Problem II 4.1(b)  Find the integral surface for
V = (x, -y, 0) containing the curve C: x = t, y = t, z = t^2; 0 < t < infinity.
The equation for the integral surface is wrong.  You need to figure it out and run it to show that the curve lies in the surface.

u1[x_,y_,z_] := x y
u2[z_, z_, z_] := z

curve1 = ParametricPlot3D[{t, t, t^2}, {t, 0, 1}]
surface1 = ContourPlot3D[z  == 0, {x, 0,1}, {y, 0, 1}, {z, 0, 1}]
Show[curve1, surface1]

Well, that was sort of boring.  How about the integral surface for the same vector field that contains the helix
x = cos(t), y = sin(t), z = t/π  for all t.

curve2 = ParametricPlot3D[{Cos[t], Sin[t], t/π}, {t, -π, π}]
surface2 = ContourPlot3D[z  == 0, {x, -1.5,1.5}, {y, -1.5, 1.5}, {z, -1, 1}]
Show[curve2, surface2]