Answer :
Looks like we're given
[tex]\vec F(x,y)=\langle-x,-y\rangle[/tex]
which in three dimensions could be expressed as
[tex]\vec F(x,y)=\langle-x,-y,0\rangle[/tex]
and this has curl
[tex]\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle[/tex]
which confirms the two-dimensional curl is 0.
It also looks like the region [tex]R[/tex] is the disk [tex]x^2+y^2\le5[/tex]. Green's theorem says the integral of [tex]\vec F[/tex] along the boundary of [tex]R[/tex] is equal to the integral of the two-dimensional curl of [tex]\vec F[/tex] over the interior of [tex]R[/tex]:
[tex]\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA[/tex]
which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of [tex]R[/tex] by
[tex]\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle[/tex]
[tex]\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt[/tex]
with [tex]0\le t\le2\pi[/tex]. Then
[tex]\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt[/tex]
[tex]=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0[/tex]