Copyright © Cengage Learning. All rights reserved. 16.6 Parametric Surfaces and Their Areas3 Parametric Surfaces and Their Areas Here we use vector functions to describe more general surfaces, called parametric surfaces, and compute their areas. Then we take the general surface area formula and see how it applies to special surfaces.4 Parametric Surfaces5 Parametric Surfaces In much the same way that we describe a space curve by a vector function r(t) of a single parameter t, we can describe a surface by a vector function r(u, v) of two parameters u and v. We suppose that r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k is a vector-valued function defined on a region D in the uv-plane.6 Parametric Surfaces So x, y, and, z, the component functions of r, are functions of the two variables u and v with domain D. The set of all points (x, y, z) in such that x = x(u, v) y = y(u, v) z = z(u, v) and (u, v) varies throughout D, is called a parametric surface S and Equations 2 are called parametric equations of S.7 Parametric Surfaces Each choice of u and v gives a point on S; by making all choices, we get all of S. In other words, the surface is traced out by the tip of the position vector r(u, v) as (u, v) moves throughout the region D. (See Figure 1.) Figure 1 A parametric surface8 Example 1 Identify and sketch the surface with vector equation r(u, v) = 2 cos u i + v j + 2 sin u k Solution: The parametric equations for this surface are x = 2 cos u y = v z = 2 sin u9 Example 1 – Solution So for any point (x, y, z) on the surface, we have x2 + z2 = 4 cos2u + 4 sin2u = 4 This means that vertical cross-sections parallel to the xz-plane (that is, with y constant) are all circles with radius 2. cont’d10 Example 1 – Solution Since y = v and no restriction is placed on v, the surface is a circular cylinder with radius 2 whose axis is the y-axis (see Figure 2). Figure 2 cont’d11 Parametric Surfaces If a parametric surface S is given by a vector function r(u, v), then there are two useful families of curves that lie on S, one family with u constant and the other with v constant. These families correspond to vertical and horizontal lines in the uv-plane.12 Parametric Surfaces If we keep u constant by putting u = u0, then r(u0, v) becomes a vector function of the single parameter v and defines a curve C1 lying on S. (See Figure 4.) Figure 413 Parametric Surfaces Similarly, if we keep v constant by putting v = v0, we get a curve C2 given by r(u, v0) that lies on S. We call these curves grid curves. (In Example 1, for instance, the grid curves obtained by letting u be constant are horizontal lines whereas the grid curves with v constant are circles.) In fact, when a computer graphs a parametric surface, it usually depicts the surface by plotting these grid curves, as we see in the next example.14 Example 2 Use a computer algebra system to graph the surface r(u, v) = 〈(2 + sin v) cos u, (2 + sin v) sin u, u + cos v〉 Which grid curves have u constant? Which have v constant?15 Example 2 – Solution We graph the portion of the surface with parameter domain 0 ≤ u ≤ 4π, 0 ≤ v ≤ 2π in Figure 5. Figure 516 Example 2 – Solution It has the appearance of a spiral tube. To identify the grid curves, we write the corresponding parametric equations: x = (2 + sin v) cos u y = (2 + sin v) sin u z = u + cos v If v is constant, then sin v and cos v are constant, so the parametric equations resemble those of the helix. Thus the grid curves with v constant are the spiral curves in Figure 5. cont’d17 Example 2 – Solution We deduce that the grid curves with u constant must be curves that look like circles in the figure. Further evidence for this assertion is that if u is kept constant, u = u0 , then the equation z = u0 + cos v shows that the z-values vary from u0 – 1 to u0 + 1. cont’d18 Example 4 Find a parametric representation of the sphere x2 + y2 + z2 = a2 Solution: The sphere has a simple representation ρ = a in spherical coordinates, so let’s choose the angles φ and θ in spherical coordinates as the parameters.19 Example 4 – Solution Then, putting ρ = a in the equations for conversion from spherical to rectangular coordinates, we obtain x = a sin φ cos θ y = a sin φ sin θ z = a cos φ as the parametric equations of the sphere. The corresponding vector equation is r(φ, θ ) = a sin φ cos θ i + a sin φ sin θ j + a cos φ k cont’d20 Example 4 – Solution We have 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π, so the parameter domain is the rectangle D = [0, π] × [0, 2π]. The grid curves with φ constant are the circles of constant latitude (including the equator). cont’d21 Example 4 – Solution The grid curves with θ constant are the meridians (semi-circles), which connect the north and south poles (see Figure 7). Figure 7 cont’d22 Parametric Surfaces Note: We saw in Example 4 that the grid curves for a sphere are curves of constant latitude and longitude. For a general parametric surface we are really making a map and the grid curves are similar to lines of latitude and longitude. Describing a point on a parametric surface (like the one in Figure 5) by giving specific values of u and v is like giving the latitude and longitude of a point.23 Surfaces of Revolution24 Surfaces of Revolution Surfaces of revolution can be represented parametrically and thus graphed using a computer. For instance, let’s consider the surface S obtained by rotating the curve y = f (x), a ≤ x ≤ b, about the x-axis, where f (x) ≥ 0. Let θ be the angle of rotation as shown in Figure 10. Figure 1025 Surfaces of Revolution If (x, y, z) is a point on S, then x = x y = f (x) cos θ z = f (x) sin θ Therefore we take x and θ as parameters and regard Equations 3 as parametric equations of S. The parameter domain is given by a ≤ x ≤ b, 0 ≤ θ ≤ 2π .26 Example 8 Find parametric equations for the surface generated by rotating the curve y = sin x, 0 ≤ x ≤ 2π, about the x-axis. Use these equations to graph the surface of revolution. Solution: From Equations 3, the parametric equations are x = x y = sin x cos θ z = sin x …
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