1. Let ${w=f(x_{1},x_{2},\dots,x_{n})}$ where ${x_{1}}$, ${x_{2}}$, …, ${x_{n}}$ are all independent variables. Then the “Domain” of ${f}$ is the set of ${n}$-tuples ${{\mathbb R}^{n}}$. Note that an ordered pair is
(1)$\displaystyle (x,y)=\mbox{2-tuple}.$
The set of corresponding values is the “Range” (or Codomain) of the function. So we have
(2)$\displaystyle f\colon\underbrace{{\mathbb R}^{n}}_{\text{domain}}\rightarrow\underbrace{{\mathbb R}}_{\text{range}}$
Sometimes we write ${\mathrm{dom}(f)}$ for the domain of ${f}$, and ${\mathrm{ran}(f)}$ or ${\mathrm{cod}(f)}$ for the range (or codomain) of ${f}$. We DO NOT write ${f({\mathbb R}^{2})}$ for the range, because this is the collection of all points mapped by ${f}$.

Example 1. Consider ${f\colon{\mathbb R}^{2}\rightarrow{\mathbb R}}$ defined by
(3)$\displaystyle f(x,y) = \sqrt{x^{2}+y^{2}}$
Notice that ${f(x,y)\geq0}$ for any ${x,y\in{\mathbb R}}$. So
(4)$\displaystyle f({\mathbb R}^{2})=\{f(x,y)\in{\mathbb R} : x,y\in{\mathbb R}\} = \{ u\in{\mathbb R} : u\geq0\}\not={\mathbb R}.$
This is not the codomain! It’s contained in the codomain, though. The image is always a subset of the codomain.

Example 2. Let ${g(x,y)=\ln(x^{2}-y)}$. For ${g}$ to be defined, we need
(5)$\displaystyle x^{2}-y>0\quad\mbox{or}\quad x^{2}>y.$
The boundary of the domain is ${x^{2}=y}$, which we can doodle:

Since the boundary is not in the domain, then the domain of ${g}$ is open.

Example 3. Consider
(6)$\displaystyle h(x,y)=\sqrt{25-x^{2}-y^{2}}.$
The domain is the set of ${(x,y)}$ such that
(7)$\displaystyle x^{2}+y^{2}\leq25.$
We can doodle this:

Observe that the boundary is the circle; the disc is the circle and everything enclosed in it.

2. So we have just discussed domains and codomains, but we have not discussed the graph of the function ${z=f(x,y)}$. What would this look like? Well, when we plot ${y=g(x)}$, it’s on the plane ${{\mathbb R}^{2}}$. So plotting ${z=f(x,y)}$ would be on the 3-space ${{\mathbb R}^{3}}$. Lets start considering examples of what this looks like.

Example 4. Consider the graph given by
(8)$\displaystyle z = f(x,y) = 9 - x^{2}-y^{2}.$
How can we draw this? Well, the first trick is to draw when ${x=0}$ and ${y=0}$:
(9)\displaystyle \begin{aligned} z &= f(0,y) = 9-y^{2}\\ z &= f(x,0) = 9-x^{2} \end{aligned}
These are parabolas in the ${yz}$ and ${xz}$ planes, respectively. Now we can start drawing “Level Curves”, i.e., curves where we fix ${z}$ to be some constant.

For example, when ${z=0}$, we have a circle
(10)$\displaystyle x^{2}+y^{2}=9$
Observe then that ${z}$ controls the radius of the circles: for nonzero ${z}$, we have
(11)$\displaystyle x^{2}+y^{2}=9-z.$
The left hand side must be non-negative, and can be zero only when ${z=9}$. So we get a surface that looks like

We call this surface a “Paraboloid” as we have parabolas along the ${x}$– and ${y}$-axes.

This is the general scheme for picturing a surface: draw level curves ${f(x,y)=c}$ for some constant ${c}$, which produces the level curve for ${f}$ corresponding to ${c}$.

Example 5. Consider ${x^{2}+y^{2}-z^{2}=0}$. What does this surface look like?

Solution: First we observe the curves along the ${x}$-axis, i.e., when ${y=0}$ is ${x^{2}=z^{2}}$. Similarly when ${x=0}$ we have ${y^{2}=z^{2}}$. What sort of curves are these? Well, ${(\pm t,0,t)}$ and ${(0,\pm t,t)}$ are the curves, for ${t\in{\mathbb R}}$.

Also note that ${x^{2}+y^{2}}$ describes a circle, whose radius happens to be ${z^{2}}$. This tells us the level curves are simply circles. So we have a cone:

This surface is precisely a “Cone”.

Example 6. What if we deform the previous example, writing
(12)$\displaystyle x^{2}+y^{2}-z^{2}=1.$
What surface does this describe?

Solution: Well, we see that the radius of the level curves deform ${z^{2}\rightarrow z^{2}+1}$. So the resulting surface looks like a cone, but along the ${x}$-axis we don’t have a straight-line: we have a hyperbola ${x^{2}-z^{2}=1}$. Similarly along the ${y}$-axis we have another hyperbola ${y^{2}-z^{2}=1}$. Thus our surface is doodled as:

This “deformed cone” is called a “One Sheeted Hyperboloid”.

Example 7. Another variation, consider the surface
(13)$\displaystyle x^{2}+y^{2}-z^{2}=-1$
What does it look like?

Solution: Well, we see that ${x^{2}+y^{2}\geq0}$ always, whereas ${z^{2}-1\geq0}$ only when ${|z|\geq1}$. So we have two “surfaces” or “Sheets” here. The level curves are again circles, and this enables us to doodle the graph:

Along the ${x}$-axis and ${y}$-axis, we have hyperbolas. For this reason, we call the surface a “Two-Sheeted Hyperbaloid”.

Example 8. Suppose we have a surface given by ${z=x^{2}-y^{2}}$. What does it look like?

Solution: Observe the level curves for ${z>0}$ gives us hyperbolas, and for ${z<0}$ we again have hyperbolas (the same as before but reflected about the line ${x=y}$). For ${z=0}$ we have a cone.

When we take ${y=\pm1,0}$ we see ${z=x^{2}-C}$ for some constant ${C}$. These curves look like smiles. For ${x=\pm1,0}$, we have ${z=C-y^{2}}$ for some constant ${C}$. These curves look like frowns.

This surface is a saddle, and we can doodle it as

Since the curves when we hold ${z}$ constant form hyperbolas, and the curves when we hold ${x}$ or ${y}$ constaant are parabolas, people sometimes call this saddle a “Hyperbolic Paraboloid”.