By Green’s theorem

$\displaystyle \int_c (Pdx+Qdy) = \iint_R \left ( \dfrac {\partial Q}{\partial x} - \dfrac {\partial P}{\partial y} \right )dx dy $

$Here \ P=x^2 -y \ Q=2y^2+x$

$\dfrac {\partial P}{\partial y} = -1, $

$\dfrac {\partial Q}{\partial x} =1$

$(a) \ Along \ C_1, y=x^2$

$\displaystyle \therefore dy = 2xx \therefore \int_{c_1} (Pdx + Qdy)$

$\displaystyle =\int^{2}_0 [(x^2 - x^2 ) + (2x^4 +x) 2x]dx$

$\displaystyle = \int^2_0 (4x^5 +2x^2 ) dx = \left [ \dfrac {2x^6}{3} + \dfrac {2x^3}{3} \right ]^2_0 = \dfrac {128}{3}+ \dfrac {16}{3} = \dfrac {144}{3}$

$Alogn \ C_2 , y=4$

$\displaystyle \therefore \ dy = 0 \therefore \int_{c_2} (Pdx + Qdy) = \int^0_2 (x^2 -4)dx = \left [ \dfrac {x^3}{3}-4 \right ]^0_2 = 0 - \left ( \dfrac {8}{3}-8 \right ) = \dfrac {16}{3}$

$Along \ C_3 ,x =0$

$\displaystyle \therefore dx =0 \therefore \int_{c_3} (Pdx+Qdy) = \int^0_4 (2y^2) dy = \left [ \dfrac {2y^3}{3} \right ]^0_4 = 0- \dfrac {2}{3} .64 = - \dfrac {128}{3}$

$\displaystyle \int_{c}(Pdx + Qdy) = \dfrac {144}{3}+ \dfrac {16}{3} - \dfrac {128}{3} = \dfrac {32}{3}$

$\displaystyle (b) \ \iint_{R} \left ( \dfrac {\partial Q} {\partial x} - \dfrac {\partial p}{\partial y} \right )dxdy = \iint (1+1) dxdy = \int^4_0 \int^{\sqrt{y}}_0 2dxdy = 2 \int^4_0 [x]^{\sqrt{y}}_0 = 2 \int^4_0 \sqrt{y}{dy} 2 \cdot \dfrac {2}{3} [y^{3/2}]^4_0 = \dfrac {4}{3} \cdot 8 =32/3$