 ### 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

# Problems on Line Integral-2

Lesson Progress
0% Complete Example : 1

Evaluate the line integral $\int\limits_c {\vec F \cdot d\vec r}$ where $\vec{F} = (x + 2y) \hat{i} + (2y - x)\hat{j}$ and C is curve in xy plane consisting of the straight lines from (0, 0) to (1, 0) and then to (3, 4).

Solution :

The curve C consists of two pieces of smooth curves $C_1$ and $C_2$. $C_1$ is straight line from (0, 0) to (1, 0) ie. y = 0 $C_2$ is straight line from (1, 0) to (3, 4)

i.e. $y - 0 = \left( {\dfrac{{4-0}}{{3-1}}} \right) \cdot (x-1)$

or, $y = 2x - 2$

So, along $C_1, y = 0, dy = 0$ (x is an independent variable) $\vec F \cdot d\vec r = x dx$

Along $C_2; y = 2x - 2, dy = 2dx$ (let us take x as independent variables). $\vec F \cdot d\vec r = (x + 2y) dx + (2y - x) dy$

on $C_2, \ \ \ \ \vec F \cdot d\vec r = (x + 2(2x - 2))dx + (2(2x - 2) - x) \cdot 2dx$ $= (11x - 12)dx$

So, $\int\limits_C {\vec F \cdot d\vec r} = \int\limits_{{C_1}} {\vec F \cdot d\vec r} + \int\limits_{{C_2}} {\vec F \cdot d\vec r}$ $= \int\limits_0^1 {xdx} + \int\limits_1^3 {(11x - 12)dx}$ $= \left. {\dfrac{{{x^2}}}{2}} \right|_0^1 + \left. {\left( {\dfrac{{11}}{2}{x^2} - 12x} \right)} \right|_1^3 = 20.5$

Example 2 :

Evaluate $\oint\limits_C {\vec F \cdot d\vec r}$ where $\vec F = ({x^2} + {y^2})\hat i - 2xy\hat j$ where curve C is a rectangle in the xy plane bounded by y = 0, x = a, y = b, x = 0.

Solution :

The curve C as shown in figure consists of four pieces of smooth curves $C_1, C_2, C_3 \ \& \ C_4.$ $\vec F \cdot d\vec r] = (x^2 + y^2)dx - 2xydy$

On $C_1, y = 0, dy = 0, \vec F \cdot d\vec r = x^2 dx$

On $C_2, x = a, dx = 0, \vec F \cdot d\vec r = - 2ay dy$

On $C_3, y = b, dy = 0, \vec F \cdot d\vec r = (x^2 + b^2)dx$

On $C_4, x = 0, dx = 0, \vec F \cdot d\vec r = 0$ $\oint {\vec F \cdot d\vec r} = \int\limits_{{C_1}} {\vec F \cdot d\vec r} + \int\limits_{{C_2}} {\vec F \cdot d\vec r} + \int\limits_{{C_3}} {\vec F\cdot d\vec r} + \int\limits_{{C_4}} {\vec F \cdot d\vec r}$ $= \int\limits_0^a {{x^2}dx + \int\limits_0^b { - 2aydy} + \int\limits_a^0 {({x^2} + {b^2})dx} } + \int\limits_b^0 {0 \cdot dy}$ $= \left. {\dfrac{{{x^3}}}{3}} \right|_0^a + \left[ { - a{y^2}} \right]_0^b + \left[ {\dfrac{{{x^3}}}{3} + {b^2}x} \right]_a^0$ $= \dfrac{{{a^3}}}{3} - a{b^2} - \dfrac{{{a^3}}}{3} - a{b^2} = - 2ab^2$