TITLE: Atlas for the closed ball.
QUESTION [6 upvotes]: Consider the manifold with boundary $$D:= \{x \in \mathbb{R}^2 : \Vert x \Vert  \leq 1\}$$
I want to calculate $$\int_D dx \wedge dy$$
using the definition of the integral.
The integral is defined this way in my book:

Assume we have a manifold $M$ of dimension $m$. We will define $\int_M
 \omega$ where $\omega$ is a compactly supported $m$-form (i.e.
  $\operatorname{supp}(\omega) := \overline{\{p \in M\mid \omega(p) \neq
 0\}}$ is compact)
(Step 1): Assume $\operatorname{supp}(\omega) \subseteq U$ for some
  chart $(U, \phi)$. Write $$(\phi^{-1})^*(\omega) = f dx_1 \land \dots
 \land dx_m \in \Omega(\phi(U))$$  where $f \in C^\infty(\phi(U))$.
  Define $$\int_M \omega := \int_{\phi(U)}f$$
where the right integral is an ordinary Riemann-integral.
(Step 2): Involves a partition of unity to define the general case and
  I don't think it is important here so I'll ommit it.

I want to use this definition to calculate the above integral, so the first issue I'm stumbling upon. What is an atlas for $D$? I guess, I can just consider the map $id_D: D \to \mathbb{R}^2$ and this is an atlas for $D$? My confusion is the following: Formally, a chart is a map $U \to V$ where $V$ is an open in some $\mathbb{R}^n$ or some $\mathbb{H}^n$, so aren't there actually two charts involved (one for the interior, another for the boundary?)
If I can just pick the identity as one chart, then $$\int_D dx \land dy := \int_D dxdy = \pi$$
Any advice?

REPLY [3 votes]: Take $U=(0,1)\times(0,2\pi)$, and $\varphi(r,\theta)=(r\cos\theta,r\sin\theta)$. Then integrating over $D$ is equivalent to integrating over $\phi(U)$ because $D\backslash\phi(U)$ has zero area, and 
$$\int_{\phi(U)}dx \land dy=\int_{U}\phi^*(dx \land dy)=\int_{U}d(r\cos\theta)\land d(r\sin\theta)\\=\int_{U}rdr\land d\theta=\int_0^1\!rdr\int_0^{2\pi}\!\!\!\!d\theta=\frac12\cdot2\pi=\pi.$$
If we integrate in Cartesian coordinates then $U=\textrm{int}D$ and $\varphi(x,y)=(x,y)$. There is no need for the boundary chart since the boundary area is $0$, and it makes no contribution to the integral. So $\int_{\phi(U)}dx \land dy=\int_Ddx\,dy$. But unless we just assume that the area is $\pi$ the actual integration is unpleasant.