Question

${\mathbf{\int }}_{\mathbf{C}}\mathbf{ }\left(x^2-y\right)\mathbf{dx}\mathbf{+}\left(x+y^3\right)\mathbf{dy}$: where $\mathbf{C}$is the parallelogram with vertices $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{1}\mathbf{)}$.

Short answer

The solution is${\text{∮}}_{\mathrm{c}}\left({\mathrm{x}}^2-\mathrm{y}\right)\mathrm{dx}+\left(\mathrm{x}+{\mathrm{y}}^3\right)\mathrm{dy}=4$

Step-by-step solution

Given Information.

The given information is,${\text{∮}}_{\mathrm{C}}\left({\mathrm{x}}^2-\mathrm{y}\right)\mathrm{dx}+\left(\mathrm{x}+{\mathrm{y}}^3\right)\mathrm{dy}$

Definition of Green’s Theorem.

Green's theorem connects a line integral around a simple closed curve 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.

${\text{∮}}_{\mathrm{c}}\left({\mathrm{x}}^2-\mathrm{y}\right)\mathrm{dx}+\left(\mathrm{x}+{\mathrm{y}}^3\right)\mathrm{dy}$

Use Green’s Theorem role="math" ${\iint }_{\mathrm{A}} \left(\frac{\mathrm{\partial }\mathrm{Q}}{\mathrm{\partial }\mathrm{x}}-\frac{\mathrm{\partial }\mathrm{P}}{\mathrm{\partial }\mathrm{y}}\right)\mathrm{dxdy}={\text{∮}}_{\mathrm{\partial }\mathrm{A}}\mathrm{Pdx}+\mathrm{Qdy}$, where is the boundary of the area $\mathrm{A}$.

It can be seen that

$\mathrm{P}={\mathrm{x}}^2-\mathrm{y}\to \frac{\mathrm{\partial }\mathrm{P}}{\mathrm{\partial }\mathrm{y}}$

$=-1$

$\mathrm{Q}=\mathrm{x}+{\mathrm{y}}^3\to \frac{\mathrm{\partial }\mathrm{Q}}{\mathrm{\partial }\mathrm{x}}$

$=1$

Since the contour is the parallelogram with the given vertices, the area $\mathrm{A}$is then the area of the paralleologram.

${\text{∮}}_{\mathrm{c}}\left({\mathrm{x}}^2-\mathrm{y}\right)\mathrm{dx}+\left(\mathrm{x}+{\mathrm{y}}^3\right)\mathrm{dy}=2\left(2\right)\left(1\right)$

$=4$

$\int \mathrm{V}\times \mathrm{dr}=5{\left(\sqrt{(1-0{)}^2+(0-1{)}^2}\right)}^2$

$=10$

Hence, the solution is ${\text{∮}}_{\mathrm{c}}\left({\mathrm{x}}^2-\mathrm{y}\right)\mathrm{dx}+\left(\mathrm{x}+{\mathrm{y}}^3\right)\mathrm{dy}=4$.