# describe the domain of f(x,y,z)=ln(z-x^2-y^2+2y+3)

• Jun 15th 2012, 04:24 PM
sjjc1993
describe the domain of f(x,y,z)=ln(z-x^2-y^2+2y+3)
hi,
the question asks for the domain of f(x,y,z)=ln(z-x^2-y^2+2y+3).

which is z-x^2-y^2+2y+3 > 0.

when I try do describe the domain, all I get is 2 > x^2 + (y-1)^2 -z. I have no idea what the shape is like.

any ideas?

thx.
• Jun 15th 2012, 04:58 PM
Reckoner
Re: describe the domain of f(x,y,z)=ln(z-x^2-y^2+2y+3)
Quote:

Originally Posted by sjjc1993
hi,
the question asks for the domain of f(x,y,z)=ln(z-x^2-y^2+2y+3).

which is z-x^2-y^2+2y+3 > 0.

when I try do describe the domain, all I get is 2 > x^2 + (y-1)^2 -z. I have no idea what the shape is like.

Consider the surface

\$\displaystyle z-x^2-y^2+2y+3=0.\$

By completing the square, we have

\$\displaystyle z+4=x^2+(y-1)^2\$

This is a circular paraboloid, centered at (0,1,-4). So \$\displaystyle z+4>x^2+(y-1)^2\$ would consist of all the points above the paraboloid.
• Jun 15th 2012, 06:43 PM
sjjc1993
Re: describe the domain of f(x,y,z)=ln(z-x^2-y^2+2y+3)
thanks a lot!