Surface integrals

Question 1

Parametrisation

Answer 1

The parametrisation of a surface S is a map r(a,b) : Ω -> |R^3 such that r(Ω) = S. Here Ω C |R^2.

Question 2

Smooth parametrisation

Answer 2

The parametrisation r is called smooth if the maps ra, rb exist and are continuous and ra x rb ≠ 0.

Question 3

Tangent vector

Answer 3

Consider r : Ω -> |R^3, r(a,b) = (x(a,b), y(a,b) , z(a,b) be a parametrisation of a surface and suppose that the derivatives of r are continuous. We define two tangent vectors
ra = ∂x/∂a, ∂y/∂a, ∂z/∂a
rb = ∂x/∂b, ∂y/∂b, ∂z/∂b

Question 4

Normal vector

Answer 4

ra x rb = | i j k |
| ∂x/∂a ∂y/∂a ∂z/∂a |
| ∂x/∂b ∂y/∂b ∂z/∂b |

Question 5

Tangent plane

Answer 5

The equation of the tangent plane to the surface at the point (a,b,c) is
( ra x rb) . (x-a, y-b, z-c) = 0.

Question 6

The area of a surface

Answer 6

Suppose that we have a smooth parametrised surface S described by r : Ω -> |R^3 for some Ω C |R^2. Then the area of the surface is then
AreaS = ∫∫(over Ω) |ra x rb| . dadb

Question 7

More precise definition of the area of a surface

Answer 7

AreaS = ∫∫ (over Ω) √(∂(Y,Z)^2/∂(a, b) + ∂(Z,X)^2/∂(a, b) + ∂(X,Y)^2/∂(a, b) . dadb

Question 8

Lagrange’s identity

Answer 8

If u, v e |R^3 then

| u x v| = √(u.u)(v.v) - (u.v)^2

Question 9

Surface integral of the first kind

Answer 9

Let Ω C |R^2 and r : Ω-> S be a smooth parametrisation of a surface S and f: S-> |R be a smooth scalar field. The integral of f on S ( a surface integral of the first kind) is
∫∫ (over S) f. dS = ∫∫ (over Ω) f(r(a,b)) | ra x rb| da db.

Question 10

Surface integral of the first kind (for piecewise smooth functions)

Answer 10

If the surface is the union of finitely many smooth parametrisation, the integral on the surface is defined as the sum of the integrals of the respective smooth pieces.

Question 11

Surface integrals of the second kind

Answer 11

Let r: Ω-> |R^3 be a smooth parametrisation of an (orientable) surface and F: S -> |R^3 a smooth vector field. The integral of F on S (also called a surface integral of the second kind) is
∫∫ (over S) F. n dS = ∫∫ (over Ω) F(r(a,b) . (ra x rb) / |ra x rb| da db = ∫∫ (over Ω) F (r(a,b)) . (ra x rb) da db

Question 12

Surface integrals of the second kind ( for piecewise smooth functions)

Answer 12

A before if the surface is the union of finitely many smooth parameterisations, the integral on the surface is defined as the sum of the integrals of the respective pieces.