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) = |ru × rv| 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

√ 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 |rφ × rθ| = 16 sin φ cos θ + 16 sin φ sin θ + 16 sin φ cos φ q 2 4 2 2 2 |rφ × rθ| = 16 sin φ + 16 sin φ cos φ q q 2 2 2 2 2 2 |rφ × rθ| = 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 √ S = |rφ × rθ| dφ dθ = 16 sin φ dφ dθ = ··· = 16π( 3 − 1) θ=0 φ=π/6 θ=0 φ=π/6