Math 251-copyright Joe Kahlig, 18B Page 1 Section 16.6: Parametric Surfaces and Their Areas A space curve is parametrized by the vector function r(t) = hx(t); y(t); z(t)i. A surface, z = f(x; y), is parametrized by a vector function of two variables. r(u; v) = hx(u; v); y(u; v); z(u; v)i with (u; v) in region D. Useful Parametrizations • Surface given by z = f(x; y). Example: z = x2 + 4y2 Example: x = z2 + y2 • Surface in cylindrical coordinates Example: x2 + y2 = 9 for 0 ≤ z ≤ 2 • Surface in spherical coordinates Example: x2 + y2 + z2 = 4 Example: z = px2 + y2 Math 251-copyright Joe Kahlig, 18B Page 2 • Surface of rotation around an axis. Assume f(x) ≥ 0 and rotated around x-axis. Example: Identify the surface with the given vector equation. r(u; v) = hu + v; 4 − v; 3 + 4u + 6vi Example: Find the tangent plane to the surface with parametric equations given below at the point (1; 4; 5). r(u; v) = u3; v2; u + 2v Math 251-copyright Joe Kahlig, 18B Page 3 Note: In the special case the surface is defined by z = f(x; y) and is parametrized by r(x; y) = hx; y; f(x; y)i. Then a normal vector is Example: Find a normal vector for the surface defined as x = f(y; z) Definition: If a smooth parametric surface S is given by the equation r(u; v) and S is covered just once as (u; v) ranges throughout the parametric domain D, then the surface area of S is ZZ A(S) = jru × rvj dA D Example: Find the surface area for the surface given by x = uv, y = u + v, and z = u − v where u2 + v2 ≤ 1 Math 251-copyright Joe Kahlig, 18B Page 4 Example: Find the surface area for the part of the plane 2x+2y+z = 8 inside the cylinder x2 +y2 = 9. Math 251-copyright Joe Kahlig, 18B Page 5 p Example: Find the surface area of the sphere x2 +y2 +z2 = 16 between the planes z = 2 and z = 2 3. x = 4 sin φ cos θ y = 4 sin φ sin θ z = 4 cos φ π π where 0 ≤ θ ≤ 2π and 6 ≤ φ ≤ 3 i j k rφ × rθ = 4 cos φ cos θ 4 cos φ sin θ −4 sin φ −4 sin φ sin θ 4 sin φ cos θ 0 2 2 2 2 rφ × rθ = 16 sin φ cos θ; −16 sin φ sin θ; 16 sin φ cos φ cos θ + 16 sin φ cos φ sin θ 2 2 rφ × rθ = 16 sin φ cos θ; −16 sin φ sin θ; 16 sin φ cos φ q 2 4 2 2 4 2 2 2 2 jrφ × rθj = 16 sin φ cos θ + 16 sin φ sin θ + 16 sin φ cos φ q 2 4 2 2 2 jrφ × rθj = 16 sin φ + 16 sin φ cos φ q q 2 2 2 2 2 2 jrφ × rθj = 16 sin φ(sin φ + cos φ) = 16 sin φ = 16 sin φ Note: sin φ > 0 on the given interval of φ. Z 2π Z π=3 Z 2π Z π=3 p S = jrφ × rθj dφ dθ = 16 sin φ dφ dθ = ··· = 16π( 3 − 1) θ=0 φ=π=6 θ=0 φ=π=6.