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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.

Conjecture
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1 Answers1

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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

Tsemo Aristide
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  • Thank you for the link. In that case, If we choose counter clockwise to be the direct direction and switch to cylindrical coordinates, I feel like the minus sign should rather go with the first integral: as $\theta$ increases counter clockwise, the normal vector points inward for $\mathbb{S}^1 \times {0}$ and outwards for $\mathbb{S}^1 \times {1}$? – Conjecture Dec 30 '18 at 21:44
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    The best way tounderstand this is to consider a continuous $f:[a,b]\rightarrow \mathbb{R}$, $\int_a^bf(t)dt=F(b)-F(a)$ where $F'=f$, the orientation of $\mathbb{R}$ points towards $+\infty$, the induced orientation on $a$ in the interval $[a,b]$ points towards $-\infty$ because it points outwards. This is exactly the same if you consider a cube in $\mathbb{R}^n$ even something more complicated by using Fubini. – Tsemo Aristide Dec 30 '18 at 21:50
  • Concerning Stokes, the cylinder is not a boundary. – Thomas Dec 31 '18 at 15:14