11) Surface Integrals

Question 1

Smooth surface

Answer 1

A surface is smooth if the derivative matrix D(F)(u,v) of F has rank 2 for all points (u,v) in D

That is…
Equivalent to

Vectors ∂F/∂u and ∂F/∂v or the columns of D(F)(u,v) are linearly independent.
3 by 2 matrix as F(u,v) =(x,y,z)
.

Question 2

Surface in surface Integrals

Answer 2

Is a smooth surface and

D is subset of R^2 and is a simple region(singly connected region with no holes) enclosed by a curve c. Let (x,y,z) =F(u,v) be a C^1 map from D to R^3.

Maps from 2d to 3D

Question 3

Image -of surface in surface Integrals

Answer 3

Image: S=F(D) ={F(u,v)| (u,v) in D} is a surface in R^3

Question 4

Curve -of surface in surface Integrals edge of s

Answer 4

Curve: ∂S=F(c) is called the edge of s

I.e. Curve c is edge of region D in R^2 maps to curve of surface in R^3

Question 5

Parameterisation of the surface

Answer 5

Map F ?

Is possible if singly connected no holes?

Question 6

Singular point in surface Integrals

Answer 6

Singular point occurs for rank of D(F)(u,v) less than 2 for this point (u,v)

Question 7

Surface is open or closed

Answer 7

Surface S is open if every pair of points in R^3 not on S can be joined by a continuous curve not passing through S

Closed if divides R^3 into distinct regions R_1 and R_2 such that every continuous curve joining a point in R_1 to a point in R_2 crosses at least once

Question 8

Simple surface

Answer 8

Surface is union of finite smooth piece wise

E.g. Cube is an example of union of 6 open smooth surfaces is closed

Question 9

Normal vector to a surface

Answer 9

Vector N = ∂Vector(r)/ ∂u cross ∂Vector(r)/ ∂v.

From the condition of S being smooth we have that the columns of D(F)(u,v) are linearly independent. With the position vector of P on the surface is vector(r).

They’re linearly independent hence form a basis of the tangent plane to a at P. Hence scalar product is not 0 and is normal to.

Question 10

Unit vector normal to surface

Answer 10

Cross products of D(F)(u,v) as linearly Independent.

^n_ = (1/|N|) vector(N)

Unit normal is independent of the chosen parameterisation of S and is unique up to a sign.

Orientation of S affects sign of ^n_

A side chosen as positive oriented for closed surface.

The unit normal vector exists then is continuous on all points for a smooth surface.
For unit normal usually well defined when compared to the non unit vector which depends on parameterisation.

Question 11

Surface area

Answer 11

Surface are of S is:
Double_integral( | ∂Vector(r)/ ∂u cross ∂Vector(r)/ ∂v | du.dv)

Over D

Where integral taken over region D so whole of S is covered.

Magnitude of vector.
I.e. Surface element and normal integral for area!

Question 12

Surface element

Answer 12

Surface element is portion of S bounded by P_0P_1 and P_3P_2.

Surface element

ds roughly equal to
| ∂Vector(r)/ ∂u cross ∂Vector(r)/ ∂v|.du.dv
Is an element of the surface area

I.e. Magnitude
As the vectors are tangent to coordinate lines and if displacements are small roughly parallelogram with area given as magnitude of the vector product
Coordinate lines in 2d map to 3D

Question 13

Surface Integrals

Answer 13

Let S be a simple surface with parametric equation vector(r) = r(u,v) and let D denote the region in the uv plane corresponding to points of S.

Is a SCALAR FIELD phi and vector field vector(v) are defined at all points on a, then on S vectors φ=φ(u,v) and v=v(u,v).

The surface Integrals of φ and vector V over a are defined

Integral( φ .ds) over s
= double_integral( φ(u,v) | ∂Vector(r)/ ∂u cross ∂Vector(r)/ ∂v|.du.dv)

Magnitude

Integral (vector(v).ds) over s
= double integral(vector(v)(u,v) • ( ∂Vector(r)/ ∂u cross ∂Vector(r)/ ∂v).du.dv).

Dot product

E.g. Can be component form as line integral, we can define the surface Integrals of any vector or scalar function from the calar vector products of the two columns
Region is D defined by u and v parametrisation

OR CAN USE NORMAL IN VECTOR

Question 14

Flux and flux integral

Answer 14

Flux of vector(v) across s is found by the flux integral. For a vector field v(x,y,z). On a simple orientable surface with unit normal ^n_(x,y,z)

φ_v = integral ( vector(v)(x,y,z)•^n_(x,y,z).ds)

E.g. Vector v represents veclociry of fluid at a point then φ_v is volume crossing surface I. The direction of its orientation ^n_(x,y,z) per unit of time.

I.e. If φ_v is bigger than 0 more fluid passes through a in the positive direction than in negative etc

Question 15

The divergence theorem

Answer 15

Integral ( vector(v)•^n_ .ds) over s
= integral( div(v) .dV) over R

Where dV is an element of Volume

E.g. if region in 3D then triple integral
We use the divergence theorem to find the. Outward flux of the vector field across the region

For region R of R^3 bounded by a closed simple surface S, oriented outwards. If v(x,y,z) is a vector field whose components have continuous partial derivatives on some open set containing R and if ^n_(x,y,z) is outward directed normal at (x,y,z) on s.

Question 16

The divergence theorem and greens

Answer 16

Divergence theorem for R^2 is greens theorem as used for finding flux if a vector field across piece wise smooth surfaces of distinct sections

The theorem states: total flux outwards from close surface S = volume integral of the divergence of the vector field throughout volume enclosed by S.

Question 17

When div(v) is 0 in R

Answer 17

Integral gives total flux which is Sum of sources -sum of sinks within S.
If vector(v) such that divergence of v =0 in R then it's called a solenoidal field vecit(v) neither net sink nor net source

Question 18

Stokes theorem

Answer 18

For S a piece wise smooth oriented surface bounded by simple closed piece wise smooth curve c with positive orientation. If vector(v(x,y,z)) is a CONTINUOUS vector field with CONTINUOUS first partial derivatives on some open set containing S.

IntegralO( vector(v). d(vector(r))) over curve c =
Integral( curl( vector(v)) • ^n_ .ds))
Over s

Where c is edge of s Curve: ∂S=F(c).

E.g. Given paraboloid and a vector field we represent c by circle and parameterise to find line integral also by dot product of curl and normal unit vector. (Cross product). E.g. Constant Z or Z=0 for xy plane

Question 19

Greens and stokes theorem

Answer 19

Vector(v)=(f,g,h) and a smooth curve c lying wholly in the xy plane / z=0.

Let at be the region enclosed by c and so a closed surface a has edge Curve: ∂S which is C.
Since s lies in the xy plane the unit normal to a is ^n_=(0,0,1) where Normal points in the positive z direction. Positive orientation of S corresponds to C being traversed in a counterclockwise direction.

Curl(v) •^n_ = ∂g/∂x - ∂f/∂y
Since c is wholly in xy plane, as dvector(r) = (.dx,.dy,0).

Hence stokes becomes greens:
IntegralO( f.dx +g.dy) over c =double over R GREENS