<s>[INST] Continuity of multivariable function $\frac{xy}{\sqrt{16-x^2-y^2}}$ How can I define where is the following function continuous? The function being: $$f(x,y) = \frac{xy}{\sqrt{16-x^2-y^2}}$$
Is it enough with getting the domain by doing $16-x^2-y^2>0$. How does it change if instead of $xy$ I have $\frac{y}{\sqrt{16-x^2-y^2}}$ or $\frac{x^2+y^2}{\sqrt{16-x^2-y^2}}$?
Thanks for the help.
 [/INST] For $\sqrt{x}$ to be defined, we must have $x\geq 0$. For $\frac{1}{\sqrt{x}}$ to be defined, we must have $\sqrt{x}\neq 0$, i.e. $x\neq 0$. Hence, for the function 
$f(x,y) = \frac{xy}{\sqrt{16-x^2-y^2}}$ to be defined, we must have $16-x^2-y^2>0$, just as you said. 
Now, $xy$ is a continuous function, and $\sqrt{16-x^2-y^2}$ is a nonzero (in fact, positive) continuous function when $16-x^2-y^2>0$. Therefore, the quotient of these two continuous functions, $f(x,y) = \frac{xy}{\sqrt{16-x^2-y^2}}$, is a continuous function when $16-x^2-y^2>0$.
The situation is the same for $\frac{y}{\sqrt{16-x^2-y^2}}$ or $\frac{x^2+y^2}{\sqrt{16-x^2-y^2}}$, since $y$ and $x^2+y^2$ are still continuous functions.  
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