18 18 Vector Forms of Green’s Theorem19 19 Vector Forms of Green’s Theorem The curl and divergence operators allow us to rewrite Green’s Theorem in versions that will be useful in our later work. We suppose that the plane region D, its boundary curve C, and the functions P and Q satisfy the hypotheses of Green’s Theorem. Then we consider the vector field F = P i + Q j.20 20 Vector Forms of Green’s Theorem Its line integral is and, regarding F as a vector field on with third component 0, we have21 21 Vector Forms of Green’s Theorem Therefore and we can now rewrite the equation in Green’s Theorem in the vector form22 22 Vector Forms of Green’s Theorem Equation 12 expresses the line integral of the tangential component of F along C as the double integral of the vertical component of curl F over the region D enclosed by C. We now derive a similar formula involving the normal component of F. If C is given by the vector equation r(t) = x(t) i + y(t) j a ≤ t ≤ b then the unit tangent vector is23 23 Vector Forms of Green’s Theorem The outward unit normal vector to C is given by (See Figure 2.) Figure 224 24 Vector Forms of Green’s Theorem Then from the equation we have25 25 Vector Forms of Green’s Theorem by Green’s Theorem. But the integrand in this double integral is just the divergence of F.26 26 Vector Forms of Green’s Theorem So we have a second vector form of Green’s Theorem. This version says that the line integral of the normal component of F along C is equal to the double integral of the divergence of F over the region D enclosed by C.27 27 Copyright © Cengage Learning. All rights reserved. 16.6 Parametric Surfaces and Their Areas28 28 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.29 29 Parametric Surfaces30 30 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.31 31 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.32 32 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 surface33 33 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 u34 34 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’d35 35 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’d36 36 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.37 37 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 438 38 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.39 39 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?40 40 Example 2 – Solution • We graph the portion of the surface with parameter domain 0 ≤ u ≤ 4π, 0 ≤ v ≤ 2π in Figure 5. Figure 541 41 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’d42 42 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’d43 43 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.44 44 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 …
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