How do you find the domain and range of #f(x,y) = sqrt(9-x^2-y^2)#?

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Answer 1 · 2017-04-07T04:06:19

D: #-3<=x<=3#
R: #-3<=y<=3#

For the function:

#f(x,y)=sqrt(9-x^2-y^2)#

What is the list of allowable #x# values - the domain?

We can't have a value under the square root sign that is less than 0, and so we can see that with #-3<=x<=3# we have acceptable values.

What then is the resulting values of #y# - the range?

With #x=-3, y=0#
With #x=0, y=pm3#
With #x=3, y=0#

And so the range is #-3<=y<=3#