DOUBLE INTEGRAL
BETA AND GAMMA FUNCTION
VOLUME
SURFACE AREA
GRADIENT, DIVERGENCE AND CURL
LINE INTEGRAL
GREENS THEOREM
SURFACE INTEGRAL
GAUSS DIVERGENCE THEOREM
STOKES THEOREM
CONSERVATIVE VECTOR FIELD
LIMITS
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
CYCLIC GROUP
PERMUTATION GROUP
ISOMORPHISM OF GROUPS
EXTRNAL DIRECT PRODUCT
HOMOMORPHISM OF GROUP
SET THEORY
REAL NUMBERS
TOPOLOGY ON REAL LINE
REAL SEQUENCES
INFINITE SERIES
POWER SERIES

Problems on Line Integral-1

video

Example 1 :

If  \vec F = 3xy\hat i - {y^2}\hat j determine the value of  \int\limits_c {\vec F.d\vec r} where C is the curve  y = 2x^2 in the xy plane from (0, 0) to (1, 2).

Solution :

The curve lies in xy plane, so, z = 0. z can never be taken as independent variable z is a dependent variable. Now, out of x and y, any one variable can be taken as independent.
Suppose x is taken as independent variable

 y = 2x^2, dy = 4xdx

 \vec F \cdot d\vec r = 3xydx - y^2dy

 = 6x^3dx - 4x^4 \cdot 4xdx

 = (6x^3 - 16x^5)dx


So, the line integral  \int\limits_C {\vec f \cdot d\vec r} reduces to a definite integral.

 \int\limits_0^1 {(6{x^3} - 16{x^5})dx} = \left. {6\dfrac{{{x^4}}}{4}} \right|_0^1\left. { - 16\dfrac{{{x^6}}}{6}} \right|_0^1

 = - \dfrac{7}{6}

If y is taken as independent variable then x can be expressed in terms of y as

 x = \sqrt {\dfrac{y}{2}}, dx = \dfrac{1}{{2\sqrt 2 }}\dfrac{1}{{\sqrt y }} \ dy

So,  \vec f \cdot d\vec r = 3xydx - y^2dy

 = 3y\sqrt {\dfrac{y}{2}} \cdot \dfrac{1}{{2\sqrt 2 }} \dfrac{1}{{\sqrt y }}dy - y^2dy

 = \left( {\dfrac{3}{4}y - {y^2}} \right)dy

So, the line integral  \int {\vec f \cdot d\vec r} reduces to a definite integral

 \int\limits_0^2 {\left( {\dfrac{3}{4}y - {y^2}} \right)dy} = \left. {\dfrac{3}{8}{y^2} - \dfrac{{{y^3}}}{3}} \right|_0^2 = - \dfrac{7}{6}

Example : 1

Evaluate  \oint {xdy - ydx} around a circle  x^2 + y^2 = r^2

Solution :

Let C denotes the circle. The parametric equations of circle is

 x = r \cos \theta

 y = r \sin \theta

Here, x and y have been expressed in terms of parameter which varies from 0 to  2 \pi as one traverses the circle.

 x = r \cos \theta \Rightarrow dx = - r \sin \theta \ d \theta

 x = r \sin \theta \Rightarrow dy = r \cos \theta \ d \theta

 x dy - y dx = r \cos \theta r \cos \theta d \theta - r sin \theta (- r sin \theta) d \theta


 = r^2d \theta

So,  \oint\limits_c {xdy - ydx = {r^2}\oint {d\theta } } = 2 \pi r^2

Here, r is a constant, because integral is carried over a circle.