I was just reading the solutions given by my teacher, and I got confused on the following domain (D) restriction:
$$z=-\sqrt{1-2x^2-4y^2},D=\{(x,y)|-\frac{1}{\sqrt{2}}\leq x \leq \frac{1}{\sqrt{2}}, -\frac{1}{2} \leq y \leq \frac{1}{2}\}$$
I'm just wondering, based on the domain D, if we let $$x=\frac{1}{\sqrt{2}}$$ and $$y=\frac{1}{2}$$ then, $$1-2x^2-4y^2 = 1-1-1=-1$$ which is a negative number thats not allowed inside the squre root, but both $x$ and $y$ are within the specified domain. So is the domain D given incorrect? If so, how should we state the domain in this case? would $D=\{(x,y)|2x^2+4y^2\leq 1\}$ be a good way to represent the domain in this case?
Thanks very much.
Your domain keeps $z$ real. The teacher's solution domain, as you showed, does not. In fact your domain represents the boundary and interior of an ellipse, whereas the teacher's represents the boundary and interior of a rectangle. There are many other points where $z$ is non-real i.e. those exterior to the ellipse. Both the ellipse and rectangle are shown in the figure.