how to find and sketch domain f(x,y)=sin inverse (x^2+y^2)+ln(y^2-x)...help me pls...

Question

e.mccormick · Answer

The domain is valid input (x values) and the range is valid output (y value) so, what do you think.

amistre64 · Answer

z is a function of x and y,  so valid x,y points are the domain and the resulting z values are your range

amistre64 · Answer

inverse sin inputs are any real number
inverse sin output are from -pi/2 to pi/2

ln inputs are any nonzero,nonnegative real number
ln outputs are from -inf to +inf

amistre64 · Answer

0 < y^2 - x

x < y^2

|dw:1427647750090:dw|

1,0 is false so the solution set of this is the other side of the curve

amistre64 · Answer

since this is the only restriction to x,y points it defines the domain