domain of function

Mar 2014
909
2
malaysia
ii.PNGthe 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
$16 - x^2 - y^2 \geq 0$ 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.
 
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