Line Integral: Overview, Questions, Preparation

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.

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

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?

Q: What are the applications of line integrals?