We can choose [tex]x=6\cos t[/tex] and [tex]y=6\cos t[/tex], so that the path [tex]C[/tex] is traversed when [tex]-\dfrac\pi2\le t\le\dfrac\pi2[/tex].
Now
[tex]\displaystyle\int_C 3xy^2\,\mathrm dS=3\int_{t=-\pi/2}^{t=\pi/2}(6\cos t)(6\sin t)^2\sqrt{x'(t)^2+y'(t)^2}\,\mathrm dt[/tex]
[tex]=\displaystyle648\int_{t=-\pi/2}^{t=\pi/2}\cos t\sin t^2\sqrt{36\cos^2t+36\sin^2t}\,\mathrm dt[/tex]
[tex]=\displaystyle3888\int_{t=-\pi/2}^{t=\pi/2}\cos t\sin t^2\,\mathrm dt[/tex]
Taking [tex]u=\sin t[/tex], we have
[tex]=\displaystyle3888\int_{u=-1}^{u=1}u^2\,\mathrm du[/tex]
[tex]=1296(1^3-(-1)^3)[/tex]
[tex]=2592[/tex]
