Rachit Kumar SaxenaManager-Editorial
What is Line integral?
The process of differentiation and integration are inverses of each other. So, d/dx g(x).dx= g(x). If the function to be integrated is determined along a curve, it is a line integral.
In an integral, if the function to be integrated is evaluated along a curve, the integral is known as the line integral. A line integral is also known as a path integral, curve integral or curvilinear integral.
Line integral for a scalar field is given as:
For a function g, g: U Rm R, the line integral along with a curve V, such that V U is stated as:
V. g(m) ds= klg[r(t)].r’tdt
Where [k,l] -> V is a parameter of the curve.
r(k) and r(l) show the values of the function at boundary point k and l, such that k
Line integral in a vector field is given as:
For a function, G: V Rm -> Rm, the line integral along the curve V such that V U in the direction r is stated as:
V G(r). dr= klG[r(t)].r’(t).dt
Where [k,l] -> V is a parameter of the curve.
r(k) and r(l) show the values of the function at boundary point k and l, such that k
About Line Integral in Class 12
The topic of line integral is a part of chapter 8 ‘Application of integrals’ in the class 12 NCERT textbook. It carries a weightage of around 5-6 marks in the examination.
Illustrated Examples on Line Integral
1. Evaluate the given function v∫ G. dr to find the line integral. Given, G(x,y,z)= [K(x,y,z), L(x,y,z), M(x,y,z)]= (z, x, y) and v is defined by x= t2, y= t3, z=t2, 0≤ t≤1.
Solution:
The given function- G(x,y,z)= [K(x,y,z), L(x,y,z), M(x,y,z)]= (z, x, y)
And the parameters are x= t2, y= t3, z=t2, 0≤ t≤1.
So, v∫ G. dr= v∫ K. dx+ L.dy + M.dz= 0∫1 z(t) x’(t)dt + x(t) y’(t)dt + y(t) z’(t)dt
= 0∫1 t2(2t).dt+ t2(3t2).dt + t3(2t).dt= 0∫1 2t3.dt+ 3t4.dt + 2t4.dt
= 0∫1 (5t4+ 2t3).dt= (5.t5/5 + 2.t4/4)01= 3/2
So, the value of the line integral is 3/2.
2. Evaluate v∫xy- 4z.ds, where v is the line segment from (1,1,0) to (2,3,-2)
Solution:
The given function is g(x)= v∫xy- 4z.ds
The representation of the line segment is:
k→= (1-t)[1,1,0] + t[2,3,-2]= [1+t, 1+2t, -2t]
|k→|= √{12+22+(-22)}= √9= 3.
So, xy- 4z= (1+ t)(1+ 2t)- 4(-2t)= 2t2+ 11t+ 1
vxy- 4z.ds= 0∫1(2t2+ 11t+ 1).3 dt= 43/2.
3. Evaluate v∫ 3x3.ds, where v is the line segment from (1,2) to(-2,-1).
Solution:
The curve is given as:
v= (1-m)(1,2)+ m(-2,-1)
v∫3x3.ds= v∫3(1-3m)3.√18ds= 9√2.(-5/4)= -15.90
FAQs on Line Integrals
Q: How can we express the functions to be integrated?
Q: What do you mean by vector calculus?
Q: Which parameter can be calculated with a line integral?
Q: What is the fundamental theorem of line integrals?
v∇s.dx= s(k)- s(l)
Q: What are the applications of line integrals?
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