Answer
$(a) f(x, y) =1-x^{2}-y^{2}$
$(b) f(x, y)=\sqrt{x^{2}+y^{2}}$
$(c) f(x, y)=x^{2}+y^{2}$
Work Step by Step
We see $z=0$, so the graph of $\sqrt{x^{2}+y^{2}}=0 \Rightarrow x^{2}+y^{2}=0$ is a point.
Also, when $z=1$, the graph of $\sqrt{x^{2}+y^{2}}=1 \Rightarrow x^{2}+y^{2}=1$ is a unit circle.
When $z=2$, the graph of $\sqrt{x^{2}+y^{2}}=2 \Rightarrow x^{2}+y^{2}=4$ is a circle of radius 2
This means the circle becomes bigger when we increase z values.
Therefore, the contour plot of $f(x, y)=\sqrt{x^{2}+y^{2}}$ is $\bar{b}$
We can write the function $f(x, y)=x^{2}+y^{2}$ as $z=x^{2}+y^{2}$
Then, $z=0$, so the graph of $x^{2}+y^{2}=0$ is a point.
Also, when $z=1$, the graph of $x^{2}+y^{2}=1$ is unit circle.
When $z=2$, the graph of $x^{2}+y^{2}=2$ is a circle of radius $\sqrt{2}$
This means the circle becomes bigger when we increase z values.
Therefore, the contour plot of $f(x, y)=x^{2}+y^{2}$ is square.
We can write the function $f(x, y)=1-x^{2}-y^{2}$ as $z=1-x^{2}-y^{2}$
The graph of $1-x^{2}-y^{2}=0 \Rightarrow x^{2}+y^{2}=1$ is a unit circle for z=0.
Also, when $z=1$, the graph of $1-x^{2}-y^{2}=1 \Rightarrow x^{2}+y^{2}=0$ is a point.
When $z=-1$, the graph of $1-x^{2}-y^{2}=-1 \Rightarrow x^{2}+y^{2}=2$ is a circle of radius $\sqrt{2}$
This means the circle become bigger when z decreases.
Therefore, the contour plot of $f(x, y)=1-x^{2}-y^{2}$ is a square.
