - x \le y \le \sqrt {1 - {x^2}} } \right\}$.
The $x$-interval implies that the domain is in the second quadrant. Also, at $x=0$, that is, along the $y$-axis, $\sqrt {1 - {x^2}} = 1$. Therefore, $0 \le y \le 1$. We conclude that $B$ is the domain of integration (please see the figure attached).
