Scalar Line Integrals

Scalar line integral is a definite integral it will be taken over a surface and integrated, so the scalar line integral is defined as the sum of all points in the surface.  Let x: [a, b] gives R3 be a path of class C and f: X subeR3 gives R be a continuous function contains the image of x.  The scalar line integral of f along x is

int_a^bf(x(t)) || x'(t) dt

Notation is usually written as int_x f ds.

Algebra is widely used in day to day activities watch out for my forthcoming posts on Definite Integrals and Anti derivative. I am sure they will be helpful.

General form of scalar line integral:

The scalar function of line integral  is

if F = F1 i + F2 j + F3 k

r = x i + y j + z k

So int_cF. dr = int_c (f_(1)dx + f_(2) dy + f_(3) dz)

Scalar line integral and parametrizations

If y is a reparametrization of x. Then

If y is orientation-preserving, then int_y F.ds =  int_x F.ds

If y is orientation-preserving, then  int_y F.ds  = -int_x F.ds


When the path is parametrized by length of arc, the natural analog of the integral has done in 1 dimension. The integral of a scalar function f along a curve r(s) is simply int f (r(s)) ds

Applications:

F is a force acting upon a the particular particle so the particle moves along a curve C in  sample space and r be the position vector of the  given particle at a point on C. Then work done by the given particle at C is F.dr and the total work is done by F along a curve C is given by the line integral  int_c F. dr



Having problem with Convergent and Divergent Boundaries keep reading my upcoming posts, i will try to help you.


Scalar Line Integrals don’t depend on parameterizations.
Scalar Line Integrals-example Problems

Evaluate I = int_e f(x, y, z) ds where f(x, y, z) = = x2 – y2 – 1 + z and e is the helix parametrized by c(t) = (cos t, sin t, t)  [o <= t <=pi]. 3

Solution:

I = int_alpha^beta f(c(t)) || c'(t) || dt

See that f (c (t)) = cos2t + sin2t - 1

Also  c' (t) = (-sin t, cos t, 1) =    sqrt(-sin t^(2) + cos^(2) t + 1^(2))

See that f(c (t)) = cos2t + sin2t - 1

I = int_0^(3pi)sqrt(2) t dt

= sqrt(2) [(1)/(2)  t^(2) ][]_0^(3pi)

= (sqrt(2))/(2) 9pi^(2).

The correct answer for scalar line integral is = (sqrt(2))/(2) 9pi^(2).

Practice Problem:

Evaluate int sin 6x cos 3x dx


Answer: -1/2 [(cos 9x / 9) + (cos 3x / 3)] + c.