$\mathbf{\int }\left(y^2-x^2\right)\mathit{d}\mathit{x}\mathbf{+}\left(2xy+3\right)\mathit{d}\mathit{y}$along the x axis from (0,0) to$\left(\sqrt{5},0\right)$ and along a circular are from$\mathbf{(}\sqrt{\mathbf{5}}\mathbf{,}\mathbf{0}\mathbf{)}$ to (1,2).

The given information is,

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

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).

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

Use Green’s Theorem.

$\int {\int }_A\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy={\oint }_{\partial A}Pdx+Qdy$

Where$\partial A$is the boundary of the area A.

It can be seen that

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

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

But,

$$\begin{gathered} {\oint }_C\left(y^2-x^2\right)dx+\left(2xy+3\right)dy={\iint }_{C1}\left(y^2-x^2\right)dx+\left(2xy+3\right)dy+{\int }_{C2}\left(y^2-x^2\right)dx+\left(2xy+3\right)dy \\ =0 \end{gathered}$$

Evaluate the integral over the contour. $C_2$

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

Hence, the Solution to the problem is

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