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 $C_4, x = 0, dx = 0, \vec F \cdot d\vec r = 0$

$\oint {\vec F \cdot d\vec r} = \int\limits_{{C_1}} {\vec F \cdot d\vec r} + \int\limits_{{C_2}} {\vec F \cdot d\vec r} + \int\limits_{{C_3}} {\vec F\cdot d\vec r} + \int\limits_{{C_4}} {\vec F \cdot d\vec r}$

$= \int\limits_0^a {{x^2}dx + \int\limits_0^b { - 2aydy} + \int\limits_a^0 {({x^2} + {b^2})dx} } + \int\limits_b^0 {0 \cdot dy}$

$= \left. {\dfrac{{{x^3}}}{3}} \right|_0^a + \left[ { - a{y^2}} \right]_0^b + \left[ {\dfrac{{{x^3}}}{3} + {b^2}x} \right]_a^0$

$= \dfrac{{{a^3}}}{3} - a{b^2} - \dfrac{{{a^3}}}{3} - a{b^2} = - 2ab^2$

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