Line integrals

Question 1

Path

Answer 1

A parametisation f a continuous path in |R^n is a continuous map r : [a,b] -> |R^n.

Question 2

Smooth

Answer 2

The path is called smooth if the derivative r’ exists and is continuous on [a,b].

Question 3

Piecewise smooth

Answer 3

The path is called piecewise smooth if one can decompose [a,b] into finitely many subintervals on which r is smooth

Question 4

Line integral

Answer 4

If r : [a,b] -> |R^n is a smooth parametrisation of a path C and F : Ω-> |R^n s a vector field so that F is defined and bounded on the image of r then we define
∫ (over C) F. dr = ∫(over a to b) F(r(t)). r’(t) .dt
If r is only piecewise smooth, the corresponding integral is the sum of the integrals on the smooth pieces.

Question 5

Conservative

Answer 5

A vector field in which the line integral only depends on its ends points.

Question 6

Path equivilance

Answer 6

Suppose s : [c, d] -> |R^n and r : [a, b] -> |R^n are two parameterisations. We say that they are equivalent if there exists u : [c,d]-> [a,b] so that the derivative of u is continuous, nowhere zero and s(t) = r(u(t)).

Therefore r and s have the image.

Question 7

Same orientations (informal)

Answer 7

The derivative of u, u’ is continuous and nowhere zero then it is either strictly positive or strictly negative.
If u’(t) >0 for all t e [c,d] then the two parameterisations have the same orientation.

Question 8

Opposite orientation (informal)

Answer 8

If u’(t)< 0 for all t e [c,d] then the two parameterisations have the opposite orientation.

Question 9

Same orientation proposition

Answer 9

Suppose F is a vector field and r, s are equivalent parameterisations of C.
If r and s have the same orientation then
∫(over C) F.dr = ∫(over C) F. ds

Question 10

Opposite orientation proposition

Answer 10

Suppose F is a vector field and r, s are equivalent parameterisations of C.
If r and s have opposite orientations then
∫(over C) F. dr = ∫ (over C) F.ds