orientation problem in computing a 2-form integral

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Let $ \{(x, y, z) ∈ \mathbb{R}^3 | x^2 + y^2 = 1, 0\le z \le 1\}$ and $α = z^2 xdy$ defined over $\mathbb{R}^3$

Let $ω = dα$. Compute $\int_{\mathcal{C}} ω$.

If we use Stokes formulae: $\int_{\mathcal{C}} ω = \int_{\partial \mathcal{C}} \alpha = \int_{\mathbb{S}^1 \times \{0\}} z^2x dy - \int_{\mathbb{S}^1 \times \{1\}} z^2 x dy$.

1- Why does the orientation (the minus before the second integral) change between the two circles at $z=0$ and $z=1$?

2- If we use Stokes formulate differently with $V$ the volume of $\mathcal{C}$:

$\int_{\mathcal{C}=\partial V} \omega = \int_{V} dω == \int_{V} d\circ dω =0 $.

There is an orientation problem in the 1- and a false reasonning in the 2- I can't see.

Thank you for your help.

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This is a question of orientation, you have to choose a normal vector which points outward.

Inducing orientations on boundary manifolds