### IIT JAM MATHS

DOUBLE INTEGRAL
BETA AND GAMMA FUNCTION
VOLUME
SURFACE AREA
LINE INTEGRAL
GREENS THEOREM
SURFACE INTEGRAL
GAUSS DIVERGENCE THEOREM
STOKES THEOREM
CONSERVATIVE VECTOR FIELD
LIMITS
CONTINUITY
Continuity
DIFFERENTIABILITY
APPLICATION OF DERIVATIVES
MEAN VALUE THEOREM
Part B: FUNCTION OF TWO VARIABLES
INTRODUCTION
DIFFERENTIAL EQUATIONS OF FIRST ORDER AND FIRST DEGREE
ORTHOGONAL TRAJECTORY
DIFFERENTIAL EQUATION WITH CONSTANT COEFFICIENTS
CAUCHY EULER EQUATIONS
DIFFERENTIAL EQUATION OF SECOND ORDER
Section 5: LINEAR ALGEBRA
VECTOR SPACE AND LINEAR EQUATIONS
ORTHOGONALITY
EIGENVALUES AND EIGENVECTORS
LINEAR TRANSFORMATION
Section 6: GROUP THEORY
Unit Group
Subgroup
CYCLIC GROUP
PERMUTATION GROUP
ISOMORPHISM OF GROUPS
EXTRNAL DIRECT PRODUCT
HOMOMORPHISM OF GROUP
SET THEORY
Set Theory
REAL NUMBERS
TOPOLOGY ON REAL LINE
Open Sets
REAL SEQUENCES
INFINITE SERIES
POWER SERIES

# Introduction to Line Integral

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Basic Concept

Oriented Curve : Let C be a curve in space from A to B.

It can be oriented by taking one of the possible two directions as positive. If the direction from A to B, is taken as positive then the direction from B to A will be negative. If the initial point A and terminal point B coincide then the curve C is called closed cuve. (Figure 4.2)

Smooth Curve : Let the curve C is represented by parametric equation.

$x = x(t); y = y(t); z = z(t)$

Then each point on the curve C is represented by position vector.

$\vec r (t) = x(t) \hat i + y(t) \hat j + z(t) \hat k$

A curve is said to be smooth if the function $\vec r (t)$ is continuous and has continuous first derivatives not equal to zero for all values of t. There is unique tangent to each point on the curve C.

Piecewise Smooth Curve : A curve C is said to be piecewise smooth if it is composed of finite number of smooth curves. For example :-

The above curve (Figure 4.3) is piece wise smooth curve because it consists of four pieces of smooth curves AB, BC, CD and DE.

The rectangle consists of four piece of smooth curves.

Line Integrals : The integral which is carried on a curve is called line integral.

Mathematically, line integral is written as

$\int_{c} \vec F \cdot d \vec r$

On the curve C, only one variable is independent, other two variables are dependent.

If $\vec F = f_1 \hat i + f_2 \hat j + f_3 \hat k$

$\vec r = x \hat i + y \hat j + z \hat k$

$d \vec r = dx \hat i + dy \hat j + dz \hat k$

Then, $\vec F \cdot d \vec r = f_1 dx + f_2 dy + f_3 dz$ where $f_1, f_2, f_3$ are functions of x, y, z.

If we take x as independent variable, then y and z can be expressed in terms of x using the equation of curve

$y = g(x) \ \ \Rightarrow \ \ dy = g'(x) dx$

$z = h(x) \ \ \Rightarrow \ \ dz = h'(x) dx$

Now, $\vec F \cdot d \vec r = f_1 dx + f_2 dy + f_3 dz$ will be reduced to f(x) dx.

So, the line integral $\int_{c} \vec F \cdot d \vec r$ reduces to a simple definite integral.

$\int_{c} \vec F \cdot d \vec r = \int_{x_{1}}^{x_{2}} f(x)dx$

which can be solved using usual techniques.

Similarly, if y is taken as independent variable then x and z can be expressed in terms of y using equation of curve

$x = g(y) \ \ \Rightarrow \ \ dx = g'(y)dy$

$z = h(y) \ \ \Rightarrow \ \ dz = h'(y)dy$

So, $\vec F \cdot d \vec r = f_1 dx + f_2 dy + f_3 dz$ will be reduced to f(y) dy.

So, line integral $\int_{c} \vec F \cdot d \vec r$ reduces to a simple definite integral $\int_{y_{1}}^{y_{2}} f(y)dy$ which can be solved using usual techniques.

Similarly, if z is taken as independent variable, then x and y can be expressed in terms of z using equation of curve.

$x = g(z) \ \ \Rightarrow \ \ dx = g'(z) dz$

$y = h(z) \ \ \Rightarrow \ \ dy = h'(z) dz$

So, $\vec F \cdot d \vec r = f_1 dx + f_2 dy + f_3 dz$ will be reduced to f(z) dz.

So, line integral $\int_{c} \vec F \cdot d \vec r$ reduces to a simple definite integral $\int_{z_{1}}^{z_{2}} f(z) dz$ which can be solved using usual techniques.

Sometimes, the equation of curve given in terms of parameters, x, y, z are expressed in terms of parameters t using parametric equation of curve

$x = g_1 (t) \ \ \Rightarrow \ \ dx = g'_1 (t) dt$

$y = g_2 (t) \ \ \Rightarrow \ \ dy = g'_2 (t) dt$

$z = g_3 (t) \ \ \Rightarrow \ \ dz = g'_3 (t) dt$

So, $\vec F \cdot d \vec r = f_1 dx + f_2 dy + f_3 dz$ will be reduced to f(t) dt.

So, line integral $\int_{c} \vec F \cdot d\vec r$ reduced to a simple definite integral $\int_{t_{1}}^{t_{2}} f(t) dt$ which can be solved using usual techniques.

So, there are the four ways of solving line integral by converting it into a definite integral.