Induced Orientation of Boundary of Manifold I am trying to work through the following problem: Let $M$ be the cylinder $S^1 \times [0,1]$ with counterclockwise orientation when viewed from the exterior. I am trying to figure out the boundary orientation on the $C_0 = S^1 \times \{0\}$ and on $C_1 = S^1 \times \{1\}$. 
I am not sure how to find an orientation form on the cylinder, as it seems to be different from the other manifolds I have worked with. Also, what would be the outward pointing vectors on the boundary? 
This is the image associated with the problem:

 A: *

*Take your cylinder $S^1 \times[0,1]$ and draw a small disk on the barrel part.

*Next, place a curved arrow on it to represent the counter clockwise direction.  

*Now just simply slide your disk to the boundary curves and you'll see the induced boundary. They will be opposite. 
The top should be clockwise with your right index finder and the bottom is counter-clockwise with your left index finger. 
A: This same exercise is 22.8 in the second edition of Loring Tu's book An Introduction to Manifolds. Here is my solution:
In order to describe the induced orientation on the boundary, we first need an outward-pointing vector field on the boundary. Consider the component $C_1 = S^1\times \{1\}$ at the top of the cylinder. If $t$ is the coordinate for $[0,1]$, then an outward-pointing vector field along $C_1$ is the vector field $\partial/\partial t$. Now, we take the interior product of $\partial/\partial t$ with the $2$-form $d\theta\wedge dt$:
\begin{align}
\iota_{\partial/\partial t}(d\theta\wedge dt) &= -\iota_{\partial/\partial t}(dt\wedge d\theta) \\
&= -d\theta.
\end{align}
The corresponding orientation for this orientation $1$-form on $C^1\cong S^1$ is the reverse of the usual orientation. Hence $C_1$ is oriented clockwise when viewed from the exterior as in the figure. Similarly, to figure out the orientation of the bottom of the cylinder, we take the interior product of the outward-pointing vector field $-\partial/\partial t$ on $C_0$ with $d\theta\wedge dt$, to get
\begin{align}
\iota_{-\partial/\partial t}(d\theta\wedge dt) &= -\iota_{\partial/\partial t}(d\theta\wedge dt) \\
&= \iota_{\partial/\partial t}(dt\wedge d\theta)\\
&= d\theta.
\end{align}
Thus the orientation on $C_0$ is the usual orientation of $S^1$, so it is counterclockwise when viewed from the exterior as in the figure.
