Math 2415 – Calculus III Section 16.6 Parametric Surfaces and Their Areas ”I wish I had paid more attention in calculus class when we were studying parametric surfaces. It sure would have helped me today.” • We can describe a surface by a vector function~r(u;v) = x(u;v)iˆ+ y(u;v) jˆ+ z(u;v)kˆ 3 • The set of all points (x;y;z) in R 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. Ex: Identify and sketch the surface with vector equation~r(u;v) = 2cosuiˆ+ v jˆ+ 2sinukˆ. • We can get different surfaces if we restrict u and v. p Ex: In previous example, let 0 ≤ u ≤ and 0 ≤ v ≤ 3. This gives 2 • There are two useful families of curves that lie on S, one with u constant, the other with v constant. Math 2415 Section 16.6 Continued • If u = u0 is constant, ~r(u0;v) is a vector function of a single parameter v and defines curve C1 lying on S. Likewise, if v = v0. These are called grid curves. (Computers graph like this). • It may be difficult to find a vector function to represent a surface. Ex: Find a vector function that represents the plane passing through the point P0 with position vector ~r0 and contains two nonparallel vectors ~a and~b. Ex: Find a parametric representation of the sphere x2 + y2 + z2 = a2. Ex: Find a parametric representation for the cylinder x2 + y2 = 4;0 ≤ z ≤ 1. 2 Math 2415 Section 16.6 Continued • In general, a surface given as a graph of a function x and y (z = f (x;y)) can be regarded as a parametric surface with equations x = x;y = y;z = f (x;y). Parameterizations are not unique. p Ex: Find a parametric representation for z = 2 x2 + y2, i.e. the top half of the cone z2 = 4x2 + 4y2. Surfaces of Revolution Can be represented parametrically. • Consider the surface S obtained by rotating y = f (x);a ≤ x ≤ b where f (x) ≥ 0 about the x−axis. Let q be the angle of rotation. Ex: Find parametric equations for the surface generated by rotating the curve y = sinx;0 ≤ x ≤ 2p about the x−axis. Graph the surface of revolution. 3 Math 2415 Section 16.6 Continued Tangent Planes Let~r(u;v) = x(u;v)iˆ+ y(u;v) jˆ+ z(u;v)kˆ be a vector function at point P0 with position vector ~r(u0;v0). Keep u constant by letting u = u0: ~r(u0;v) is a vector function of a single parameter v and defines a grid curve C1 lying on S. ¶x ¶y ¶z • Tangent vector to C at P : ~r = (u ;v )iˆ+ (u ;v ) jˆ+ (u ;v )kˆ 1 0 v ¶v 0 0 ¶v 0 0 ¶v 0 0 • Let v = v0. We get a grid curve C2 given by~r(u;v) that lies on S and its tangent vector at P0 ¶x ¶y ¶z is ~r = (u ;v )iˆ+ (u ;v ) jˆ+ (u ;v )kˆ. u ¶u 0 0 ¶u 0 0 ¶u 0 0 • If a surface is smooth, the tangent plane exists at all points. Since ~ru ×~rv is the normal vector, if ~ru ×~rv =~0, there can be no tangent plane at that point. Thus the surface is not smooth at that point. (if ~ru ×~rv 6=~0, the surface is smooth.) Then write the equation using the normal vector and a point. Ex: Find the tangent plane to the surface with parametric equations x = u2;y = v2;z = u + 2v at the point (1;1;3). 4 Math 2415 Section 16.6 Continued s 0 s 1 ZZ ¶z 2 ¶z 2 Z b dy2 • Surface area: A(S) = 1 + + dA. @Very similar to arc length L = 1 + dxA D ¶x ¶y a dx Ex: Find the area of the part of the paraboloid z = x2 + y2 that lies under the plane z = 9: 5.