MAT236H5F, University of Toronto, Mississauga. Instructor: Ke Zhang.

**Definition** Say a region $D \subset \mathbb{R}^2$ has *piecewise smooth boundary* if its boundary $\partial D$ can be decomposed into finitely many pieces, each piece is either of the type $\{(x, g(x)) : a \le x \le b \}$, or $\{(h(y), y) : c \le y \le d\}$.

The area of a region can be understood using the single-variable Riemann integral. Here we provide an alternative point of view using double integrals.

**Definition** Let $D$ be a region with piecewise smooth boundary, we define
$$
F_D(x, y) =
\begin{cases}
1 & (x, y) \in D \\
0 & (x, y) \notin D,
\end{cases}
$$
called the *indicator function* of $D$.

**Proposition** Let $D$ be a closed region with piecewise smooth boundary, and suppose $R = [a, b] \times [c, d]$ is a rectangle which contains $D$. Then the function $F_D(x, y)$ is integrable on $R$, and we define the area of $D$
$$
Area(D) = \iint_R F_D(x, y) dA.
$$

Suppose we partition the rectangle $[a, b] \times [c, d]$ into $n \times m$ grid, and let $(x_{ij}^*, y_{ij}^*)$ be a sample point in the $(i, j)$ block. We note:

- If the rectangle $[x_{i-1}, x_i] \times [y_{j-1}, y_j] \subset D$, then $f(x_{ij}^*, y_{ij}^*) = 1$ regardless of how the point is chosen.
- If the rectangle $[x_{i-1}, x_i] \times [y_{j-1}, y_j]$ is disjoint from $D$, then $f(x_{ij}^*, y_{ij}^*) = 0$ regardless of how the point is chosen.
- When $[x_{i-1}, x_i] \times [y_{j-1}, y_j]$ is neither disjoint or contained in $D$, $f(x_{ij}^*, y_{ij}^*) = 0$ may take either $0$ or $1$. Moreover, the upper value $M_{ij} = 1$, the lower value $m_{ij} = 0$.

The three types of blocks for the domain $D = \{(x,y): x^2 + 2y^2 - xy \le 1\}$ is shown in the following picture.

The following picture illustrate why the difference between the upper and lower sums should converge to $0$.

We have define the area of a domain as
$$ Area(D) = \iint_R F_D(x,y) dA $$
where $R$ is a rectangle containing $D$, and $F_D(x, y)$ is the *indicator function* of $D$. On the other hand, we can view the area of a domain as the integral of the constant function $1$ on a domain $D$: $Area(D) = \iint_D 1 dA$, which suggest the following relation:
$$ \iint_R F_D(x,y) dA = \iint_D 1 dA. $$

**Definition** If $f(x,y)$ is a function on $R$, and $D \subset R$ is a domain, define
$$ F(x,y) = \begin{cases} f(x,y) & (x,y) \in D \\ 0 & (x,y) \notin D \end{cases} = F_D(x, y) \cdot f(x,y), $$
and define
$$ \iint_D f(x,y) dA = \iint_R F(x,y) dA = \iint F_D(x,y) \cdot f(x,y)dA. $$

Similarly, we have:

**Thoerem** If $f(x,y)$ is contininuos, then the function $F(x,y) = f(x,y) \cdot F_D(x,y)$ is Riemann integrable.

The proof is a combination of the rectangle case, and the indicator function case.