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The vector function $F\left ( r \right )=-x\hat{i}+y\hat{j}$ is defined over a circular arc $C$ shown in the figure.

The line integral of $\int _{C} F\left ( r \right ).dr$ is

1. $\frac{1}{2}$
2. $\frac{1}{4}$
3. $\frac{1}{6}$
4. $\frac{1}{3}$

$\nabla \times F = \begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\{\frac {\partial }{\partial x}}&{\frac {\partial }{\partial y}}&{\frac {\partial }{\partial z}}\\-x & y & 0 \end{vmatrix} = 0$

Therefore, it is path independent integral and can be integrated along path y = 0 and x = y also.

For path y = 0, dy = 0

$\int _{1} ^{0} (-xdx + 0) = \dfrac{1}{2}$

For path x = y , dx = dy

$\int _{0} ^{\frac{1}{\sqrt{2}}} (-xdx +xdx ) = 0$

Add both path integrals to get $\frac{1}{2}$
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