5. Integration Over Curves and Surfaces Flashcards Preview

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Flashcards in 5. Integration Over Curves and Surfaces Deck (34)
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1
Q

Curve

Definition

A

-a curve is a path through space described parametrically by writing the position vector as a function of a parameter t that varies along the curve

2
Q

Equations of Curves

General Form of a Straight Line

A

|r (t) = |a + t |b

-where |a and |b are constant

3
Q

Equations of Curves

General Form of a Parabola

A

|r(t) = (at, bt² ,0)

4
Q

Equations of Curves

Helix of Radius 1

A

|r(t) = (cos t , sin t , t)

5
Q

Line Integral

Description

A
  • the domain of a one dimensional definite integral does not have to lie on the x axis
  • instead we can integrate along any one dimensional curve embedded in three dimensional space
  • an integral of this kind is called a line integral
6
Q

Line Integral

Arc Length Equation

A

arc length = ∫ ds

-where ds measures the distance moved in an infinitesimal step from |r to |r+d|r , ds = | d|r |

7
Q

Line Integrals

|r)(t) in parametric form

A

|r(t) = ( x(t) , y(t) , z(t) )

8
Q

Line Integrals

d|r

A

d|r = d|r/dt * dt
= ( dx/dt dt , dy/dt dt , dz/dt dt)
= (dx\dt , dy/dt , dz/dt) dt

9
Q

Line Integrals

ds

A

ds = | d|r |
ds² = ((dx/dt)² , (dy/dt)² , (dz/dt)²) dt²
so:
ds = [ (dx/dt)² + (dy/dt)² + (dz/dt)² ]^(1/2) * dt

10
Q

Properties of Line Integrals and Closed Curves

Addition

A

-if Q is an intermediate point on curve C between points P and R then:
the integral of the line between P and Q plus the integral of the line between Q and R, is equal to the integral of the line between P and R

11
Q

Properties of Line Integrals and Closed Curves

Closed Curves

A

-if C is a simple closed curve (ie is made of one non-intersecting loop), then the line integral around the curve is written:
∮ f(|r)
and is independent of the starting point

12
Q

The Vector Line Integral

A

-suppose we have a curve C given by |r(t) and a vector field F defined on the curve C
-we can integrate |F(|r) along the curve with respect to arc length giving:
∫ |F(|r) ds = (∫F1 ds , ∫F2 ds , ∫F3 ds)

13
Q

The Scalar Line Integral

A

-suppose we have a curve C given by |r(t) and a vector field F defined on the curve C
-formed by integrating the component of |F in the direction of unit tangent vector ^T, which points in the direction of d|r
∫|F . ^T
-and since ^T ds = d|r (as ^T is the direction of d|r and ds is the magnitude of ds
-so the scalar line integral is usually written:
∫|F . d|r

14
Q

Work

A

-if |F is a force field acting on a particle moving along the curve C from A to B,
-then the work done in displacing the particle an infinitesimal distance d|x along the curve is |F.d|x
-thus the total work done in moving the particle from A to B is given by:
W = A,B ∫ |F . d|x
(d|x = d|r)

15
Q

Circulation

A

-the circulation of a vector field |F around a closed curve C is defined as:
∮ |F . d|x
-note that circulation changes sign under reversal of direction

16
Q

Conservative Field

Two Definitions

A

-a vector field |F is defined as conservative if its circulation around any closed loop is zero:
∮ |F . d|x = 0 around loop C for all C
-an equivalent definition is that a vector field is conservative if the line integral between any two points P and Q is independent of the path taken, this follows from the addition rule

17
Q

Properties of a Conservative Field Theorem

A

-if and only if
i) |F is a conservative field,
then:
ii) there exists a scalar field φ such that |F = |∇ φ
this is so if and only if:
iii) |∇ x |F = 0 everywhere

18
Q

Potential

A

-any conservative field |F can be written in terms of the gradient of a scalar field:
|F = |∇ φ
-the function φ is called the potential of |F

19
Q

Gravitational Potential

A
-for a gravitational field
|g  =  -^j g
-the potential is
φ = -mgy
-taking the gradient of this potential:
|∇ φ = (∂/∂x,∂/∂y,∂/∂z)(-mgy)
= ( 0 , -mg , 0 )
= -mg ^j
= m |g
= |F
-we obtain the gravitational force on a body in a uniform gravitational field
20
Q

Potential of a Physical Field

A
  • for a physical force field, the scalar potential φ (where |F = |∇ φ) gives the physical potential energy at that point in the field
  • note that we can always add an arbitrary constant to a potential since |∇(constant) = 0
  • this is just like measuring the potential energy relative to a different point in the field
21
Q

Examples of Line Integrals That Give Scalar Results

A

∫ f(|x) ds
and
∫ |F(|x) . d|x

22
Q

Examples of Line Integrals That Give Vector Results

A

C,∫ f(|x) d|x , where f is scalar
and
C,∫ |F(|x) x d|x

23
Q

Surface Integral

Equation

A

I = ∬ f(|x) dS

24
Q

Surface Integral

Definition

A

-scalar and vector fields can be integrated over a surface S even if S is curved
-a surface integral is defined via the Riemann sum:
I = N->∞ lim (k=1->N)Σ f(|xk) 𝛿Sk
-where the sum is over all are elements (labelled by k) covering the surface, each of area 𝛿Sk and containing the point with position vector |xk

25
Q

Surface Integral

Parameterisation

A

-generally the area elements have different sizes, 𝛿Sk
-to integrate over S, use parameterisation to convert this to an integral over a 2D plane
-the position vector of a general point on the surface S is given in terms of two parameters (u,v) so that:
|x = ( x(u,v) , y(u,v) , z(u,v) )
-note that 𝛿Sk is NOT given by dudv in general because the surface elements may be tilted and non-rectangular

26
Q

Surface Integral

Jacobian

A

-the tangent vectors to the plane of the surface are given by |tu = ∂|x/∂u and |tv = ∂|x/∂v
-the surface element d|S is given by:
-the Jacobian is given by:
|J| = | |tu x |tv |
-so, dS = |J| du dv = | |tu x |tv | du dv

27
Q

How to find a surface integral over a surface S

A

1) sketch the surface
2) parameterise in terms of two parameters (if necessary switch to a different coordinate system)
3) find the two tangent vectors, one for each parameter
4) find the cross product of the two tangent vectors
5) find the magnitude of the cross product, this is the Jacobian
6) dS = |J| du dv , where u and v are the two parameters
7) integrate with respect to dS using limits which define the boundaries of the surface S.

28
Q

Flux

Equation

A

Flux = ∬ |F . |n dS = ∬ |F . d|S

29
Q

Rate of Fluid Volume Flow Through a Surface

A

∬ |u . |n dS

  • where S is the surface
  • |u(|x,t) is the velocity of the fluid
  • |n is the unit normal to the surface
30
Q

Vector Surface Element

A

-since the cross product of the tangent vectors is directed normal to the surface, the unit normal is given by:
^n = ( |tu x |tv ) / | |tu x |tv |
-thus |n dS is given by:
|n dS = ( |tu x |tv ) / | |tu x |tv | * | |tu x |tv | du dv
|n dS = (|tu x |tv ) du dv
-which is the vector surface element:
d|S = |n dS = (|tu x |tv ) du dv

31
Q

Orientated Surfaces

A

both ^n and -(^n) are normal to the surface S

  • by convention, ^n points from the positive side of the surface
  • in calculating ( |tu x |tv ) , ensure that ^n points out from the positive side
  • if it does not you can reverse the order of the cross product or simply multiply by -1
32
Q

Surface Integrals Over Closed Surfaces

A

-surface integrals over closed surfaces are written:
∯ = |F . d|S
-where ^n points out of the enclosed volume
-by convention the positive side of a closed surface is the outside

33
Q

Other Surface Integrals

vector results

A

-we can also define surface integrals that give vector results, such as:
∬ |F dS
∬ |F x d|S
∬ f d|S

34
Q

Net Force Applied to a Surface by Pressure in a Gas

A

|F = ∬ d|F = - ∬ P d|S

-where P(|r) is the pressure field in the gas