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.