Parametric Surfaces

A parametric surface is defined by a function \(\mathbf{r}:\mathbb{R}^2 \to \mathbb{R}^3\). The idea is exactly the same as for a parametric curve: we are taking a (piece of) plane and inserting it into the space, bending, stretching a compressing in during the process.

Let’s look at several examples:

\(\mathbf{r}(u,v) = \left\langle \cos(u), \sin(u), v\right\rangle\) for \(-\pi \le u \le \pi\) and \(0 \le v \le 1\).

This function maps the rectangle \([-\pi,\pi]\times[0,1]\)

to the cylinder:

The horizontal segment at \(v = \frac{1}{2}\) (green) gets mapped to the green circle, while the vertical segment at \(u = 0\) (blue) gets mapped to the blue vertical segment.
\(\mathbf{r}(u,v) = \left\langle v\cos(u), v\sin(u), v\right\rangle\) for \(-\pi \le u \le \pi\) and \(0 \le v \le 1\).

This function maps the rectangle \([-\pi,\pi]\times[0,1]\)

to the cone:
\(\mathbf{r}(u,v) = \left\langle v^2\cos(u), v^2\sin(u), v\right\rangle\) for \(-\pi \le u \le \pi\) and \(0 \le v \le 1\).

This function maps the rectangle \([-\pi,\pi]\times[0,1]\)

to the following surface:
\(\mathbf{r}(u,v) = \left\langle v\cos(u), v\sin(u), v^2\right\rangle\) for \(-\pi \le u \le \pi\) and \(0 \le v \le 1\).

This function maps the rectangle \([-\pi,\pi]\times[0,1]\)

to a paraboloid:
\(\mathbf{r}(u,v) = \left\langle \cos(v)\cos(u), \cos(v)\sin(u), v\right\rangle\) for \(-\pi \le u \le \pi\) and \(-\frac{\pi}{2} \le v \le \frac{\pi}{2}\).

This function maps the rectangle \([-\pi,\pi]\times\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)

to a lemon:
\(\mathbf{r}(u,v) = \left\langle \sin(v)\cos(u), \sin(v)\sin(u), v\right\rangle\) for \(-\pi \le u \le \pi\) and \(-\frac{\pi}{2} \le v \le \frac{\pi}{2}\).

This function maps the rectangle \([-\pi,\pi]\times\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)

to an hourglass:
The function

\[\begin{aligned} x(u,v) &= (3 - \cos(v))\cos(u)\\ y(u,v) &= (3 - \cos(v))\sin(u)\\ z(u,v) &= \sin(v) \end{aligned}\]

for \(-\pi \le u \le \pi\) and \(-\pi \le v \le \pi\) maps the rectangle \([-\pi,\pi]\times\left[-\pi,\pi\right]\)

to a torus:
The function

\[\begin{aligned} x(u,v) &= u(3 - \cos(v))\cos(u)\\ y(u,v) &= u(3 - \cos(v))\sin(u)\\ z(u,v) &= u\sin(v) + 6\pi - 2u \end{aligned}\]

for \(0 \le u \le 3\pi\) and \(-\pi \le v \le \pi\) maps the rectangle \([0,3\pi]\times\left[-\pi,\pi\right]\)

to this surface:

A video on parametric surfaces

Illustration for an “area unit”

showing how the area of a tiny part of a surface can be approximated by an area of a parallelogram:

The yellow region is an image of a small rectangle in the \(uv\)-plane with corner at \((u, v)\) and sides \(du\) and \(dv\). Its area is approximated by the purple parallelogram with sides \(\mathbf{r}_u\; du\) and \(\mathbf{r}_v\; dv\). The area of the parallelogram is \(\left\lvert\mathbf{r}_u)u,v)\times \mathbf{r}_v(u,v) \right\rvert\;du\;dv\).

Normal vector

Suppose \(\mathbf{r}(u,v) = \left\langle x(u,v), y(u,v), z(u,v)\right\rangle\) is a vector function of two variables with differentiable components at \((u,v)\), and suppose that \(\mathbf{r}_u(u,v)\) and \(\mathbf{r}_v(u,v)\) are not zero vectors.

Then the vectors \(\mathbf{r}_u(u,v)\) and \(\mathbf{r}_v(u,v)\) belong to the tangent plane to the parametric surface at \(\mathbf{r}(u,v)\), and their cross product is a normal vector to this plane. We call this vector a normal vector to the surface at \(\mathbf{r}(u,v)\).

Dividing a normal vector to a surface by its magnitude, we get a unit normal vector. At each point where there is a normal vector to a surface, there are two unit normal vectors, pointing in opposite directions. We can think about these as pointing to opposite sides of the surface.

Suppose a surface has a normal vector at all of its points with possibly finitely many exceptions (for example, the “lemon” surface above has no normal vector at the two “tips”). At each point where a normal vector exists, we can choose one of the two unit normal vectors as the one that points to the outside of the surface. The question is whether we can do this in a consistent way. One way to define what consistent means is, can we make the choice of the unit normal vectors in such a way that the resulting vector field is continuous?

If we can, we say the surface is orientable, and the choice of one of the two continuous unit normal vector fields is called the orientation of the surface.

Examples

All the surfaces in the examples above are orientable. The following surface is orientable, showing one of the two choices of orientation:

The following is an example of perhaps the most famous non-orientable surface, so called Mőbius band. You can make one yourself by taking a strip of paper, twisting one end by \(\pi\), and glue it to the other end. You can see that the unit normal vector field is discontinuous along the blue line segment on the surface:

Another well known example of non-orientable surface is so called cross-cap. This is a self-intersecting surface: there is a segment such that each point on this segment lies on the surface twice, for two different values of the parameter \(u\). This illustration is colored in such a way that you can see the purple part of the surface crossing the yellow part. Separately, each part has a tangent plane and therefore a normal vector at each point of the intersection segment. At the “bottom” of the surface, the two parts (purple and yellow) come seamlessly together.

The following shows a choice of a unit normal vector field on the cross-cap. The vectors are colored in two colors, red and green, that correspond to the two colors on the coloring of the cross-cap shown above. You can see how the green vectors pointing “out” of the cross-cap point “into” the cross-cap after passing through the intersection. There is a discontinuity of the field along the blue curve on the surface.

You can see that the whole region that appears “surrounded” by the surface is in fact inside the surface from the point of view of the field on one half of the surface, but outside of the surface from the point of view of the field on the other half of the surface. There is no way to choose a continuous unit normal vector field on the surface.