Question

$\mathbf{\text{∮}}\mathbf{V}\mathbf{\times }\mathbf{dr}$around the boundary of the square with vertices$\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{)}$ , if$\mathbf{V}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{i}\mathbf{+}\mathbf{5}\mathbf{xj}$

Short answer

The solution is$\int \mathrm{V}\times \mathrm{dr}=10$

Step-by-step solution

Given Information.

The given information is,$\mathrm{V}={\mathrm{x}}^2\mathrm{i}+5\mathrm{xj}$

Definition of Green’s Theorem.

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.

For, $\mathrm{V}={\mathrm{x}}^2\mathrm{i}+5\mathrm{xj}$

$\int \mathrm{V}\times \mathrm{dx}=\int {\mathrm{x}}^2\mathrm{dx}+5\mathrm{xdy}$

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

It can be seen that

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

$=0$

role="math" $\mathrm{Q}=5\mathrm{x}\to \frac{\mathrm{\partial }\mathrm{Q}}{\mathrm{\partial }\mathrm{x}}$

$=5$

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

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

$=10$

Hence, the solution is $\int \mathrm{V}\times \mathrm{dr}=10$.