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Parametrized Surfaces · Tutorial 849 of 999

Choosing Coordinates For A Surface Patch

Choose coordinates that make a surface patch and its parameter domain easier to describe, while checking coverage, duplication, and singular points.

Intermediate 10 min read

What You'll Learn

  • Match parameter choices to planar, circular, and rotational geometry
  • Describe a cylindrical patch with angular and axial coordinates
  • Compare Cartesian and polar parameters for a graph over a disk
  • Select radial and angular parameters for a cone patch
  • Build a surface-of-revolution parametrization from a generating curve
  • Check a parametrization for coverage, duplication, and boundary behavior

Why the Choice of Parameters Matters

A surface can often be described in more than one way. The coordinates you choose affect how simply you can state the patch, its boundary, and the range of parameter values. A good choice follows the geometry: use angular parameters when a surface wraps around an axis, Cartesian parameters when the patch is naturally a graph over a planar region, and a curve parameter when the surface is generated by moving a curve.

Here the goal is to describe patches, not to compute their areas or integrate over them. The central question is: which parameters give a clear, accurate description of exactly the desired part of the surface?

Definition: A parametrization of a surface patch is a map \(\mathbf{X}:D\to\mathbb{R}^3\), where \(D\) is a parameter domain in the plane and \(\mathbf{X}(u,v)\) gives the point on the surface associated with each \((u,v)\in D\). The image \(\mathbf{X}(D)\) is the patch described by the parametrization.

A useful parametrization should cover the intended patch. Ideally, each point of the patch should come from one parameter pair, apart from harmless duplication along a boundary such as the seam at angles \(0\) and \(2\pi\). At interior points, the two parameter directions should also describe two independent directions along the surface; otherwise the parametrization is singular there. Some familiar parametrizations, such as polar coordinates at the origin, necessarily have a singular point.

Match the Parameters to the Geometry

Start by identifying what makes the surface recognizable. A fixed distance from an axis suggests an angular coordinate. A patch described as \(z=f(x,y)\) over a region in the \(xy\)-plane suggests using \(x\) and \(y\) directly. If a curve is rotated around an axis, one parameter can follow the curve while another specifies the rotation angle.

These choices are not competing definitions of different surfaces. They can be alternative descriptions of the same patch. The practical differences are how easy it is to express the domain, how clearly the patch is covered, and whether the chosen parameters introduce duplication or singularities.

Selection principle: Choose parameters whose ranges express the patch’s geometric constraints directly. Then verify that every intended point is attained and that the parameter ranges do not introduce unintended points or repeated coverage.

Worked Example: A Quarter-Cylinder Patch

Worked Example: A Quarter-Cylinder Patch

Describe the part of the cylinder \(x^2+y^2=4\) with \(x\geq 0\), \(y\geq 0\), and \(0\leq z\leq 3\).

Step 1: Identify the geometric constraints. The equation \(x^2+y^2=4\) says that the distance from the \(z\)-axis is \(2\). The nonnegative conditions on \(x\) and \(y\) select the first-quadrant quarter of each circular cross-section. The height ranges from \(0\) to \(3\).

Step 2: Use an angle around the axis and height along it. On a circle of radius \(2\), the coordinates are \(x=2\cos u\) and \(y=2\sin u\). The first-quadrant condition corresponds to \(0\leq u\leq \frac{\pi}{2}\). Let the second parameter \(v\) be the height.

$$ \mathbf{X}(u,v)=(2\cos u,\,2\sin u,\,v), \qquad 0\leq u\leq \frac{\pi}{2},\quad 0\leq v\leq 3. $$

Step 3: Check that the image lies on the required patch. For every parameter pair in the domain,

$$ x^2+y^2 =(2\cos u)^2+(2\sin u)^2 =4(\cos^2 u+\sin^2 u) =4. $$

Also, \(\cos u\geq 0\) and \(\sin u\geq 0\) on the stated angle interval, and \(z=v\) lies between \(0\) and \(3\). Thus the parametrization produces no points outside the requested patch.

Step 4: Check coverage with a point on the patch. The point \((\sqrt{2},\sqrt{2},\frac{3}{2})\) has radius \(2\), lies in the first quadrant, and has height \(\frac{3}{2}\). Its angle is \(u=\frac{\pi}{4}\), so the parametrization gives

$$ \mathbf{X}\left(\frac{\pi}{4},\frac{3}{2}\right) = \left(2\cdot\frac{\sqrt{2}}{2}, 2\cdot\frac{\sqrt{2}}{2}, \frac{3}{2}\right) = \left(\sqrt{2},\sqrt{2},\frac{3}{2}\right). $$

More generally, every point on this quarter-cylinder has a first-quadrant angle \(u\) and a height \(v=z\) in the stated ranges, so it is attained. The two edge angles give the two vertical boundary lines, and \(v=0\) and \(v=3\) give the bottom and top edges.

A Cartesian alternative is to use \(x\) and \(z\), setting \(y=\sqrt{4-x^2}\), with \(0\leq x\leq 2\) and \(0\leq z\leq 3\). This describes the same quarter-cylinder, but the square root and its endpoint behavior make the angular description more directly aligned with the circular geometry.

Worked Example: A Paraboloid Patch Over a Disk

Worked Example: A Paraboloid Patch Over a Disk

Parametrize the portion of \(z=x^2+y^2\) lying above the disk \(x^2+y^2\leq 4\).

Step 1: Use the graph description. Since the surface is given as \(z\) in terms of \(x\) and \(y\), a direct Cartesian choice is

$$ \mathbf{X}(u,v)=(u,v,u^2+v^2), \qquad u^2+v^2\leq 4. $$

For every pair in this disk, the third coordinate is \(u^2+v^2\), so the equation of the surface is satisfied, and the first two coordinates lie over the required disk. Conversely, a point of the specified patch has \(u=x\), \(v=y\), so this parametrization covers it.

Step 2: Choose polar coordinates when the disk’s circular boundary is central. Set \(u=r\cos\theta\) and \(v=r\sin\theta\), where \(0\leq r\leq 2\) and \(0\leq\theta\leq 2\pi\). Substituting into the graph equation gives

$$ \mathbf{X}(r,\theta) = (r\cos\theta,\,r\sin\theta,\,r^2). $$

Step 3: Check the surface equation and the domain. The first two coordinates satisfy

$$ x^2+y^2 =r^2\cos^2\theta+r^2\sin^2\theta =r^2. $$

The third coordinate is \(z=r^2\), so \(z=x^2+y^2\). Since \(0\leq r\leq 2\), the projection satisfies \(x^2+y^2=r^2\leq 4\), as required.

Step 4: Check a point in both descriptions. Take \(x=1\), \(y=\sqrt{3}\). Then \(x^2+y^2=1+3=4\), so the corresponding point on the surface is \((1,\sqrt{3},4)\). In polar coordinates, \(r=2\) and \(\theta=\frac{\pi}{3}\). Substitution gives

$$ \mathbf{X}\left(2,\frac{\pi}{3}\right) = \left(2\cdot\frac12,\, 2\cdot\frac{\sqrt{3}}{2},\, 2^2\right) = (1,\sqrt{3},4). $$

At \(r=0\), every value of \(\theta\) gives the origin. This is a parameter duplication at the center, not a gap in the patch. If distinct parameter pairs at every point are important, the Cartesian description avoids that particular duplication.

This example shows how the patch’s projection can guide the choice. Cartesian parameters make the graph formula especially direct. Polar parameters make the circular boundary and radial distance especially direct. Both are correct; the better choice depends on what needs to be expressed simply.

Worked Example: A Truncated Cone Patch

Worked Example: A Truncated Cone Patch

Describe the part of the cone \(z=\sqrt{x^2+y^2}\) between heights \(1\) and \(3\), including a full rotation around the \(z\)-axis.

Step 1: Relate height to radial distance. Write \(r=\sqrt{x^2+y^2}\). On this cone, \(z=r\). The height restriction therefore gives \(1\leq r\leq 3\). A full rotation is represented by \(0\leq\theta\leq 2\pi\).

Step 2: Use radial and angular parameters. Since \(x=r\cos\theta\), \(y=r\sin\theta\), and \(z=r\), take

$$ \mathbf{X}(r,\theta) =(r\cos\theta,\,r\sin\theta,\,r), \qquad 1\leq r\leq 3,\quad 0\leq\theta\leq 2\pi. $$

Step 3: Verify that the parametrization satisfies the cone equation.

$$ \sqrt{x^2+y^2} = \sqrt{r^2\cos^2\theta+r^2\sin^2\theta} = \sqrt{r^2} = r = z. $$

The equality \(\sqrt{r^2}=r\) holds because the parameter range requires \(r\geq 1\), so \(r\) is positive. The bounds on \(r\) also ensure \(1\leq z\leq 3\). Thus every generated point lies in the requested patch.

Step 4: Verify a sample point. At \(r=2\) and \(\theta=\frac{\pi}{3}\),

$$ \mathbf{X}\left(2,\frac{\pi}{3}\right) = \left(2\cdot\frac12,\, 2\cdot\frac{\sqrt{3}}{2},\, 2\right) = (1,\sqrt{3},2). $$

Checking directly, \(\sqrt{1^2+(\sqrt{3})^2}=\sqrt{4}=2\), which equals the point’s \(z\)-coordinate. Conversely, a point on the specified cone has \(r=z\in[1,3]\) and an angle around the axis, so these parameter ranges cover the whole patch. The angle endpoints describe the same seam.

The restriction to positive \(r\) is important here. It lets \(z=r\) express the cone’s height and ensures the square-root check above is valid. For a patch that includes the apex, \(r=0\) would be included; the apex would then be represented for every angle, just as the center of a polar disk is.

Worked Example: Choosing a Curve Parameter for a Surface of Revolution

Worked Example: Choosing a Curve Parameter for a Surface of Revolution

Rotate the curve \((\rho,z)=(2+t,t^2)\), for \(0\leq t\leq 1\), around the \(z\)-axis. Parametrize the resulting surface patch.

Step 1: Interpret the generating curve. At each value of \(t\), the curve has distance \(\rho=2+t\) from the axis and height \(z=t^2\). Rotating that point around the axis gives a circle of radius \(2+t\) at height \(t^2\).

Step 2: Add an angle parameter. The angle \(\theta\) selects a point on each circle. A parametrization is

$$ \mathbf{X}(t,\theta) = \bigl((2+t)\cos\theta,\, (2+t)\sin\theta,\, t^2\bigr), \qquad 0\leq t\leq 1,\quad 0\leq\theta\leq 2\pi. $$

Step 3: Verify the generating-curve constraints. For any parameter pair, the distance from the \(z\)-axis is

$$ \sqrt{x^2+y^2} = \sqrt{(2+t)^2\cos^2\theta+(2+t)^2\sin^2\theta} = 2+t, $$

because \(2+t>0\). The height is \(z=t^2\). Thus each point lies on the circle of radius \(2+t\) and height \(t^2\) specified by the generating curve.

Step 4: Check a sample point and recover its parameters. For \(t=\frac12\) and \(\theta=\frac{\pi}{3}\), the radius is \(2+\frac12=\frac52\), and the height is \(\frac14\). Hence

$$ \mathbf{X}\left(\frac12,\frac{\pi}{3}\right) = \left(\frac52\cdot\frac12,\, \frac52\cdot\frac{\sqrt{3}}{2},\, \frac14\right) = \left(\frac54,\frac{5\sqrt{3}}{4},\frac14\right). $$

For this point, \(\sqrt{x^2+y^2}=\sqrt{\frac{25}{16}+\frac{75}{16}}=\sqrt{\frac{100}{16}}=\frac52\). Therefore \(t=\frac52-2=\frac12\), and \(z=t^2=\frac14\), confirming the curve and parameter values. The angle is \(\frac{\pi}{3}\), as seen from the ratio \(y/x=\sqrt{3}\) in the first quadrant. Since \(2+t\) determines \(t\) uniquely on this patch, and the angle specifies a point on its circle, the parametrization covers the generated patch, with the usual angle seam duplicated.

Why the Coordinate Formulas Work

The angular and curve-based choices follow directly from the geometry of rotation. At radius \(\rho\) from the \(z\)-axis, a point with angle \(\theta\) has horizontal coordinates \((\rho\cos\theta,\rho\sin\theta)\). This is because these coordinates have squared distance \(\rho^2(\cos^2\theta+\sin^2\theta)=\rho^2\) from the axis, and their direction is specified by \(\theta\).

For a surface of revolution, let the generating curve be given by a radius \(\rho(t)\geq 0\) and height \(z(t)\). Applying the circle formula at each curve point gives the general parametrization

$$ \mathbf{X}(t,\theta) = \bigl(\rho(t)\cos\theta,\, \rho(t)\sin\theta,\, z(t)\bigr). $$

Its correctness can be checked directly: \(x^2+y^2=\rho(t)^2\) and the height is \(z(t)\), precisely the radius-height data of the generating curve. The parameter \(t\) need not be arc length; it is simply a parameter that runs through the chosen curve. The formula assumes the curve is being revolved about the \(z\)-axis. A different axis requires the rotation coordinates to be adjusted accordingly.

A Common Parameter-Domain Mistake

A correct coordinate formula can still describe the wrong patch if its parameter domain is too small. For example, the formula \((2\cos u,2\sin u,v)\) does lie on the cylinder of radius \(2\). But if one chooses only \(0\leq u\leq\frac{\pi}{2}\), it describes the first-quadrant quarter, not the full cylinder.

Common mistake: Treating a formula that satisfies the surface equation as proof that the entire intended patch is covered. The equation verifies where the parametrized points lie; the parameter ranges determine which of those points are actually reached.

To describe the full cylinder between heights \(0\) and \(3\), use the same formula with \(0\leq u\leq 2\pi\) and \(0\leq v\leq 3\). For instance, the point \((-2,0,1)\) requires \(u=\pi\) and \(v=1\). It cannot be produced when \(u\) is restricted to the first quadrant. Always check coverage against the intended boundary and test points from different parts of the patch.

A Practical Choice Checklist

  • For a graph over a planar region: use the two planar coordinates as parameters when the projection region is easy to state.
  • For circular or rotational geometry: use an angle, often together with a radius or height.
  • For a surface generated by a curve: let one parameter run along the curve and another specify the rotation or other motion.
  • For every choice: state the parameter domain, verify the surface constraints, and check that the image covers the intended boundaries and interior.
  • Check for duplication or singular points: full-turn angle intervals repeat a seam, and angular coordinates collapse at zero radius.

Choosing coordinates is not merely a notational preference. It is part of describing the patch accurately. A well-matched choice makes the geometry visible in the formula and makes the parameter domain easier to verify.

Check Your Understanding

For each question, consider both the parameter formula and its domain.

  1. For the cylinder \(x^2+y^2=9\), what angular and axial parameters describe the patch with \(y\leq 0\) and \(1\leq z\leq 4\)?
  2. For \(z=x^2+y^2\) above the disk \(x^2+y^2\leq 1\), state a Cartesian parametrization and a polar parametrization.
  3. Why does using \(0\leq\theta\leq\frac{\pi}{2}\) fail to cover a full circular cylinder?
  4. Write the surface-of-revolution parametrization for a generating curve \((\rho(t),z(t))\) about the \(z\)-axis.
  5. What point is duplicated when polar parameters include \(r=0\) and a full range of angles?

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