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Math Help - multivariable function- domain and range

  1. #1
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    multivariable function- domain and range

    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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  2. #2
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    Quote Originally Posted by dataspot View Post
    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?
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by dataspot View Post
    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 (0,0)

    For the range think the max and min of f(x,y)=\sqrt{4-(x^2+y^2)} bounded by x^2+y^2\leq{4}
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