91Ó°ÊÓ

The given information is,

${âˆ\circledR }_c\left(y^2-x^2\right)dx+\left(2xy+3\right)dy$

The Green's theorem connects a line integral around a simple closed curve C to a double integral over the plane region D circumscribed by C in vector calculus. Stokes' theorem has a two-dimensional special case.

Consider the closed contour C that consists of two parts.

Part 1: $C_1$, which is given in the problem.

Part 2: $C_2$, which is a straight line from (1,2) to (0,0).

${âˆ\circledR }_c\left(y^2-x^2\right)dx+\left(2xy+3\right)dy$

Use Green’s Theorem.

$∫{∫}_A\left(\frac{∂Q}{∂x}-\frac{∂P}{∂y}\right)dxdy={âˆ\circledR }_{∂A}Pdx+Qdy$

Where$∂A$is the boundary of the area A.

It can be seen that

$$\begin{gathered} P=y^2-x^2→\frac{∂P}{∂y} \\ =2y \\ Q=2xy+3→\frac{∂Q}{∂x} \\ =2y \end{gathered}$$

${âˆ\circledR }_C\left(y^2-x^2\right)dx+\left(2xy+3\right)dy={âˆ\neg }_A\left(2y-2y\right)dxdy$

But,

$$\begin{gathered} {âˆ\circledR }_C\left(y^2-x^2\right)dx+\left(2xy+3\right)dy={âˆ\neg }_{C1}\left(y^2-x^2\right)dx+\left(2xy+3\right)dy+{∫}_{C2}\left(y^2-x^2\right)dx+\left(2xy+3\right)dy \\ =0 \end{gathered}$$

Evaluate the integral over the contour. $C_2$

$$\begin{gathered} {âˆ\circledR }_C\left(y^2-x^2\right)dx+\left(2xy+3\right)dy={∫}_1^0\left(4x^2-x^2\right)dx\left(4x^2-3\right)\left(2dy\right) \\ ={∫}_1^0\left(\frac{11}{3}{x}^3+6x\right){\overline{)}}_1^0 \\ =\frac{29}{3} \end{gathered}$$

Hence, the Solution to the problem is

${âˆ\circledR }_C\left(y^2-x^2\right)dx+\left(2xy+3\right)dy=\frac{29}{3}$.