Given that
[tex]\vec G(x,y,z)=xy\,\vec\imath+z\,\vec\jmath+3y\,\vec k[/tex]
has a fairly simple curl,
[tex]\nabla\times\vec G(x,y,z)=2\,\vec\imath-x\,\vec k[/tex]
we can take advantage of Stokes' theorem by transforming the line integral of [tex]\vec G[/tex] along the boundary of the square (call it [tex]S[/tex]) to the integral of [tex]\nabla\times\vec G[/tex] over [tex]S[/tex] itself. Parameterize [tex]S[/tex] by
[tex]\vec s(u,v)=u\,\vec\jmath+v\,\vec k[/tex]
with [tex]-\dfrac92\le u\le\dfrac92[/tex] and [tex]-\dfrac92\le v\le\dfrac92[/tex]. Then take the normal vector to [tex]S[/tex] to be
[tex]\vec s_u\times\vec s_v=\vec\imath[/tex]
so that
[tex]\displaystyle\int_{\partial S}\vec G\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec G)\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle\int_{-9/2}^{9/2}\int_{-9/2}^{9/2}(2\,\vec\imath)\cdot(\vec\imath)\,\mathrm du\,\mathrm dv=\boxed{162}[/tex]
We have,the circulation of [tex]G=xyi+zi+3yk[/tex] 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
[tex]\int_{\theta} G.dr=162[/tex]
From the Question we have the equation to be
[tex]G=xyi+zi+3yk[/tex]
Therefore
[tex]\triangle *G=\begin{Bmatrix}i & j & k\\\frac{d}{dx} & \frac{d}{dy} & \frac{d}{dz}\\xy&z&3y\end{Bmatrix}[/tex]
Where
For i
[tex]i(\frac{d}{dy}*3y-(\frac{d}{dz}*xy))\\\\2i[/tex]
For j
[tex]j(\frac{d}{dx}*3y-(\frac{d}{dz}*xy))[/tex]
[tex]0j[/tex]
For z
[tex]z(\frac{d}{dy}*xy-(\frac{d}{dx}*z))[/tex]
[tex]xk[/tex]
Therefore
[tex]\triangle *G=2i-xk[/tex]
Generally considering [tex]\theta[/tex] as the origin
We apply Stoke's Theorem
[tex]\int_{\theta} G.dr=\int_{\theta}(\triangle *G) i ds[/tex]
[tex]\int_{\theta} G.dr=\int_{\theta}(=2i-xk) i ds[/tex]
[tex]\int_{\theta} G.dr=\int2ds[/tex]
[tex]\int_{\theta} G.dr=2*9^2[/tex]
[tex]\int_{\theta} G.dr=162[/tex]
Therefore
The circulation of [tex]G=xyi+zi+3yk[/tex] 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
[tex]\int_{\theta} G.dr=162[/tex]
