Question

For a simple closed curve Cin the plane show by Green’s theorem that the area inclosed is

$\mathit{A}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\oint }}_{\mathbf{C}}\mathbf{(}\mathit{x}\mathit{d}\mathit{y}\mathbf{-}\mathit{y}\mathit{d}\mathit{x}\mathbf{)}$

Short answer

The solution to this problem is that the condition is satisfied and the area is inclosed.

Step-by-step solution

Given Information.

The given information is $A=\frac{1}{2}{\oint }_{\mathrm{C}}(xdy-ydx)$.

Definition of Green’s Theorem

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.

Find the solution.

Area of a simple curve is obtained by using the formula below.

$\iint dydx$

Compare the area formula’s equation to Green’s Theorem.

role="math" localid="1659153988451" $\frac{\partial Q}{\partial z}-\frac{\partial P}{\partial y}=1$

Find the differentiation with respect to.

$P=\frac{-y}{x}Q=\frac{x}{2}$

Use the value of P and Q and find the value of their derivatives with respect to z and y, respectively.

$$\begin{gathered} \frac{\partial Q}{\partial z}=\frac{1}{2} \\ \frac{\partial P}{\partial y}=\frac{-1}{2} \end{gathered}$$

The values of the derivatives satisfy the condition.