How to sketch $y = \left\lfloor \sqrt{2-x^2} \right\rfloor$? How to sketch $y = \left\lfloor \sqrt{2-x^2} \right\rfloor$, where $\lfloor \cdot \rfloor$ denotes the greatest integer function?
Please help. I have no idea about this.

First, notice that the domain of $x$ is 
$$2-x^2\ge0\iff-\sqrt2\le x\le \sqrt 2.$$
And the floor function $\lfloor x\rfloor$ is defined as
$$\lfloor x\rfloor=t\iff t\le x\lt t+1.$$
So, in your question, we have
$$\lfloor \sqrt{2-x^2}\rfloor =t\iff t\le\sqrt{2-x^2}\lt t+1$$
Now note that $t\ge 0 \in\mathbb Z$ (This is because $\sqrt{2-x^2}\ge0$). So we have
$$t^2\le 2-x^2\lt (t+1)^2\iff 2-(t+1)^2\lt x^2\le 2-t^2.$$
The fact that $\lfloor \sqrt{2-0^2}\rfloor=1, \lfloor \sqrt{2-({\sqrt 2})^2}\rfloor=0$ tells us that $t=0,1$.
Hence, we know 
$$2-(0+1)^2\lt x^2\le 2-0^2\Rightarrow t=0,$$
$$2-(1+1)^2\lt x^2\le 2-1^2\Rightarrow t=1.$$
