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# 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.