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.

• did you choose an orientation? – Xipan Xiao Nov 6 '16 at 2:38
• Orientation is what I'm not sure about. The question says that $M$ has the standard orientation. Does my parametrisation give the wrong orientation? – IAlreadyHaveAKey Nov 6 '16 at 2:43

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,