4.1 Functions of several variables

1. In [link] , we determined that the domain of $g\left(x,y\right)=\sqrt{9-x^2-y^2}$ is $\left\{\left(x,y\right)\in {ℝ}^2|x^2+y^2\le 9\right\}$ and the range is $\left\{z\in {ℝ}^2|0\le z\le 3\right\}.$ When $x^2+y^2=9$ we have $g\left(x,y\right)=0.$ Therefore any point on the circle of radius $3$ centered at the origin in the $x,y\text{-plane}$ maps to $z=0$ in ${ℝ}^3.$ If $x^2+y^2=8,$ then $g\left(x,y\right)=1,$ so any point on the circle of radius $2\sqrt{2}$ centered at the origin in the $x,y\text{-plane}$ maps to $z=1$ in ${ℝ}^3.$ As $x^2+y^2$ gets closer to zero, the value of z approaches 3. When $x^2+y^2=0,$ then $g\left(x,y\right)=3.$ This is the origin in the $x,y\text{-plane}.$ If $x^2+y^2$ is equal to any other value between $0\text{and}9,$ then $g\left(x,y\right)$ equals some other constant between $0\text{and}3.$ The surface described by this function is a hemisphere centered at the origin with radius $3$ as shown in the following graph.
   

   

2. This function also contains the expression $x^2+y^2.$ Setting this expression equal to various values starting at zero, we obtain circles of increasing radius. The minimum value of $f(x,y)=x^2+y^2$ is zero (attained when $x=y=0.).$ When $x=0,$ the function becomes $z=y^2,$ and when $y=0,$ then the function becomes $z=x^2.$ These are cross-sections of the graph, and are parabolas. Recall from Introduction to Vectors in Space that the name of the graph of $f(x,y)=x^2+y^2$ is a paraboloid . The graph of $f$ appears in the following graph.
   

   

This function is a polynomial function in two variables. The domain of $f$ consists of $\left(x,y\right)$ coordinate pairs that yield a nonnegative profit:

This is a disk of radius $4$ centered at $\left(3,2\right).$ A further restriction is that both $x\text{and}y$ must be nonnegative. When $x=3$ and $y=2,$ $f\left(x,y\right)=16.$ Note that it is possible for either value to be a noninteger; for example, it is possible to sell $2.5$ thousand nuts in a month. The domain, therefore, contains thousands of points, so we can consider all points within the disk. For any $z<16,$ we can solve the equation $f\left(x,y\right)=16\text{:}$

Since $z<16,$ we know that $16-z>0,$ so the previous equation describes a circle with radius $\sqrt{16-z}$ centered at the point $\left(3,2\right).$ Therefore. the range of $f\left(x,y\right)$ is $\left\{z\in ℝ|z\le 16\right\}.$ The graph of $f\left(x,y\right)$ is also a paraboloid, and this paraboloid points downward as shown.