If $C=M \times [0,1]$, what is an integral over $\partial C$? Let $M$ be a compact Riemannian manifold. Define $C=M \times [0,1]$ so that $\partial C = M \times \{0\} \cup M \times \{1\}$.
If $f:\partial C \to \mathbb{R}$, is then the integral over $\partial C$ of $f$ defined as
$$\int_{\partial C}f = \int_M f(\cdot, 0) + \int_M f(\cdot, 1)?$$
 A: This is integration over a manifold with boundary that we are looking at here. Hence, you should be integrating a differential $ n $-form over $ \partial C $, where $ n = \dim(M) $. A smooth function $ f: \partial C \to \Bbb{R} $ is only a differential $ 0 $-form, so your expression is well-defined if and only if $ M $ is $ 0 $-dimensional. This means that $ M $ is a finite set of points, as you have assumed $ M $ to be compact. If you want to know how to integrate a differential $ n $-form, please refer to this Wikipedia article.
Note: You need to assume that $ M $ is oriented in order to do integration in the manner explained by the Wikipedia article; otherwise you need to use the concept of a density.
As a working example, suppose that $ M $ is a finite collection of points. Then for any smooth function $ f: C \to \Bbb{R} $, we have
$$
\int_{\partial C} f = \sum_{p \in M} [f(p,1) - f(p,0)],
$$
where the natural orientation of $ [0,1] $ is being used.
Now, the definition of the boundary of a manifold (with boundary) requires that the boundary be a subset of the manifold itself. Hence, it would be preferable if you write $ C = M \times [0,1] $. I suppose that you were thinking of the topological boundary of $ M \times (0,1) $ inside $ M \times \Bbb{R} $.
