$\displaystyle f(x, y)=\sqrt{(4-x^2-y2)}$ domain 4x-x^2-y^2 has to be > or equal to 0, so x^2+y^2 < or equal to 4. range: I have no idea how to find the range here.
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Originally Posted by dataspot x^2+y^2 < or equal to 4 Domain: Well, yes, but what is that? Do you see the border of this region as a circle? The region contained within the circle is a ...? Range: What happens for x = y = 0? Is that significant? What do you know about the "Principle" square root? Positive? Negative? Zero?
Originally Posted by dataspot $\displaystyle f(x, y)=\sqrt{(4-x^2-y2)}$ domain 4x-x^2-y^2 has to be > or equal to 0, so x^2+y^2 < or equal to 4. range: I have no idea how to find the range here. Yes you are right, the range is the interior and edge of the circle of radius of two centered at $\displaystyle (0,0)$ For the range think the max and min of $\displaystyle f(x,y)=\sqrt{4-(x^2+y^2)}$ bounded by $\displaystyle x^2+y^2\leq{4}$
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