The area of the surface is
[tex]\displaystyle2\pi\int_0^{\pi/6}|y(t)|\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt[/tex]
[tex]t\cos t\ge0[/tex] for all [tex]0\le t\le\dfrac\pi6[/tex], and
[tex]\dfrac{\mathrm dx}{\mathrm dt}=t\cos t+\sin t[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dt}=-t\sin t+\cos t[/tex]
[tex]\implies\left(\dfrac{\mathrm dx}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy}{\mathrm dt}\right)^2=(t\cos t+\sin t)^2+(t\sin t-\cos t)^2=t^2+1[/tex]
The integral is then
[tex]\displaystyle2\pi\int_0^{\pi/6}t\cos t\sqrt{t^2+1}\,\mathrm dt\approx0.8547[/tex]
