Question

Evaluate the circulation of G⃗ =xyi⃗ +zj⃗ +3yk⃗ around a square of side length 9, centered at the origin, lying in the yz-plane, and oriented counterclockwise when viewed from the positive x-axis.

Answer 1

Given that

`\vec G(x,y,z)=xy\,\vec\imath+z\,\vec\jmath+3y\,\vec k`

has a fairly simple curl,

`\\abla*\vec G(x,y,z)=2\,\vec\imath-x\,\vec k`

we can take advantage of Stokes' theorem by transforming the line integral of
`\vec G` along the boundary of the square (call it
`S`) to the integral of
`\\abla*\vec G` over
`S` itself. Parameterize
`S` by

`\vec s(u,v)=u\,\vec\jmath+v\,\vec k`

with
`-\frac92\le u\le\frac92` and
`-\frac92\le v\le\frac92`. Then take the normal vector to
`S` to be

`\vec s_u*\vec s_v=\vec\imath`

so that

`\displaystyle\int_(\partial S)\vec G\cdot\mathrm d\vec r=\iint_S(\\abla*\vec G)\cdot(\vec s_u*\vec s_v)\,\mathrm du\,\mathrm dv`

`=\displaystyle\int_(-9/2)^(9/2)\int_(-9/2)^(9/2)(2\,\vec\imath)\cdot(\vec\imath)\,\mathrm du\,\mathrm dv=\boxed{162}`

Answer 2

We have,the circulation of
`G=xyi+zi+3yk` around a square of side 3, centered at the origin, lying in the yz-plane, and oriented counterclockwise when viewed from the positive x-axis is

`\int_(\theta) G.dr=162`

From the Question we have the equation to be

`G=xyi+zi+3yk`

Therefore

`\triangle *G=\begin{Bmatrix}i & j & k\\(d)/(dx) & (d)/(dy) & (d)/(dz)\\xy&z&3y\end{Bmatrix}`

Where

For i

`i((d)/(dy)*3y-((d)/(dz)*xy))\\\\2i`

For j

`j((d)/(dx)*3y-((d)/(dz)*xy))`

`0j`

For z

`z((d)/(dy)*xy-((d)/(dx)*z))`

`xk`

Therefore

`\triangle *G=2i-xk`

Generally considering
`\theta` as the origin

We apply Stoke's Theorem

`\int_(\theta) G.dr=\int_(\theta)(\triangle *G) i ds`

`\int_(\theta) G.dr=\int_(\theta)(=2i-xk) i ds`

`\int_(\theta) G.dr=\int2ds`

`\int_(\theta) G.dr=2*9^2`

`\int_(\theta) G.dr=162`

Therefore

The circulation of
`G=xyi+zi+3yk` around a square of side 3, centered at the origin, lying in the yz-plane, and oriented counterclockwise when viewed from the positive x-axis is

`\int_(\theta) G.dr=162`