Is every top-dimensional manifold-with-boundary orientable? If this is not true, is there an easy-to-understand counterexample?
EDIT: Some context in case this question was unclear. The hypotheses of Stokes' theorem on a manifold-with-boundary $M$ obviously requires $M$ to be orientable. But most statements of Green's theorem (where $M$ is a 2-manifold-with-boundary in $\mathbb{R}^2$) or the divergence theorem (where $M$ is a 3-manifold-with-boundary in $\mathbb{R}^3$) do not explicitly require $M$ to be orientable. Presumably, they already are orientable because they are top-dimensional, and although this seems intuitively obvious, I can't seem to come up with an easy proof of this, so maybe it isn't even true.
Interestingly, if one strictly sticks with following definition (from Spivak's Calculus on Manifolds), for example that $x$ belongs to $\partial M$ iff there is an open set $U$ containing $x$, and open set $V$, and a diffeomorphism $h : U \to V$ where $h(M \cap U) = V \cap (\mathbb{R}^{k-1} \times \mathbb{R}_+ \times \{0\})$ and $h_k(x) = 0$, then not even the interval $[a, b]$ is orientable. As was mentioned in another post, Loring Tu's book fixes this anomaly by allowing only in the one-dimensional case for the right half-space to be replaced by the left half-space.
OK, I think I have answered my own question.
Let $M \subset \mathbb{R}^n$ be an $n$-manifold-with-boundary, where $n \geq 2$. Then $M$ is orientable.
Proof: We first note that since $M$ is a $n$-manifold in $\mathbb{R}^n$, then points in $M - \partial M$ are actually interior points of $M$. For note that $x \in M - \partial M$ means that there is some open set $W \subset \mathbb{R}^n$ and an open set $U$ containing $x$ and a coordinate system $f: W \to U$ where $f(W) = M \cap U$. But by the inverse function theorem, $f(W)$ must contain an open set containing $x$, so in fact $x$ is an interior point of $M$.
Now let $I^n$ be the identity function from the interior of $M$ to itself. Obviously, $I^n(\text{int } M) = \text{int } M$, so $I^n$ is a coordinate system of $M$, and every point in $\text{int } M$ is part of $M - \partial M$.
We now construct an oriented atlas of $M$ whose volume form is just the determinant on $\mathbb{R}^{n \times n}$. Of course, $I^n$ takes care of all points of $M - \partial M$, with \begin{equation*} ((I^n)^*\det)(e_1, \dots, e_n) = \det\begin{bmatrix} e_1 & \dots & e_n \end{bmatrix} = 1 \end{equation*} Now for any $x \in \partial M$, there is an open set $U$ containing $x$, and open set $W$, and a coordinate system $f : W \to U$ where $f(W_+) = M \cap U$ (here we define $W_+$ be the set of all $w \in W$ where $w_n \geq 0$). In fact, we can restrict $f$ to an open rectangle $W_0 \subset W$ containing $a \equiv f^{-1}(x)$. By the inverse function theorem again, $U_0 \equiv f(W_0)$ is an open set containing $x$, and moreover $f((W_0)_+) = M \cap U_0$. Now note that \begin{equation*} (f^*\det)_p(e_1, \dots, e_n) = \det Df(p) \neq 0 \end{equation*} for all points $p \in W_0$. In fact, by the intermediate value theorem, the above quantity must either be all positive or all negative for all $p \in W_0$. If the quantity is negative, then we can flip the sign of the first coordinate (we can do this since $n \geq 2$), thus changing the sign of the column $D_1 f(p)$ so that the determinant above becomes always positive. In this way, we can construct an atlas of coordinate systems of $M$ whose orientations with respect to the determinant is always positive.