a. Take
[tex]\vec r(t)=\sqrt3\cos t\,\vec\imath+\sqrt3\sin t\,\vec\jmath[/tex]
with [tex]0\le t\le\pi[/tex].
b. The line integral reduces to
[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_0^\pi(3\cos^2t\,\vec\imath+3\sin^2t\,\vec\jmath)\cdot(-\sqrt3\sin t\,\vec\imath+\sqrt3\cos t\,\vec\jmath)\,\mathrm dt[/tex]
[tex]=\displaystyle3\sqrt3\int_0^\pi(\sin^2t\cos t-\cos^2t\sin t)\,\mathrm dt[/tex]
c.
[tex]\displaystyle3\sqrt3\left(\int_0^\pi\sin^2t\cos t\,\mathrm dt-\int_0^\pi\cos^2t\sin t\,\mathrm dt\right)[/tex]
[tex]=\displaystyle3\sqrt3\left(\int_0^0u^2\,\mathrm du+\int_1^{-1}v^2\,\mathrm dv\right)[/tex]
(where [tex]u=\sin t[/tex] and [tex]v=\cos t[/tex])
[tex]=\displaystyle-6\sqrt3\int_0^1v^2\,\mathrm dv=\boxed{-2\sqrt3}[/tex]
