The domain of is the surface that is the circle centered at the origin and which radius is ... that's not ! The level curve should be a circle lying in the plane centered at and which radius is .The function is the following:
The solution I've attempted and which I'm not sure it's correct is:
I've drawn a level curve of level 4,because it's within the domain of f(x,y)(which is ) and it's the point where the function changes to the other branch.
Does this make sense?
What is "the condition" ? The only condition I can see is that because of the square root. As this inequality is true for such that , working with the "if" clause is enough.Just another question,to determine the domain of the second "piece" of the function,why do we also use the sqrt(32-x^2-y^2) condition and not only just the if clause?