Finding the interval of which a multivariable function is definedFind the interval in which f ( x , y , z ) = z + l n ( 1 − x 2 − y 2 ) is definedSo all that is need to to check for which values l n ( 1 − x 2 − y 2 ) ≥ 0That mean 1 − x 2 − y 2 ≥ 1 → x 2 + y 2 ≥ 0But how do I find the specific x terms and y terms?Or is x 2 + y 2 ≥ 0 is sufficient?

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Finding the interval of which a multivariable function is definedFind the interval in which   f  (  x  ,  y  ,  z  )  =  z  +  l  n  (  1  −      x    2    −      y    2    ) is definedSo all that is need to to check for which values   l  n  (  1  −      x    2    −      y    2    )  ≥  0That mean   1  −      x    2    −      y    2    ≥  1  →      x    2    +      y    2    ≥  0But how do I find the specific   x terms and   y terms?Or is       x    2    +      y    2    ≥  0 is sufficient?

Lilyana Simon · Accepted Answer

You are only partly right: if the only thing that you want is for   f to be defined, then   ln must be defined, which means that its argument must be   &amp;gt;  0. In formulae:   1  −      x    2    −      y    2    &amp;gt;  0, which means       x    2    +      y    2    &amp;lt;  1, which is the interior of the circle of radius   1 and center   (  0  ,  0  ). There is absolutely no reason why the logarithm should be positive, as you say.

Urijah Zhang · Accepted Answer

it is sufficient, you don&#039;t need separate limits on x and y

Finding the interval of which a multivariable function is defined Find the interval in which f(x,y,z)=z+ln(1−x^2−y^2) is defined So all that is need to to check for which values ln(1-x^2-y^2)>=0 That mean 1-x^2-y^2>= 1 x^2+y^2>= 0 But how do I find the specific x terms and y terms?

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