Expand/Contract All

Line Integrals

 
 

Tangential Line Integral

Consider a vector function #~v #: &reals.&powthree. -> V_3 , _ i.e. #~v ( ~t ) &in. V&powthree. , ~t &in. &reals.. _ Let &Gamma. be a curve in &reals.^3, whose points are given by the function _ #~r _ = _ #~r ( ~s ) , _ ~s being the arc length along &Gamma. from a given reference point.

The #~{tangential line integral} of #~v along &Gamma. from point ~A to point ~B is defined as

I _ _ = _ _ int{#~v (~s) &dot. ~#t (~s) ,~s(~A),~s(~B), d~s}

Where ~#t (~s) is the tangent to &Gamma. at the point corresponding to arc lenght ~s.

But _ ~#t (~s) _ = _ #~r' ( ~s ). _ i.e:

int{array{ _ /#~v (~s) &dot. ~#t (~s)/ _ } ,~s(~A),~s(~B), d~s} _ _ = _ _ int{#~v (~s) &dot. fract{d~#r,d~s} (~s) ,~s(~A),~s(~B), d~s}

which we write as

int{#~v (~s) &dot. ,~s(~A),~s(~B), d~#s} _ _ _ _ _ _ or simply _ _ _ _ _ _ int{#~v &dot. ,~A,~B, d~#s}

General Parameters

Suppose the curve &Gamma. can also be represented by the vector function #~q of some general parameter &theta. , i.e. _ ~#r _ = _ #~q ( &theta. ) , _ then in terms of change of integration variable:

int{array{ _ /#~v (~s) &dot. ~#t (~s)/ _ } ,~s(~A),~s(~B), d~s} _ _ = _ _ int{ fract{#~v (~s) &dot. ~#t (~s), d&theta./d~s} ,&theta.(~A),&theta.(~B), d&theta.}

= _ _ _ _ int{ fract{#~v (~s) &dot. ( d~#q/d&theta. # d&theta./d~s ), d&theta./d~s} , &theta.(~A),&theta.(~B), d&theta.} _ _ = _ _ int{#~v &dot. fract{d~#q,d&theta.} ,&theta.(~A),&theta.(~B), d&theta.}

Note that we have introduced the notation _ ~#r = #~q ( &theta. ) _ here to try to clarify what is happening, it is more usual to write

~I _ = _ ~{∫} #~v &dot. (d#~r/d&theta.) d&theta.

 

Example

Integrate the function _ ~#u = ~x~y~#i , _ over the first quadrant of a circle radius ~a center (0,0).

The quadrant has equation _ #~r ( &theta. ) _ = _ ~a cos &theta. #~i + ~a sin &theta. #~j , _ 0 =< &theta. =< &pi./2 .

so

d#~r/d&theta. _ = _ -~a sin &theta. #~i + ~a cos &theta. #~j

On the circle, _ ~x _ = _ ~a cos &theta. , _ ~y = ~a sin &theta. , _ so _ ~#u = ~x~y~#i _ = _ ~a^2 sin &theta. cos &theta. #~i

int{#~u &dot. ,~s(~A),~s(~B), d~#s} _ = _ int{#~u &dot. d#~r/d&theta. ,0,&pi./2, d&theta.}

_ _ _ _ _ = _ int{ - ~a&powthree. sin&powtwo. &theta. cos &theta. ,0,&pi./2, d&theta.} _ = _ - ~a^3 script{sqrb{fract{sin&powthree. &theta., 3}},,,&pi./2,0} _ = _ - fract{~a^3,3}

 

 

Integral of a Gradient

&phi. #: &reals.^3 -> &reals. . _ &Gamma. is a curve in &reals.^3 given by the equation _ #~r _ = _ #~v ( ~s ) , _ where ~s represents arc length, or distance along the curve.

int{&nabla.&phi. &dot. ,~s(~A),~s(~B), d~#s} _ _ _ _ #:= _ _ _ _ int{&nabla.&phi. &dot. ~#t (~s) ,~s(~A),~s(~B), d~s}

_ _ _ _ _ = _ int{&nabla.&phi. &dot. fract{d#~r,d~s} ,~s(~A),~s(~B), d~s} _ _ _ = _ _ _ int{fract{d&phi.,d~s} ,~s(~A),~s(~B), d~s}

_ _ _ _ _ = _ int{ _ ,&phi.(~A),&phi.(~B), d&phi.} _ _ _ = _ _ _ &phi.(~B) - &phi.(~A)

i.e. the integral is only dependent on the value of &phi. at the end points, and not of the path of the curve.