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Section 7.1 Parametric Surfaces

Subsection 7.1.1 Motivation

For the last few sections, I've focused heavily on the force example of fields. Now, let me return to the idea of a fluid flow. Conisder a field \(F\) that describe the flow of water or air currents. If I have a net or other porus object in the field, then unless the flow is perfectly across the edge of the object, some amount of the medium will be flowing through the net. Is there a way to measure such a thing?

This seems like a difficult project. The net can have a complicated geometry. At different points on the net, the interaction with the fluid can vary. If could be flowing through one part of the net in one direction, but in the opposite direciton through another part. I'd like a way to put all this together into some kind of collective sense of the flow through the net.

I'll look back at line integrals for motivation. For line integrals, I looked at the interaction of a parametric curve and a field. To do that, I needed the tangent to the curve, so that I could take the dot product to measure the interaction. That tangent came from the parametrization.

I'm going to need something similar to describe the net -- I need a two dimensional analogue of a parametric curve. Such a thing is called a parametric surface (with a formal definition to follow below. For this section, I'll be defining parametric surfaces and giving some examples. Section 8.1 will then use parametric surfaces to try to answer the motivation question of flow through a net.

Subsection 7.1.2 Definitions

Definition 7.1.1.

Let \(D \subset \RR^2\) be a simply connected set. A parametric surface in \(\RR^n\) is a continuous function

\begin{equation*} \sigma: D \rightarrow \RR^n \end{equation*}

Even though the definition is given for any \(\RR^n\text{,}\) for the purpose of this course, I'll assume that all surface are in \(\RR^3\text{.}\)

Like parametric curves, we visualize surfaces only by their outputs. The input set \(D\) isn't visualized; rather, it is thought of as the parameter space. The difference, compared to curves, is that we now have a 2-dimensional parameter space. This allows two independent directions of movement on the surface. I lose the notion of travelling in time with the parameter \(t\text{.}\) Instead, there are two parameters (with conventional variable \(u\) and \(v\)). I can move independently in either variable. Between moving in either parameter, I can range over all the points in the output.

Again similar to a curve, each coordinate is a scalar function of the parameter.

\begin{equation*} \sigma(u,v) = \left( \sigma_1(u,v), \sigma_2(u,v), \sigma_3(u,v) \right) \end{equation*}

Typically, parametric surfaces are use to give a parametrization to an already known 2-dimensional object: planes, a hollow spheres, cones, cylinders, etc. Let me work through some examples of this.

In Calculus III, I analyzed graphs of scalar fields. In particular, for a scalar field on two variables \(f(x,y)\text{,}\) the graph is a surface in \(\RR^3\text{.}\) In Calculus III, I used gradients to interpret these surfaces as altitudes; contour plots were the equivalent of topographical maps. Now, I can also treat these as parametric surfaces.

Assume \(f(x,y)\) is a scalar field on a simply-connected domain \(D\text{.}\) Then using the same \(D\) as the parameter domain, the surface \(\sigma (u,v) = (u,v,f(u,v))\) is a parametric description of the graph of the function.

In calculating volumes in Calculus II, I introduced the idea of a surface of revolution. Such a surface is formed by taking the graph of a single variable function \(y = f(x)\text{,}\) and spinning it around the \(x\) axis to create a surface in \(RR^3\text{.}\) Such surfaces always have circular cross-section, resembling something that might be formed on a lathe. I can also describe them as parametric surfaces

The axis of the surface of revolution is unchanges. The other two axes have rotation; parametrically, that is given by sine and cosine. Let \(f(x)\) be a continuous function on \([a,b]\text{.}\) Let \(D = [a,b] \times [0,2\pi]\text{.}\) Then the surface of revolution is describe as a parametric surface on the parameter domain \(D\text{.}\)

\begin{equation*} \sigma(u,v) = (u, f(u) \cos v, f(u) \sin v) \end{equation*}

I don't have to use \(u\) and \(v\) as the variable of the paremtrization. If I want to be reminded that one of the parameter is the angle of a rotation, I'll use \(\theta\text{.}\) If one of the parameter perfectly aligns with the \(x\text{,}\) \(y\) or \(z\) axis in on the surface, it's reasonable to call that the parameter. In that was, a surface of revolution about the \(x\) axis on the parameter domain \(D = [a,b] \times [2,\pi]\) can be described this way.

\begin{equation*} \sigma(x,\theta) = (x, f(x) \cos \theta, f(x) \sin \theta) \end{equation*}

The sphere is a useful and important shape to have a parametrization. The sphere has a fixed radisu \(R\text{.}\) Location on the sphere can be described, then, by the same logitude and colatitude used for spherical coordnations. This should be the parameter domain: \(D = [0, 2\pi] \times [0, \pi]\text{.}\) Then the sphere is described as a parametric surface in the following way.

\begin{equation*} \sigma (\theta, \phi) = (R \sin \phi \cos \theta, R \sin \phi \sin \theta, R \cos \phi) \end{equation*}

Very similar to the sphere, I can copy the setup of cylindrical coordinates to give a parametric description of a cylinder of length \(h\) and radiius \(R\text{.}\) The parameter domain is \(D = [0, 2\pi] \times [0,h]\text{.}\) I use \(\theta\) as one parameter for the angle, and \(z\) as the other parameter, since it is mapped direclty to the normal \(z\) coordinate.

\begin{equation*} \sigma (\theta, z) = (R \cos \theta, R \sin \theta, z) \end{equation*}

I can adjust the cylinder slightly to make a parametric decription of the cone of base radius \(R\) and height \(h\text{.}\) The cylinder has fixed radius, but the radius of the cone decreases linearly as the height increase. On the same parameter domain, \(D = [0, 2\pi] \times [0,h] \text{,}\) the cone is described by this parametric surface.

\begin{equation*} \sigma(\theta, z) = \left( \frac {R(h-z)}{h} \cos \theta, \frac{R(h-z)}{h} \sin \theta, z \right) \end{equation*}

Subsection 7.1.3 Calculus of Parametric Surfaces

The calculus of parametric curves started with the tangent vector. All the other constructions (arclength, curvature, torsion) were based on that starting point. For parametric surface, there isn't a unique tangent, since there isn't a unique parameter. Like the graphs of multivariable functions, there is a higher-dimensional tangent object. Since I am restricting my attention to surface in \(\RR^3\) in this course, instead of a tangent line, a surface (if it is differentiable) has a tangent plane. In \(\RR^3\text{,}\) a plane is describe by a normal. Therefore, the important piece of information for a parametric surface is the normal to the tangent plane. Like the tangent vector for a parametric curve, the normal will be the key to the calculus of parametric curve. Let me move on the definition.

Definition 7.1.7.

A parametric surface \(\sigma: D \rightarrow \RR^3\) is called a \(C^1\) surface if all the partial derivtive of both components of \(\sigma\) exist and are continuous. For a \(C^1\) parametric surface \(\sigma\text{,}\) write \(\sigma_u\) for the vector of partial \(\left( \frac{\del \sigma_1}{\del u}, \frac{\del \sigma_2}{\del u}, \frac{\del \sigma_3}{\del u} \right)\) and likewise for \(\sigma_v\) with partials in the variable \(v\text{.}\) For such a surface, the normal vector is the vector \(\sigma_u \times \sigma_v\text{.}\) A parametric surface is called non-singular if its normal is never the zero vector.

Let me make a couple of points aobut this definition. First, because I'm using the cross product, this definition only works in \(\RR^3\text{.}\) There are notions of tangents and normals to parametric surfaces in higher dimensions, but more machinery is needed to define these tangents and normals. To do so moves in to the study of differentiable manifolds. Second, I can think of the \(C^1\) condition saying that the parametric curve has no sharp corners (and no singularities of the parametrization, which I'll discuss in examples later). A natural term for having no sharp concers would be ‘smooth’. Unfortunately, this has a slightly different definition. A smooth parametric surface (alternatively a \(C^\infty\) parametric surface) is a parametric surface where all higher partial derivatives, of any degree, exist and where the normal is never zero. For the purpose of this course, most of the surface I introduce will be smooth; I could use this terminology if I wanted, saying ‘smooht’ instead of ‘\(C^1\) and non-singular’. However, only \(C^1\) is required for the definition of the normal.

I'll calculate the normals for the five examples I use above.

Here is the graph of a scalar field in two dimensions on a simply connected domain \(D\text{.}\)

\begin{equation*} \sigma (u,v) = (u,v,f(u,v)) \end{equation*}

I'll calculate the partial derivative and take the cross product to calculate the normal.

\begin{align*} \sigma_u \amp = \left( 1, 0, \frac{\del f}{\del u} \right) \\ \sigma_v \amp = \left( 0, 1, \frac{\del f}{\del v} \right) \\ \sigma_u \times \sigma_v \amp = \left( -\frac{\del f}{\del u}, -\frac{\del f}{\del v} , 1 \right) \end{align*}

This is precisely the same result found in Calculus III for the normal to the tangent plane. The construcitons of the tangent options are the same, but now described by the parameters \(u\) and \(v\text{.}\)

Here is the surface of revolution about the \(x\) axis for \(f: [a,b] \rightarrow \RR\text{.}\)

\begin{equation*} \sigma (x,\theta) = (x, f(x) \cos \theta, f(x), \sin \theta) \end{equation*}

I'll calculate the partial derivative and take the cross product to calculate the normal.

\begin{align*} \sigma_x \amp = \left(1, f^\prime(x) \cos \theta, f^\prime(x) \sin \theta \right) \\ \sigma_{\theta} \amp = \left( 0, -f(x) \sin \theta, f(x) \cos \theta \right) \\ \sigma_x \times \sigma_{\theta} \amp = \left( f(x) f^\prime(x), -f(x) \cos \theta, -f(x) \sin \theta \right) \end{align*}

The normal still has the same circular symmetry, as would be expected for this shape.

Here is the parametric description of the sphere.

\begin{equation*} \sigma (\theta, \phi) = (R \sin \phi \cos \theta, R \sin \phi \sin \theta, R \cos \phi) \end{equation*}

I'll calculate the partial derivative and take the cross product to calculate the normal.

\begin{align*} \sigma_{\theta} \amp = \left(-R \sin \phi \sin \theta, R \sin \phi \cos \theta, 0 \right) \\ \sigma_{\phi} \amp = \left( R \cos \phi \cos \theta, R \cos \phi \sin \theta, -R \sin \phi \right) \\ \sigma_{\theta} \times \sigma_{\phi} \amp = \left( -R^2 \sin^2 \phi \cos \theta, -R^2 \sin^2 \phi \sin \theta, -R^2 \sin \phi \cos \phi \right)\\ \amp = -R \sin \phi \left( R \sin \phi \cos \theta, R \sin \phi \sin \theta, R \cos \phi) \right) \end{align*}

I factored out the common term in the normal. After factoring out this, I see that the resulting normal is precisely same as the original position. So, the normal is a multiple of the position. This makes sense: for a sphere centred at the origin, the vector that points out of the sphere at point (as a local direction vector) has the same direction as the vector from the origin to that point.

The normal is \((0,0,0)\) when \(\phi = 0\text{,}\) which happens at the top and bottom poles of the sphere. These points are singularities of the parametrization. This was the same problem for spherical coordinates: the longitude is not propely defined at the north and south poles. The sphere itself is a surface without sharp corners, but the system of parametrization is the problem. It breaks down at the poles.

Here is the parametric description of the cylinder.

\begin{equation*} \sigma (\theta,z) = (R \cos \theta, R \sin \theta, z) \end{equation*}

I'll calculate the partial derivative and take the cross product to calculate the normal.

\begin{align*} \sigma_{\theta} \amp = \left(-R \sin \theta, R \cos \theta, 0 \right) \\ \sigma_z \amp = \left(0, 0, 1 \right) \\ \sigma_{\theta} \times \sigma_z \amp = ( R \cos \theta, R \sin \theta, 0) \end{align*}

The normals to the cylinder point outwards from the cylinder, with no vertical components.

Here is the parametric description of the cone.

\begin{equation*} \sigma (\theta,z) = \left( \frac{R(h-z)}{h} \cos \theta, \frac{R(h-z)}{h} \sin \theta z \right) \end{equation*}

I'll calculate the partial derivative and take the cross product to calculate the normal.

\begin{align*} \sigma_{\theta} \amp = \left( -\frac{R(h-z)}{h} \sin \theta, \frac{R(h-z)}{h} \cos \theta, z \right) \\ \sigma_z \amp = \left( \frac{-R}{h} \cos \theta, \frac{-R}{h} \sin \theta, 1 \right) \\ \sigma_{\theta} \times \sigma_z \amp = \left( \frac{R(h-z)}{h} \cos \theta + \frac{Rz}{h} \sin \theta, \frac{-Rz}{h} \cos \theta + \frac{R(h-z)}{h} \sin \theta, \frac{R^2(h-z)}{h^2} \right) \end{align*}

This is a complicated normal. It does have a positive vertical component, which is expected. Unlike the cylinder, the vector point out from the cylinder should point updwards.