Problems on Line Integral-1


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.