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Parametrized Surfaces · Tutorial 846 of 999

Workshop On Building Parametrizations From Descriptions

Build surface parametrizations systematically and verify their domains, tangent data, normals, and boundary curves.

Intermediate 12 min read

What You'll Learn

  • Turn geometric descriptions of plane, graph, cylinder, and sphere patches into parametrization maps.
  • Choose parameter domains that represent exactly the specified patches.
  • Compute tangent vectors and cross-product normals, and check regularity.
  • Find and verify the boundary curves of rectangular parameter domains.
  • Distinguish parameter coefficients from physical distances.

From a Geometric Description to a Parametrization

A surface description often gives a shape, location, and extent in words: a portion of a plane between four edges, a graph above a rectangle, or a sector of a sphere between specified angles. A complete parametrization turns that description into a map from a parameter domain into three-dimensional space.

For a two-parameter surface, write the map as \(\mathbf{r}(u,v)\), where \((u,v)\) ranges over a specified set \(D\) in the parameter plane. Constructing a useful parametrization means specifying both the formula and the domain. The formula determines where parameter values go; the domain determines which part of the surface is included.

Definition: A parametrized surface patch is a map \(\mathbf{r}:D\to\mathbb{R}^3\), where \(D\subseteq\mathbb{R}^2\). At a parameter value \((u,v)\), the vectors \(\mathbf{r}_u\) and \(\mathbf{r}_v\) are tangent vectors. If \(\mathbf{r}_u\times\mathbf{r}_v\neq\mathbf{0}\), the parametrization is regular there, and this cross product is a normal vector to the surface.

For a description that specifies edges, curves, or angular limits, write down the parameter intervals explicitly. Then compute the tangent vectors, choose the desired normal direction, and obtain each boundary curve by holding one parameter at a boundary value while the other varies. These checks help confirm that the parametrization covers the intended patch.

Why Tangents and Cross-Product Normals Work

Fix \(v\) and vary \(u\). The points \(\mathbf{r}(u,v)\) trace a curve on the surface, whose velocity is \(\mathbf{r}_u\). Varying \(v\) with \(u\) fixed gives a second curve with velocity \(\mathbf{r}_v\). Both velocities lie in the tangent plane when the parametrization is regular. A vector perpendicular to both is therefore perpendicular to that plane.

$$ (\mathbf{r}_u\times\mathbf{r}_v)\cdot\mathbf{r}_u=0, \qquad (\mathbf{r}_u\times\mathbf{r}_v)\cdot\mathbf{r}_v=0. $$

These equalities follow from the defining property of the cross product: its result is perpendicular to each factor. Thus, where the cross product is nonzero, it supplies a normal. Reversing its order reverses the direction: \(\mathbf{r}_v\times\mathbf{r}_u=-(\mathbf{r}_u\times\mathbf{r}_v)\).

Worked Example: A Parallelogram in a Plane

Worked Example: A Parallelogram in a Plane

Parametrize the parallelogram starting at \(P=(1,-1,2)\), with edge vectors \(\mathbf{a}=(2,1,0)\) and \(\mathbf{b}=(0,0,1)\), using coefficients \(0\leq u\leq2\) and \(0\leq v\leq1\).

The point at parameters \((u,v)\) is the starting point plus \(u\) copies of the first edge vector and \(v\) copies of the second:

$$ \mathbf{r}(u,v) =P+u\mathbf{a}+v\mathbf{b} =(1+2u,-1+u,2+v). $$

The parameter domain is the rectangle \(D=\{(u,v):0\leq u\leq2,\ 0\leq v\leq1\}\). Its corners map to

$$ \begin{aligned} \mathbf{r}(0,0)&=(1,-1,2),& \mathbf{r}(0,1)&=(1,-1,3),\\ \mathbf{r}(2,1)&=(5,1,3),& \mathbf{r}(2,0)&=(5,1,2). \end{aligned} $$

The tangent vectors are constant:

$$ \mathbf{r}_u=(2,1,0), \qquad \mathbf{r}_v=(0,0,1), \qquad \mathbf{r}_u\times\mathbf{r}_v=(1,-2,0). $$

The nonzero cross product confirms regularity everywhere on this patch. The four boundary curves, in the order around the parameter rectangle, are

$$ \begin{aligned} C_1(t)&=\mathbf{r}(0,t)=(1,-1,2+t), &&0\leq t\leq1,\\ C_2(t)&=\mathbf{r}(t,1)=(1+2t,-1+t,3), &&0\leq t\leq2,\\ C_3(t)&=\mathbf{r}(2,1-t)=(5,1,3-t), &&0\leq t\leq1,\\ C_4(t)&=\mathbf{r}(2-t,0)=(5-2t,1-t,2), &&0\leq t\leq2. \end{aligned} $$

For a join check, \(C_1(1)=(1,-1,3)=C_2(0)\), \(C_2(2)=(5,1,3)=C_3(0)\), \(C_3(1)=(5,1,2)=C_4(0)\), and \(C_4(2)=(1,-1,2)=C_1(0)\). Thus the four curves close up in the stated order.

The parameters \(u\) and \(v\) here are coefficients along the edge vectors, not distances traveled. For example, \(\|\mathbf{a}\|=\sqrt{5}\), so a displacement \(u\mathbf{a}\) has length \(u\sqrt{5}\) for \(u\geq0\), not length \(u\).

Worked Example: A Patch of a Graph

Worked Example: A Patch of a Graph

Consider the graph \(z=x^2+y\) above the rectangle \(0\leq x\leq1\), \(-1\leq y\leq1\). Use \(u=x\) and \(v=y\), so that

$$ \mathbf{r}(u,v)=(u,v,u^2+v), \qquad 0\leq u\leq1,\quad -1\leq v\leq1. $$

The formula satisfies the graph equation because its coordinates are \(x=u\), \(y=v\), and \(z=u^2+v=x^2+y\). Its tangent vectors and cross-product normal are

$$ \mathbf{r}_u=(1,0,2u), \qquad \mathbf{r}_v=(0,1,1), \qquad \mathbf{r}_u\times\mathbf{r}_v=(-2u,-1,1). $$

The third component of the cross product is \(1\), so it is nonzero throughout the domain. The patch is regular. Its boundary curves, following the rectangle in parameter space, are

$$ \begin{aligned} C_1(t)&=\mathbf{r}(0,t)=(0,t,t), &&-1\leq t\leq1,\\ C_2(t)&=\mathbf{r}(t,1)=(t,1,t^2+1), &&0\leq t\leq1,\\ C_3(t)&=\mathbf{r}(1,t)=(1,t,1+t), &&1\geq t\geq-1,\\ C_4(t)&=\mathbf{r}(t,-1)=(t,-1,t^2-1), &&1\geq t\geq0. \end{aligned} $$

The endpoint checks are \(C_1(1)=(0,1,1)=C_2(0)\), \(C_2(1)=(1,1,2)=C_3(1)\), \(C_3(-1)=(1,-1,0)=C_4(1)\), and \(C_4(0)=(0,-1,-1)=C_1(-1)\). Each boundary point also satisfies \(z=x^2+y\), as required.

Worked Example: A Quarter of a Cylinder

Worked Example: A Quarter of a Cylinder

Take the lateral patch of the cylinder of radius \(2\) around the \(z\)-axis, from height \(z=0\) to \(z=3\), and from angle \(0\) to \(\pi/2\). Its parametrization and domain are

$$ \mathbf{r}(\theta,z)=(2\cos\theta,2\sin\theta,z), \qquad 0\leq\theta\leq\frac{\pi}{2},\quad 0\leq z\leq3. $$

Indeed, \(x^2+y^2=4\cos^2\theta+4\sin^2\theta=4\), and the stated parameter bounds select the requested heights and quarter of the cylinder. The tangent vectors are

$$ \mathbf{r}_\theta=(-2\sin\theta,2\cos\theta,0), \qquad \mathbf{r}_z=(0,0,1). $$

Their cross product is \(\mathbf{r}_\theta\times\mathbf{r}_z=(2\cos\theta,2\sin\theta,0)\). It is nonzero throughout the domain and points radially outward. The four boundary curves are

$$ \begin{aligned} C_1(t)&=\mathbf{r}(0,t)=(2,0,t), &&0\leq t\leq3,\\ C_2(t)&=\mathbf{r}(t,3)=(2\cos t,2\sin t,3), &&0\leq t\leq\frac{\pi}{2},\\ C_3(t)&=\mathbf{r}\left(\frac{\pi}{2},3-t\right)=(0,2,3-t), &&0\leq t\leq3,\\ C_4(t)&=\mathbf{r}\left(\frac{\pi}{2}-t,0\right) =(2\sin t,2\cos t,0), &&0\leq t\leq\frac{\pi}{2}. \end{aligned} $$

The joins are \((2,0,3)\), \((0,2,3)\), \((0,2,0)\), and \((2,0,0)\), before returning to \((2,0,0)\) at the end of \(C_4\). This patch includes only the curved lateral portion; the top and bottom quarter-disks are not part of its parameter domain.

Worked Example: A Sector of a Sphere

Worked Example: A Sector of a Sphere

Parametrize the part of the sphere of radius \(3\) centered at the origin with polar angle \(\phi\) between \(\pi/3\) and \(2\pi/3\), and azimuthal angle \(\theta\) between \(0\) and \(\pi/2\). Use

$$ \mathbf{r}(\phi,\theta) =(3\sin\phi\cos\theta,3\sin\phi\sin\theta,3\cos\phi), \quad \frac{\pi}{3}\leq\phi\leq\frac{2\pi}{3},\quad 0\leq\theta\leq\frac{\pi}{2}. $$

The point lies on the sphere because

$$ x^2+y^2+z^2 =9\sin^2\phi(\cos^2\theta+\sin^2\theta)+9\cos^2\phi =9. $$

The tangent vectors are

$$ \begin{aligned} \mathbf{r}_\phi&=(3\cos\phi\cos\theta,3\cos\phi\sin\theta,-3\sin\phi),\\ \mathbf{r}_\theta&=(-3\sin\phi\sin\theta,3\sin\phi\cos\theta,0). \end{aligned} $$

Taking them in the stated parameter order gives

$$ \mathbf{r}_\phi\times\mathbf{r}_\theta =(9\sin^2\phi\cos\theta,9\sin^2\phi\sin\theta,9\sin\phi\cos\phi). $$

This equals \(3\sin\phi\,\mathbf{r}(\phi,\theta)\). Since \(\sin\phi>0\) on the domain, the cross product is nonzero and points outward. The boundary curves are the four parameter edges:

$$ \begin{aligned} C_1(t)&=\mathbf{r}\left(\frac{\pi}{3},t\right) =\left(\frac{3\sqrt{3}}{2}\cos t,\frac{3\sqrt{3}}{2}\sin t,\frac32\right), &&0\leq t\leq\frac{\pi}{2},\\ C_2(t)&=\mathbf{r}\left(t,\frac{\pi}{2}\right) =(0,3\sin t,3\cos t), &&\frac{\pi}{3}\leq t\leq\frac{2\pi}{3},\\ C_3(t)&=\mathbf{r}\left(\frac{2\pi}{3},\frac{\pi}{2}-t\right) =\left(\frac{3\sqrt{3}}{2}\sin t,\frac{3\sqrt{3}}{2}\cos t,-\frac32\right), &&0\leq t\leq\frac{\pi}{2},\\ C_4(t)&=\mathbf{r}\left(\frac{2\pi}{3}-t,0\right) =(3\sin(\tfrac{2\pi}{3}-t),0,3\cos(\tfrac{2\pi}{3}-t)), &&0\leq t\leq\frac{\pi}{3}. \end{aligned} $$

The endpoints join in sequence: \(C_1(\pi/2)=C_2(\pi/3)=(0,3\sqrt{3}/2,3/2)\); \(C_2(2\pi/3)=C_3(0)=(0,3\sqrt{3}/2,-3/2)\); \(C_3(\pi/2)=C_4(0)=(3\sqrt{3}/2,0,-3/2)\); and \(C_4(\pi/3)=C_1(0)=(3\sqrt{3}/2,0,3/2)\). The angular intervals stay away from the poles, so neither angular boundary collapses to a point.

A Reliable Construction Sequence

For a specified patch, a short sequence of decisions keeps the geometry and the formulas connected.

1
Identify the geometric data.
Record the shape, its position or axis, and every stated edge, height, or angular limit.
2
Choose parameters with a clear meaning.
For a plane, coefficients along two independent directions are convenient. For a graph, use the two coordinates in its domain. For a cylinder or sphere, angles naturally describe circular directions.
3
State the domain and verify the image.
Write the parameter bounds explicitly, then check that the resulting points satisfy the surface equation and cover the intended portion.
4
Compute tangents, a normal, and boundary curves.
Differentiate with respect to each parameter, form a cross product, and substitute each boundary parameter value to obtain the edge curves.

Common Pitfall: Treating a Coefficient as a Distance

In a plane parametrization \(\mathbf{r}(u,v)=P+u\mathbf{a}+v\mathbf{b}\), the parameters are coefficients of vectors. They equal distances only when the corresponding direction vectors have unit length. For the plane example, the tempting claim “\(u=1\) moves one unit along \(\mathbf{a}\)” is false:

$$ \|1\mathbf{a}\|=\|(2,1,0)\|=\sqrt{2^2+1^2+0^2}=\sqrt{5}\neq1. $$

The correction is to call \(u\) a coefficient along \(\mathbf{a}\). If a parameter \(s\) is meant to be distance traveled in that direction, use the unit vector \(\mathbf{a}/\|\mathbf{a}\|\); then the displacement \(s\mathbf{a}/\|\mathbf{a}\|\) has length \(s\) for \(s\geq0\). Keeping this distinction explicit prevents a geometrically incorrect interpretation of an otherwise valid parametrization.

What a Complete Description Should Contain

A parametrization is more than a coordinate formula. A complete description identifies its parameter domain, explains how its parameters correspond to the geometric features, and supplies tangent and normal data when needed. Boundary curves are obtained from the boundary of the domain, but their endpoints should still be checked: algebraic errors in a restricted parameter substitution can leave apparent gaps or produce incorrect edges.

Regularity also matters. If \(\mathbf{r}_u\times\mathbf{r}_v=\mathbf{0}\), the two parameter directions fail to provide two independent tangent directions at that point. The map may still describe part of a surface, but the chosen parametrization is singular there. In the examples above, each cross product was checked on the entire stated domain, not merely at a single sample point.

Check Your Understanding

For each question, identify the relevant map, domain, or calculation before writing your answer.

  1. In a plane parametrization \(P+u\mathbf{a}+v\mathbf{b}\), what condition on \(\mathbf{a}\) and \(\mathbf{b}\) makes the two parameter directions independent?
  2. For the graph \(\mathbf{r}(u,v)=(u,v,u^2+v)\), what point corresponds to \((u,v)=(1,-1)\)?
  3. Why is \(\mathbf{r}_u\times\mathbf{r}_v\) a normal vector when it is nonzero?
  4. For a cylindrical patch, which parameter values describe the boundary at the top height?
  5. What feature of the sphere example ensures that its angular parametrization does not reach a pole?

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