Orientation of parametrization

Question

This is what I have so far for part (a). I am not entirely sure if this is how I should approach this or where to go from this point.

Answer 1

$T_u =\frac {\partial}{\partial u} (u \cos v, u^2, u\sin v) = (\cos x, 2u, \sin x)\\ T_v = \frac {\partial}{\partial v} (u \cos v, u^2, u\sin v) = (-u\sin v, 0, u\cos v)\\ T_u\times T_v = (2u^2\cos v, -u, 2u^2 \sin v) = u(2x, -1, 2z)$

Which is the reverse of the desired orientation.

$\iint F\cdot dS\\ \int_0^{2\pi}\int_0^1 (0, u^2,u^2\sin v\cos v)\cdot (-2u^2\cos v, u, -2u\sin v)\ du\ dv\\ \int_0^{2\pi}\int_0^1 u^3 - 2u^3\sin^2 v\cos v \ du\ dv\\ $

And I will let you take if from here.