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Section 14.6: Parametric Surfaces and Their Areas ~ A space curve can be described by a vector function R(t) of one parameter. Similarly, a ~ surface can be described by a vector function R(u, v) of two parameters. Definition: A surface S described by a vector function ~ R(u, v) = hx(u, v), y(u, v), z(u, v)i (u, v) ∈ D is called a parametric surface. Example: Identify the surface defined by ~ R(u, v) = hu cos(v), u sin(v), ui. Note that the parametric equations satisfy z 2 = x2 + y 2 or z = described by this vector function is a cone. p x2 + y 2 . The surface Example: Find a parametric representation of the cylinder x2 + y 2 = 9, 0 ≤ z ≤ 5. The cylinder has a simple representation of r = 3 in cylindrical coordinates. This cylinder can be parameterized by ~ z) = h3 cos θ, 3 sin θ, zi R(θ, for 0 ≤ θ ≤ 2π and 0 ≤ z ≤ 5. Example: Find a parametric representation of the part of the sphere x2 + y 2 + z 2 = 36 that p lies above the cone z = x2 + y 2 . The sphere has a simple representation of ρ = 6 in spherical coordinates. Moreover, the π equation of the cone in spherical coordinates is φ = . This part of the sphere can be 4 parameterized by ~ φ) = h6 sin φ cos θ, 6 sin φ sin θ, 6 cos φi R(θ, π for 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ . 4 ~ Example: Find an equation of the tangent plane to the parametric surface given by R(u, v) = 2 hu + v, 3u , u − vi at the point (2, 3, 0). The tangent vectors are ~ u = h1, 6u, 1i, R ~ v = h1, 0, −1i. R Thus, the normal vector is ~i ~j ~k ~u × R ~ v = 1 6u 1 R 1 0 −1 = h−6u, 2, −6ui. Note that the point (2, 3, 0) corresponds to the parameter values u = v = 1. So the normal vector at (2, 3, 0) is ~ = h−6, 2, −6i. N Therefore, the tangent plane is −6(x − 2) + 2(y − 3) − 6(z − 0) = 0 −6x + 2y − 6z = −6 3x − y + 3z = 3. Definition: If a smooth parametric surface S is given by the equation ~ R(u, v) = hx(u, v), y(u, v), z(u, v)i (x, y) ∈ D and S is covered once as (u, v) ranges throughout D, then the surface area of S is ZZ ~u × R ~ v ||dA. A= ||R D ~ v) = huv, u + v, u − vi, for Example: Find the area of the parametric surface given by R(u, 2 2 u + v ≤ 1. The tangent vectors are ~ u = hv, 1, 1i, R ~ v = hu, 1, −1i. R The cross product of these vectors is ~i ~u × R ~v = v R u ~j ~k 1 1 1 −1 = h−2, u + v, v − ui. Thus ~u × R ~ v || = ||R p 4 + (u + v)2 + (v − u)2 = √ 4 + 2u2 + 2v 2 = Therefore, the area of this surface is Z 2π Z A = 0 0 1 1 √ r 2r2 + 4drdθ √ r 2r2 + 4dr 0 1 π2 2 3/2 = (2r + 4) 23 0 √ π √ (6 6 − 2 2). = 3 Z = 2π √ 2r2 + 4.