[/quote]Hint: Show that the path lies on the surface .[/QUOTE]
Have you done the "hint"? Do you know any identity for "sin(2t)"?
Stokes' theorem, since you title this thread "Stokes' Theorem", in its simplest form says that is a vector valued function then .
Here, and is the differential tangent vector, .
Since you know that the path lies entirely in the plane z= 2xy, you can write that as the position vector . The two derivative vectors, and are in the tangent plane to the surface (which here is the plane itself) and their cross product, is normal to the plane and its length gives the "differential of surface area". Assuming that the path integral was to be from 0 to (rather than from 0 to ), the normal is to be oriented by positive z so .
You still need to determine the bounds on x and y.