Area of a surface Problem: Find the area of the surface S - - PowerPoint PPT Presentation

Area of a surface

Problem: Find the area of the surface S parameterised by r(u, v) = (x(u, v), y(u, v), z(u, v)), a ≤ u ≤ b, c ≤ v ≤ d. Solution: Area of the surface = b

a

d

c

|ru × rv|dudv where ru(u, v) = (xu(u, v), yu(u, v), zu(u, v)) rv(u, v) = (xv(u, v), yv(u, v), zv(u, v)) BIG IDEA 1: The cross product of the partial derivatives is a vector perpendicular (normal) to the surface. BIG IDEA 2: The length of the cross product of the partial derivatives serves as infinitesimal area.

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Area of a surface

Example: Find the area of the surface of a sphere of radius R. Solution: Parametrisation (spherical coordinates) r(u, v) = (R cos u sin v, R sin u sin v, R cos v), 0 ≤ u < 2π, 0 ≤ v ≤ π Then ru(u, v) = (−R sin u sin v, R cos u sin v, 0) rv(u, v) = (R cos u cos v, R sin u cos v, −R sin v) giving ru × rv = (−R2 cos u sin2 v, R2 sin u sin2 v, −R2 sin v cos v) |ru × rv| = R2 sin v and thus π 2π R2 sin vdudv = 2π π sin vdv = 4πR2

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Surface integral of a scalar function

Let f : R3 → R a scalar function. The integral of f over the surface S parameterised by r(u, v) = (x(u, v), y(u, v), z(u, v)), a ≤ u ≤ b, c ≤ v ≤ d. is defined as

• S

f dS = b

a

d

c

f (r(u, v))|r u × rv|dudv

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Surface integral of a scalar function

Example: Calculate the surface integral of f (x, y, z) = x over the first

• ctant of the surface of the sphere of radius 1.

Solution: Parametrisation r(u, v) = (cos u sin v, sin u sin v, cos v), 0 ≤ u ≤ π 2 , 0 ≤ v ≤ π 2 As before |ru × rv| = sin v and thus

• π

2

• π

2

(cos u sin v) sin vdudv = ... = π 4

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