5)Line Integrals Flashcards
Scalar and vector fields and direction of line integral
Scalar field line integral is independent of direction of C
Vector field: if C is reversed line integral changes sign
What is a smooth curve ( line Integrals)
Smooth if position vector r
dr/dot is continuous and magnitude never 0
|dr/dt| not equal to 0
No stationary points
Helix centred on z-axis
x(t)= cost y(t)= Sint z(t)= kt
Line integral of a scalar field
Int( f(x,y,z).ds)
c
t_b = int( f( x(t), y(t), z(t) ) |dr/dt| .dt t_a
MAGNITUDE SQRT OF components squared dr/dt
Given Smooth curve C from A to B and scalar field f(x,y,z) defined in some region containing c
Closed curve line Integral
Closed curve C line integral
IntO( f(vector (x)).ds
c
Value is independent of point from which arc length measured
Line integral of a vector field
Def
Given a vector field BOLD v(x,y,z) defined over some region containing c.
^
Unit tangent vector T = d(r)/dt
-
I = int ( bold(v)(x,y,z).d(bold(r))) c ^ = int( v(x,y,z) • T.ds) c - ^ = int( v_1.dx + v_2.dy +v_3.dz) c
Unit tangent = derivative of position vector
Changes sign and is reversed if c reversed!
Line integral of vector field evaluating
PARAMETRICALLY WRT t
DOT PRODUCT
I = int( vector(v) • ( d(vector(r))/dt) .dt )
c
FUNCTIONs of x y=f(x), z=g(x) WRT x
I= int( v_1 + v_2 f’(x) + v_3 g’(x)) .dx
c
y,z must be singly values over domain split-> invertible
Circulation:
If in conservative field/ if in path independent field
Circulation k =
IntO( vector(v) •d (vector(r)))
Of vector(v) around c
Is the line integral of vector field vector(v) around a simple closed curve
For path independent variable circulation k=0
Closed curve line integral =0 for any simple closed curve In Conservative field
Length of curve
Length of curve is line Integral
Int(.ds)
c
Positional vector derivative
dr/dt vector
.ds = |dr/dt| .dt
Greens theorem
Calculating the area enclosed by simple closed
IntO( f(x,y) .dx + g(x,y).dy) c = closed intO( (f(x,y), g(x,y) ) • (dx/dt, dy/dt) .dt ) c
= double int(part g wrt x - part f wrt y) dx.dy
R
f(x,y) and g(x,y) continuous and have continuous first partial derivatives on some open set containing R.
Change of variable if in t
Use greens when we can’t parameterise and functions of both x and y.
Subregion (line integral)
Subregions are simply connected regions split into subregions.
Must have boundary curves that are cut by any line parallel to either coordinate axes.
Opposites cancel.
Area enclosed by smooth piece wise curve: by greens
A = 0.5 intO( -y.dx +x.dy)
For area of region enclosed by curve
As greens gives double of partials take away to give one, area
Alternatively:
g_x - f_y =1 or other value
E.g. f(x,y) =0 g(x,y) =x Or f(x,y) =-y g(x,y) =0
Direction matters in this
Path independence of line integral of c over vector field
I = int( vector(v) . d(vector(r)))
c
^ = int( vector(v) • vector(T) .dt) c. - Is path independent on R if the value of I is independent of path for a smooth curve from A to B
Circulation k=0 if Integral is path I dependent
Conservative field
If a line integral is path independent on R
Then for any closed curve we also have IntO( vector(v) . dvector(r)) =0 For any simple closed curve in R Then the field Vector(v) Is a CONSERVATIVE FIELD V