Generalised Stokes Theorem Contradiction I have a homework problem in which I am asked to verify the Generalised Stokes' Theorem for the manifold $M = \{(x,y,z)\in\mathbb R^3 : x^2 + y^2 = z, z\leq 1\}$ and the form $\omega = 4xy^2 dy + z^2 dz$. Here is my attempted solution:
$d\omega = (4y^2 dx + 8xy dy)\wedge dy + 2x dz\wedge dz = 4y^2 dx\wedge dy$. I parametrise $M$ via $\phi : (u,v) \mapsto (v\cos u, v\sin u, v^2)$ for $0\leq u\leq 2\pi$ and $0\leq v\leq 1$ and $\partial M$ via $\psi : t\mapsto (\cos t, \sin t, 1)$ for $0\leq t\leq 2\pi$.
So we have $$\phi^*(d\omega) = 4v^2\sin^2u(-v\sin u du + \cos u dv)\wedge(v\cos u du + \sin u dv)$$ $$= 4v^2\sin^2 u(-v\sin^2 du\wedge dv + v\cos^2 u dv\wedge du) = -4v^3\sin^2 u du\wedge dv.$$ We also have $$\psi^*\omega = 4\cos^2 t\sin^2 t dt = \sin^2 2t dt$$
Thus, $$\int_{\partial M} \omega = \int_0^{2\pi} \sin^2 2t dt = \pi$$
and $$\int_M d\omega = \int_0^1\int_0^{2\pi} - 4v^3\sin^2 u dudv = -\int_0^1 4v^3 dv\int_0^{2\pi} \sin^2 u du = -\pi.$$
Clearly I have done something wrong but I cannot for the life of me figure it out. I'm pretty sure the answer is supposed to be $\pi$ so I've made a mistake integrating over the surface itself but I can't figure it out. Any help would be greatly appreciated.
 A: The orientation on the boundary doesn't agree with that of the surface. The boundary integral should be $\int_{2\pi}^0$. See here for example,

"Around the edge of this surface we have a curve C.  This curve is called the boundary curve.  The orientation of the surface S will induce the positive orientation of C.  To get the positive orientation of C think of yourself as walking along the curve.  While you are walking along the curve if your head is pointing in the same direction as the unit normal vectors while the surface is on the left then you are walking in the positive direction on C."
Your surface is an upside-down version of the above picture, and the same orientation (outwardly pointing normals) would have you walk in the opposite direction along the boundary C otherwise you'd have the surface on your right.
If you don't like normals, think in terms of little circulating arrows instead as in this page: they'd be touching the boundary circle inducing an orientation which is opposite for the upward and downward pointing cups.
