How do I graph this inequality?

Question

$$y\le x\le \sqrt{1-y^2}$$

Graph: http://www.wolframalpha.com/input/?i=y%3C%3Dx%3C%3Dsqrt%281-y%5E2%29

How would I approach this problem?

I graphed $0\le x\le \sqrt{1-y^2}$ easily, but this one is tougher.

Answer 1

The equation $x=\sqrt{1-y^2}$ is the equation of the righthand half of the unit circle; the region in which $x\le\sqrt{1-y^2}$ is therefore the region extending horizontall to the left from this semicircle. The inequality $y\le x$ specifies the points lying to the right of the line $x=y$. Put the two together, and you have the blue region shown below:

The points satisfying $x\le\sqrt{1-y^2}$ are those in the $\supset$-shaped region; those satisfying $y\le x$ are those to the right of the diagonal line; and the blue region is the intersection of those two sets.

Answer 2

Start by graphing $y \le x$. This is fairly easy -- it's just the region under and including the line $y = x$.

Next, graph $x \le \sqrt{1-y^2}$. This is also pretty easy, and you have already done it.

The overlapping region is the result.