From a Parameter-Domain Edge to a Surface Curve
A parametrized surface patch is described by a map \(\mathbf{r}:D\to\mathbb{R}^3\), where \(D\) is a region in the parameter plane with coordinates \(u\) and \(v\). An edge of \(D\) is a curve in that plane. To find the corresponding curve on the surface, substitute a parametrization of the edge into \(\mathbf{r}\).
This procedure turns a two-parameter description into a one-parameter curve. Repeating it for every edge gives the boundary trace of the patch. The endpoints of those edge curves are obtained by mapping the vertices of the parameter domain; these are the surface patch’s corner points when the domain has corners.
For a rectangular or polygonal domain, each edge can usually be parametrized by letting one coordinate vary and fixing the other, or by interpolating between its endpoints. The essential step is to substitute both edge coordinates into the surface map. An edge is not traced by setting a coordinate to zero unless that edge actually has that coordinate equal to zero.
Why Mapping the Edges Gives the Boundary Trace
On the boundary of the parameter domain, every parameter point lies on one of its edges. If those edges are parametrized by \(\boldsymbol{\alpha}_j\), then their union covers \(\partial D\). Applying \(\mathbf{r}\) to those points therefore produces the image of the parameter-domain boundary.
This is a direct consequence of composition: a point on an edge has parameter coordinates \((u_j(t),v_j(t))\), so its surface point is precisely \(\mathbf{r}(u_j(t),v_j(t))\). This argument identifies the boundary trace, not necessarily the topological boundary of the image set in every possible case. If a map folds the patch over itself, identifies distinct boundary points, or collapses an edge to a point, the image can have more complicated topology. For a regular, one-to-one patch extending to its boundary, the edge traces describe the patch boundary in the usual geometric sense.
When the boundary of \(D\) is traversed counterclockwise, the resulting list of surface curves records the boundary traversal inherited from the parameter domain. At a corner, the endpoint of one edge parametrization must agree with the starting point of the next. This endpoint check is a useful way to catch errors.
Worked Example: Tracing a Rectangular Domain
Worked Example: Tracing a Rectangular Domain
Let \(D=[0,2]\times[0,1]\), and consider \(\mathbf{r}(u,v)=(u,v,u^2+v)\). Find the four boundary curves in counterclockwise order, starting at \((0,0)\), and find the images of the four corners.
Traverse the rectangle from its lower-left corner. On the bottom edge, \(v=0\) and \(u=t\), where \(0\le t\le2\). Thus
On the right edge, \(u=2\) and \(v=t\), where \(0\le t\le1\). Therefore
On the top edge, the counterclockwise traversal goes from \(u=2\) back to \(u=0\). Set \(u=2-t\), \(v=1\), with \(0\le t\le2\). Then
Finally, the left edge is traversed from \(v=1\) down to \(v=0\). Set \(u=0\), \(v=1-t\), with \(0\le t\le1\). This gives
Map each parameter-domain corner to obtain its surface point:
Check the joins. The first curve starts at \((0,0,0)\) and ends at \((2,0,4)\); the second starts at \((2,0,4)\) and ends at \((2,1,5)\). The third starts at \((2,1,5)\) and ends at \((0,1,1)\), and the fourth returns to \((0,0,0)\). Thus the four edge traces join in the stated order and close.
Worked Example: Tracing a Triangular Domain
Worked Example: Tracing a Triangular Domain
Let \(D=\{(u,v):u\ge0,\ v\ge0,\ u+v\le1\}\), with vertices \((0,0)\), \((1,0)\), and \((0,1)\). For \(\mathbf{r}(u,v)=(u,v,u+2v)\), find the boundary traces in counterclockwise order and map the vertices.
The first edge runs from \((0,0)\) to \((1,0)\). Set \((u,v)=(t,0)\), where \(0\le t\le1\). Substitution gives
The second edge runs from \((1,0)\) to \((0,1)\). Its coordinates are \((u,v)=(1-t,t)\), for \(0\le t\le1\). Hence
The third edge runs from \((0,1)\) to \((0,0)\). Put \((u,v)=(0,1-t)\). Then
The corner images are
Check the edge endpoints: \(\boldsymbol{\gamma}_1(0)=(0,0,0)\), \(\boldsymbol{\gamma}_1(1)=(1,0,1)\), and \(\boldsymbol{\gamma}_2(0)=(1,0,1)\), \(\boldsymbol{\gamma}_2(1)=(0,1,2)\). Also, \(\boldsymbol{\gamma}_3(0)=(0,1,2)\) and \(\boldsymbol{\gamma}_3(1)=(0,0,0)\). Each edge ends where the next begins, so the boundary trace closes.
Shared Curves Between Adjacent Patches
Two patches share a boundary curve when an edge of one parameter domain and an edge of the other map onto the same curve in space. Their parameter coordinates need not be the same. To verify a shared curve, compare the resulting space curves; they may be written with different parameters or traversed in opposite directions.
Worked Example: Finding a Shared Edge
Consider two patches on adjacent squares. Let \(\mathbf{r}_1(u,v)=(u,v,0)\) for \(0\le u,v\le1\), and let \(\mathbf{r}_2(s,t)=(1+s,t,0)\) for \(0\le s,t\le1\). Find their shared edge, check its traversal on each patch, and identify the outer boundary of their union.
For the first square, the right edge is \(u=1\), with \(v=q\) for \(0\le q\le1\). Its image is
For the second square, the left edge is \(s=0\), with \(t=q\) for \(0\le q\le1\). Its image is also
Thus the shared geometric curve is the line segment from \((1,0,0)\) to \((1,1,0)\). The parameter domains’ counterclockwise traversals go in opposite directions along this shared edge: the first patch traverses its right edge from bottom to top, while the second traverses its left edge from top to bottom. In formulas, the second traversal is \(\mathbf{r}_2(0,1-q)=(1,1-q,0)\), which has the same image but reversed direction.
The four outer edges of the union are the bottom edge from \((0,0,0)\) to \((2,0,0)\), the right edge from \((2,0,0)\) to \((2,1,0)\), the top edge from \((2,1,0)\) to \((0,1,0)\), and the left edge from \((0,1,0)\) to \((0,0,0)\). The shared segment lies inside the combined region, so it is a boundary curve of each individual patch but not part of the outer boundary of their union.
As an endpoint check, the shared segment has endpoints \(\mathbf{r}_1(1,0)=\mathbf{r}_2(0,0)=(1,0,0)\) and \(\mathbf{r}_1(1,1)=\mathbf{r}_2(0,1)=(1,1,0)\). Both patches therefore meet along the entire same segment, not merely at a corner.
A Reliable Edge-Tracing Procedure
For a polygonal parameter domain, the following process keeps the parameter-plane geometry separate from the surface-space calculation.
Record the parameter coordinates of each corner and decide which direction the boundary is traversed.
Use a parameter interval and write both coordinates \(u(t)\) and \(v(t)\) for that edge.
Substitute the edge coordinates into every component of \(\mathbf{r}(u,v)\) to obtain a space curve.
Evaluate \(\mathbf{r}\) at the vertices and confirm that consecutive edge traces meet at the expected points.
Map candidate edges from both domains and check whether their space curves have the same image, even if their parameters run in opposite directions.
Common Mistake: Fixing the Wrong Parameter
Suppose the patch is \(\mathbf{r}(u,v)=(u,v,u+v)\) on the rectangle \(0\le u\le2\), \(0\le v\le1\). The bottom edge has \(v=0\), not \(u=0\). A mistaken calculation might set \(u=0\) and claim the bottom edge is \(\mathbf{r}(0,t)=(0,t,t)\). That is actually the left edge, because its parameter points have \(u=0\).
For the bottom edge, let \(u=t\), \(v=0\), with \(0\le t\le2\). The correct trace is
The endpoint check confirms the correction: the bottom edge must start at \(\mathbf{r}(0,0)=(0,0,0)\) and end at \(\mathbf{r}(2,0)=(2,0,2)\). The mistaken curve ends at \(\mathbf{r}(0,1)=(0,1,1)\), so its endpoints reveal that it traces a different edge.
A second issue is confusing a correct geometric edge with its traversal direction. On the top edge of a counterclockwise rectangle, setting \(u=t\) traces the same set of points as setting \(u=2-t\), but the first runs left to right and the second right to left. When the direction matters, specify the edge parametrization and check its start and end points.
When Edge Traces Need Extra Care
Boundary edges need not remain distinct under the map. A parametrization can send an entire edge to a single point, or two different edges can have the same image. Distinct corners can also map to the same point. Therefore, mapping the vertices and edges tells you what the boundary trace is, but it does not by itself guarantee that the trace is a simple closed curve.
When several patches meet, compare their edge images rather than their parameter coordinates. A shared edge can be represented by different formulas on the two domains; checking that both cover the same points and comparing endpoints establishes the match. If their traversal directions are relevant, compare the start and end points as well. These checks are useful before treating a collection of patches as one larger surface, because internal shared edges should not be mistaken for parts of the union’s outer boundary.
Check Your Understanding
For each question, first identify the relevant parameter-domain edge, then reason about its image under the surface map.
- For \(\mathbf{r}(u,v)=(u,v,u-v)\) on \(0\le u\le1\), \(0\le v\le2\), write a parametrization of the image of the edge \(u=1\).
- What point on the surface corresponds to the parameter-domain corner \((2,0)\) under \(\mathbf{r}(u,v)=(u+v,u-v,uv)\)?
- For the triangular edge from \((1,0)\) to \((0,1)\), give a parametrization of its parameter coordinates for \(0\le t\le1\).
- Two edges from adjacent patches have images \((1,t,0)\) and \((1,1-t,0)\), both for \(0\le t\le1\). Do they represent the same geometric segment, and how do their traversal directions compare?
- Why might the image of the parameter-domain boundary fail to be a simple closed curve even when the parameter domain itself has a simple polygonal boundary?