Correct Answer - Option 3 :
\(-\frac{\pi }{2}\)
Concept:
\(\mathop \oint \limits_c \left( {ydx - xdy} \right)\)
where C is \({x^2} + {y^2} = \frac{1}{4}\) \(\mathop \smallint \limits_2^1 gh = 0\)
By Green’s Theorem,
\(\mathop \smallint \nolimits \left( {\phi~dx + \psi~dy} \right) = \mathop \int\!\!\!\int \nolimits \left( {\frac{{\partial \psi }}{{\partial x}} - \frac{{\partial \phi }}{{\partial y}}} \right)dxdy\)
\(\phi = y\,and\,\psi = - x\)
\(\therefore \frac{{\delta \phi }}{{\delta y}} = 1\,and\,\frac{{\delta \psi }}{{\delta x}} = - 1\)
\(I = \mathop \int\!\!\!\int \nolimits \left( {\frac{{\partial \psi }}{{\partial x}} - \frac{{\partial \phi }}{{\partial y}}} \right)dxdy = \mathop \smallint \limits_0^{\frac{1}{2}} \mathop \smallint \limits_0^{2\pi } \left( { - 1 - 1} \right)~r~~d\theta~~dr\)
\( = - 2\left( {\mathop \smallint \limits_0^{\frac{1}{2}} r.\left( \theta \right)_0^{2\pi }} \right)dr\)
\( = - 2\mathop \smallint \limits_0^{\frac{1}{2}} r.\left( {2\pi } \right)dr\)
\( = - 2\left( {2\pi } \right)\left( {\frac{{{r^2}}}{2}} \right)_0^{\frac{1}{2}} = - 2\pi \left( {\frac{1}{2}} \right) = \frac{{ - \pi }}{{2}}\)
