X xl5899 Mar 2014 909 2 malaysia Aug 15, 2016 #1 the domain of this fucntion is given by (x^2) +(y^2) ≤ 16 , but i think it's wrong .... It should be (x^2) +(y^2)< 16 , right ? Since if sqrt (16- (x^2) -(y^2) ) = 0 , the whole equation will become 1/ 0 , which is undefined

the domain of this fucntion is given by (x^2) +(y^2) ≤ 16 , but i think it's wrong .... It should be (x^2) +(y^2)< 16 , right ? Since if sqrt (16- (x^2) -(y^2) ) = 0 , the whole equation will become 1/ 0 , which is undefined

romsek MHF Helper Nov 2013 6,836 3,079 California Aug 15, 2016 #2 $16 - x^2 - y^2 \geq 0$ in order for this to be a valid argument to the square root function. however we cannot have $0$ in the denominator either so the valid domain is $x^2 + y^2 < 16$ so you are correct. Reactions: 1 person

$16 - x^2 - y^2 \geq 0$ in order for this to be a valid argument to the square root function. however we cannot have $0$ in the denominator either so the valid domain is $x^2 + y^2 < 16$ so you are correct.