# What is the domain of $\sqrt\frac{x+y+z-1}{x^2+y^2+z^2-1}$?

${x^2+y^2+z^2-1}$ this part remembers me to the equation of a sphere, and I can rewrite it as $x^2+y^2+z^2=1$ so the radius of the sphere will be 1. And of coure because of the square root $$\frac{x+y+z-1}{x^2+y^2+z^2-1} \ge0$$

How can I continue it? Do I have to examine where is the numerator and denominator positive/negative?

Here and as follows, I tried to figure one part of the region, i.e. $x+y+z\geq 1,~~x^2+y^2+z^2>1$. Sice we have serious difficulty to show the later region, I could to regard the box as you see. I am sure you can extend the right region by your self. In
We can see the part of space in which $x+y+z\geq1$. Of coure you may and should extend the red region properly to have it. In
We have the right region in which $x^2+y^2+z^2>1$ and you know why it must not be equal to $1$. Now intersect these two region simultaniously to get one of the region which Ross pointed. I made them by Maple18.
Yes, the fraction will be positive where the numerator and denominator are both positive or both negative. As you say, $x+y+z-1=0$ is the equation of a plane. The $x+y+z-1$ will be positive on one side of the plane, negative on the other. The denominator represents the unit sphere. It is positive outside the sphere. These divide space into four regions. You need to figure out which two constitute the domain.
• For the plane, you can just let $x,y,z$ be large and positive, then large and negative. For the sphere, the origin and some place far away are good points. I did the origin above, so the denominator is negative within the sphere and positive outside. – Ross Millikan Apr 22 '17 at 14:59