I'm having trouble verifying Stokes' Theorem, and I'm just wondering if someone's able to find my mistake in the calculation. Let $\omega=x^2\text{d}w-2yz \text{d}x$ be a one-form on $\mathbb{R}^4$, and let $M\subset\mathbb{R}^4$ be the image of the map $\varphi:[0,1]\times[0,2\pi]\to\mathbb{R}^4, (u,v)\mapsto (u\cos v,u\sin v,u,-1)$.
First, I'll calculate $\int_{M} \text{d}\omega$:
$\text{d}\omega=2x\text{d}w\wedge\text{d}x-2z\text{d}x\wedge\text{d}y-2y\text{d}x\wedge\text{d}z,$
Thus, we have
$$\text{d}\omega_{\varphi(u,v)}\left(\frac{\partial\varphi}{\partial u}, \frac{\partial\varphi}{\partial v}\right) = \omega_{\langle u\cos v, u\sin v,u,-1\rangle}\left( \langle \cos v, \sin v,1,-1\rangle,\langle -u\sin v, u\cos v,0,0\rangle \right)$$
$$=2u\cos v \text{d}t\wedge\text{d}x - 2zdx\wedge dy-2ydx\wedge dz$$
$$=2u\cos v\begin{vmatrix} 1 & \cos v\\ 0 & -u\sin v \end{vmatrix} +-2u\begin{vmatrix} \cos v & \sin v\\ -u\sin v & u\cos v \end{vmatrix} -2u\sin v \begin{vmatrix} \cos v & 1\\ -u\sin v & 0 \end{vmatrix}$$
$$=2u\cos v (-\sin v)-2u^2 + 2u^2 \sin^2 v$$
$$=2u^2 (\cos^2 v - \sin v\cos v)$$
And so to calculate the integral, we see
$$\int_M \text{d}\omega=\int_{0}^{2\pi}\int_0 ^1 2u^2(\cos^2 v - \sin v\cos v)\text{d}u\text{d}v=\frac{2\pi}{3}$$
And now we will calculate $\int_{\partial M} \omega$. Notice $\partial M=\text{im}(\psi),$ where $\psi(t)=(\cos t, \sin t, 1,-1),t\in[0,2\pi].$
$$\omega_{\psi(t)} \left(\frac{\partial \psi}{\partial t}\right)$$
$$=\omega_{\langle \cos t,\sin t, 1,-1\rangle} \left( \langle -\sin t,\cos t,0,0\rangle \right)$$
$$=\cos^2 t \cdot 0 - 2(1)(-1)(-\sin t)$$
$$=-2\sin t$$
And next, we can calculate the integral:
$$\int_{\partial M} \omega = \int_0^{2\pi}-2\sin t\text{d}t =0$$
I've clearly made an error in here somewhere, because I got two different answers. Can someone help?