Parametrized Surfaces as Maps From Parameter Domains

Introduction

A surface parametrization describes points in three-dimensional space by using two parameters. The parameters vary over a region in a plane, and a rule assigns a point in space to each permitted parameter pair.

This viewpoint separates three objects that are easy to confuse: the parameter domain, the map’s coordinate functions, and the image traced out in space. Identifying each one is the first step in working with parametrized surfaces.

Central idea: A parametrization is a function from a region of the parameter plane into three-dimensional space. Its image is the set of spatial points produced by all parameter pairs in its domain.

From a Parameter Pair to a Spatial Point

Let $D$ be a region in the $uv$-plane. A surface parametrization is a map $\mathbf{r}$ that assigns a vector in $\mathbb{R}^3$ to every pair $(u,v)$ in $D$:

\mathbf{r}:D\to\mathbb{R}^3,\qquad \mathbf{r}(u,v)=\bigl(x(u,v),y(u,v),z(u,v)\bigr).

The pair $(u,v)$ is an input. Its two entries are called parameters. The output is a point in space, whose three coordinates are the values $x(u,v)$, $y(u,v)$, and $z(u,v)$.

Definition: The parameter domain is the set $D$ of permitted parameter pairs. The coordinate functions are the three scalar functions $x(u,v)$, $y(u,v)$, and $z(u,v)$. The image of the parametrization is $$ \mathbf{r}(D)=\{\mathbf{r}(u,v):(u,v)\in D\}\subseteq\mathbb{R}^3. $$

The image is a set of points in space, not a set of parameter pairs. A point belongs to the image precisely when at least one permitted pair of parameters maps to it. This definition does not require different parameter pairs to produce different points.

In applications, additional smoothness or geometric conditions may be required of a parametrization. Here the focus is on reading the map and identifying its domain, coordinate functions, and image. The image of an arbitrary map need not automatically be a well-behaved smooth surface.

How to Read the Domain and Coordinates

When a parametrization is written as a vector, each component gives one coordinate function, in order: first the $x$-coordinate, then the $y$-coordinate, then the $z$-coordinate. The parameter restrictions are stated separately, either as a set $D$ or as inequalities such as $0\leq u\leq 1$ and $u^2+v^2\leq 1$.

To describe the image, translate the parameter restrictions through the coordinate functions. One reliable method is to start with a point $(x,y,z)$ in the image, write the equations connecting it to $u$ and $v$, and use the domain restrictions. If possible, also reverse the argument: show that every spatial point satisfying the proposed conditions comes from an allowed parameter pair.

Read the domain.
Record exactly which parameter pairs are allowed.

Read the coordinate functions.
Match the first, second, and third components to the spatial coordinates.

Form the image.
Collect the outputs for all allowed parameter pairs, and express the resulting set in spatial coordinates when useful.

Worked Examples

Worked Example: A Disk as the Parameter Domain

Consider the map

\mathbf{r}(u,v)=(u,v,u^2+v^2), \qquad D=\{(u,v):u^2+v^2\leq 1\}.

Step 1: Identify the domain. The allowed inputs are the points in the unit disk in the $uv$-plane, including its boundary. Thus the parameter domain is $D$, not a subset of three-dimensional space.

Step 2: Identify the coordinate functions. Comparing components gives

x(u,v)=u,\qquad y(u,v)=v,\qquad z(u,v)=u^2+v^2.

Step 3: Express the image in spatial coordinates. Since $x=u$ and $y=v$, the domain condition becomes $x^2+y^2\leq 1$. The third coordinate satisfies $z=u^2+v^2=x^2+y^2$. Therefore the image is

\mathbf{r}(D) = \{(x,y,z)\in\mathbb{R}^3: x^2+y^2\leq 1,\ z=x^2+y^2\}.

Check both directions. Every parameter pair in $D$ gives $x=u$, $y=v$, so $x^2+y^2=u^2+v^2\leq 1$, and $z=x^2+y^2$. Conversely, if a point satisfies the displayed spatial conditions, choose $u=x$ and $v=y$. Then $(u,v)\in D$ and $\mathbf{r}(u,v)=(x,y,z)$. Hence the description includes exactly the image.

Worked Example: A Map That Repeats Some Points

Let

\mathbf{r}(u,v)=(u^2,v,0), \qquad D=\{(u,v):-1\leq u\leq 1,\ 0\leq v\leq 2\}.

Step 1: Identify the domain and coordinates. The domain is the stated rectangle in the $uv$-plane, and the coordinate functions are

x(u,v)=u^2,\qquad y(u,v)=v,\qquad z(u,v)=0.

Step 2: Determine the spatial restrictions. Because $-1\leq u\leq 1$, squaring gives $0\leq u^2\leq 1$, so $0\leq x\leq 1$. The restriction on $v$ gives $0\leq y\leq 2$, and the third coordinate is always $z=0$. Thus every image point lies in

\{(x,y,z):0\leq x\leq 1,\ 0\leq y\leq 2,\ z=0\}.

Step 3: Check that every point in this set occurs. Given any $(x,y,0)$ satisfying those inequalities, choose $u=\sqrt{x}$ and $v=y$. Then $0\leq \sqrt{x}\leq 1$, so $(u,v)\in D$, and $\mathbf{r}(u,v)=(x,y,0)$. The proposed set is exactly the image.

This map does not assign a unique parameter pair to every image point. For example, $u=1$ and $u=-1$, with the same $v$, both give the point $(1,v,0)$. The image is still a set of spatial points; repeated outputs do not add new points to it.

Worked Example: A Triangular Parameter Domain

Define

\mathbf{r}(u,v)=(u+v,2u-v,3), \qquad D=\{(u,v):u\geq 0,\ v\geq 0,\ u+v\leq 1\}.

Step 1: Record the domain and coordinate functions. The parameter domain is the triangle described by its three inequalities. The coordinate functions are

x=u+v,\qquad y=2u-v,\qquad z=3.

Step 2: Translate the domain inequalities. Adding the first two coordinate equations gives $x+y=3u$, so $u=(x+y)/3$. Also $2x-y=3v$, so $v=(2x-y)/3$. Therefore $u\geq 0$ is equivalent to $x+y\geq 0$, and $v\geq 0$ is equivalent to $2x-y\geq 0$. Since $u+v=x$, the last domain restriction becomes $x\leq 1$. The coordinate $z$ is fixed at $3$.

These calculations suggest the image description

\{(x,y,z):z=3,\ x+y\geq 0,\ 2x-y\geq 0,\ x\leq 1\}.

Step 3: Verify the description. For any parameter pair in $D$, the equations above show that its image satisfies all four spatial conditions. Conversely, suppose $(x,y,z)$ satisfies those conditions. Set $u=(x+y)/3$ and $v=(2x-y)/3$. The inequalities give $u\geq 0$ and $v\geq 0$, while $u+v=x\leq 1$. Thus $(u,v)\in D$. Substitution gives $u+v=x$, $2u-v=y$, and $z=3$, so the point is in the image. This proves the description exactly.

Why the Image Description Needs Two Directions

A calculation that shows every output satisfies some spatial conditions proves only that the image is contained in the set described by those conditions. It does not yet prove equality. The proposed set might contain points that cannot be produced by any allowed parameter pair.

The complete argument has two parts:

Forward direction.
Start with an arbitrary $(u,v)\in D$. Show that $\mathbf{r}(u,v)$ satisfies the proposed spatial conditions.

Reverse direction.
Start with an arbitrary spatial point satisfying those conditions. Find parameters in $D$ that map to it.

This also explains why the image is formally defined by the set-builder expression $\{\mathbf{r}(u,v):(u,v)\in D\}$: membership means that at least one allowed input produces the point. In examples, the forward and reverse checks establish that a more convenient description means exactly the same thing.

One Image Can Have Different Parametrizations

The map and its image are related but are not identical objects. The map records how inputs produce outputs; the image records only which outputs occur. In the repeated-point example, changing the sign of $u$ can leave the output unchanged. Other choices of parameter domain or coordinate functions can also describe the same spatial set.

Consequently, a description of the image alone does not specify how it is parametrized. Conversely, a formula for a map must always be read together with its domain: restricting the allowed parameter pairs can change the image even when the coordinate functions stay the same.

Key takeaway: Keep the input set and output set separate. The domain tells which parameter pairs are allowed; the coordinate functions tell where each pair goes; the image is the collection of all resulting points.

Common Mistake: Confusing Parameters with Spatial Coordinates

A frequent error is to treat the symbols $u$ and $v$ as though they were automatically the spatial coordinates $x$ and $y$. That is only justified when the coordinate functions make it so. For example, in $\mathbf{r}(u,v)=(u^2,v,0)$, the spatial $x$-coordinate is $u^2$, not $u$.

A wrong calculation would take $-1\leq u\leq 1$ and conclude that $-1\leq x\leq 1$ because “$x=u$.” But the first component actually states $x=u^2$, and therefore $0\leq x\leq 1$. The correction is to read the map component by component before translating the domain restrictions.

Why This Viewpoint Matters

Many later calculations involving surfaces begin with a map from a parameter domain. The domain says which portion is being described, while the coordinate functions specify the spatial location corresponding to each parameter pair. Keeping these roles distinct makes it possible to interpret formulas reliably and to check whether a proposed spatial description is complete.

For now, the essential task is not to infer geometric properties beyond the formula. It is to identify the map’s inputs, outputs, and image precisely. Whenever an image is rewritten using spatial coordinates, checking both directions ensures that the rewrite has neither lost image points nor added points that the map cannot reach.

Check Your Understanding

For each question, distinguish the parameter-domain statement from the spatial-image statement.

For $\mathbf{r}(u,v)=(u,v,u-v)$ with $0\leq u\leq 1$ and $-1\leq v\leq 1$, what is the parameter domain and what are the three coordinate functions?
For $\mathbf{r}(u,v)=(u,v,u^2+v^2)$ on $u^2+v^2\leq 4$, what equation relates $x,y,z$, and what restriction on $x,y$ comes from the domain?
In the map $\mathbf{r}(u,v)=(u^2,v,0)$, can two distinct parameter pairs produce the same point? Give one example.
Why does showing that all outputs satisfy a proposed image description not, by itself, prove that the description is exact?
If the coordinate functions stay fixed but the parameter domain is restricted, must the image stay the same? Explain.