5. Integration Over Curves and Surfaces

Question 1

Curve

Definition

Answer 1

-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

Question 2

Equations of Curves

General Form of a Straight Line

Answer 2

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

-where |a and |b are constant

Question 3

Equations of Curves

General Form of a Parabola

Answer 3

|r(t) = (at, bt² ,0)

Question 4

Equations of Curves

Helix of Radius 1

Answer 4

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

Question 5

Line Integral

Description

Answer 5

• 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

Question 6

Line Integral

Arc Length Equation

Answer 6

arc length = ∫ ds

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

Question 7

Line Integrals

|r)(t) in parametric form

Answer 7

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

Question 8

Line Integrals

d|r

Answer 8

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

Question 9

Line Integrals

ds

Answer 9

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

Question 10

Properties of Line Integrals and Closed Curves

Addition

Answer 10

-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

Question 11

Properties of Line Integrals and Closed Curves

Closed Curves

Answer 11

-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

Question 12

The Vector Line Integral

Answer 12

-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)

Question 13

The Scalar Line Integral

Answer 13

-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

Question 14

Work

Answer 14

-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)

Question 15

Circulation

Answer 15

-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

Question 16

Conservative Field

Two Definitions

Answer 16

-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

Question 17

Properties of a Conservative Field Theorem

Answer 17

-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

Question 18

Potential

Answer 18

-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

Question 19

Gravitational Potential

Answer 19

-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

Question 20

Potential of a Physical Field

Answer 20

• 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

Question 21

Examples of Line Integrals That Give Scalar Results

Answer 21

∫ f(|x) ds
and
∫ |F(|x) . d|x

Question 22

Examples of Line Integrals That Give Vector Results

Answer 22

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

Question 23

Surface Integral

Equation

Answer 23

I = ∬ f(|x) dS

Question 24

Surface Integral

Definition

Answer 24

-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

Question 25

Surface Integral

Parameterisation

Answer 25

-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

Question 26

Surface Integral

Jacobian

Answer 26

-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

Question 27

How to find a surface integral over a surface S

Answer 27

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.

Question 28

Flux

Equation

Answer 28

Flux = ∬ |F . |n dS = ∬ |F . d|S

Question 29

Rate of Fluid Volume Flow Through a Surface

Answer 29

∬ |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

Question 30

Vector Surface Element

Answer 30

-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

Question 31

Orientated Surfaces

Answer 31

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

Question 32

Surface Integrals Over Closed Surfaces

Answer 32

-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

Question 33

Other Surface Integrals

vector results

Answer 33

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

Question 34

Net Force Applied to a Surface by Pressure in a Gas

Answer 34

|F = ∬ d|F = - ∬ P d|S

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