Use Green’s theorem (Section 9) to do Problem 8.2.

It has been asked to use Green theorem.

A quantity that has magnitude as well as direction is called a vector. It is typically denoted by an arrow in which the head determines the direction of the vector and the length determines it magnitude.

For problem 8.2, Use Green's Theorem in plane. It states.

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

Here$\partial \mathrm{A}$is the boundary of the area A. So in problem 8.2 the changes observed is mentioned below.

$$\begin{gathered} \partial \mathrm{AP}=\mathrm{x}+2\mathrm{y}\to \frac{\partial \mathrm{P}}{\partial \mathrm{y}}=2\mathrm{and} \\ \mathrm{Q}=-2\mathrm{x}\to \frac{\partial \mathrm{Q}}{\partial \mathrm{x}}=-2 \end{gathered}$$

Thus, the integral over some contour C that encloses an area A.

$$\begin{gathered} {\oint }_{\mathrm{C}}(\mathrm{x}+2\mathrm{y})\mathrm{dx}-2\mathrm{xdy}=\int {\int }_{\mathrm{A}}(-2-2)\mathrm{dxdy} \\ =-4\int {\int }_{\mathrm{A}}(-2-2)\mathrm{dxdy} \\ =-4\mathrm{A} \end{gathered}$$

(a) The area A enclosed by the circle ${\mathrm{x}}^2+{\mathrm{y}}^2=1$is $\mathrm{\pi }{\left(1\right)}^2$ so the value of the line integral is -.

(b) The area enclosed by the square whose corners at (1,1), (-1,1), (-1,-1), (1,-1) is (2)(2) so the value of the line integral is -16.

(c) The area A enclosed by the square whose corners at (0,1), (-1,0), (0,-1), (1,0) is, $\left(1^2+1^2\right)=2$so the value of the line integral is.

Hence, the derived value is mentioned below.

$$\begin{gathered} \mathrm{l}=-4\mathrm{\pi } \\ \mathrm{l}=16 \\ \mathrm{l}=8 \end{gathered}$$