MHB Mr.Ask's question at Yahoo Answers (Two variables, domain and range)

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The function f(x,y) = 1/√(16 - x² - y²) is defined only when 16 - x² - y² > 0, which corresponds to the domain being the open disk D((0,0),4). Therefore, the domain of f is the set of all points (x,y) such that x² + y² < 16. The range of the function varies from 1/4 to infinity, with 1/4 included and infinity excluded. Thus, the range is [1/4, +∞). This analysis provides a clear understanding of the function's domain and range.
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There exists $f(x,y)=\dfrac{1}{\sqrt{16-x^2-y^2}}$ if and only if $16-x^2-y^2>0$ or equivalently if and only if $x^2+y^2<4^2$ (open disk $D((0,0),4)$). That is, $$\boxed{\:\mbox{Dom } (f)=D((0,0),4)\:}$$ On the other hand, for every $(x,y)$ such that $x^2+y^2=r^2$ with $0\leq r<4$ we have $f(x,y)=\dfrac{1}{\sqrt{16-r^2}}$ so, $f(x,y)$ varies from $1/4$ (included) to $+\infty$ (excluded). That is, $$\boxed{\:\mbox{Range } (f)=\left[1/4,+\infty\right)\:}$$
 
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