What is the domain and range of #f(x,y) = sqrt(9-x^2-y^2)#?

1 Answer
Konstantinos Michailidis
Feb 28, 2016

Because #f(x,y)=sqrt(9-x^2-y^2)# we must have that

#9-x^2-y^2>=0=>9>=x^2+y^2=>3^2>=x^2+y^2#

The domain of #f(x,y)# is the border and the interior of the circle
#x^2+y^2=3^2#
or

The domain is represented by the disc whose center is the origin of the coordinates system and the radius is 3.

Now hence #f(x,y)>=0# and #f(x,y)<=3# we find that the range of the function is the interval #[0,3]#

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