Double Integrals over General Regions

In the previous set of notes we looked at how we can integrate a function of two variables over a rectangle. However, we want to be able to integrate over other shapes as well. Suppose we have some shape on the $xy$-plane which we will call $D$, then there is an rectangle $R$ that will enclose $D$. Next we define a new function $F$ with domain $R$ by
$$
F(x, y) = \begin{cases} f(x,y)\ \quad &\text{if}\ (x, y)\ \text{is in}\ D \\ 0\ \quad &\text{if}\ (x, y)\ \text{is in}\ R\ \text{but not in}\ D  \end{cases}.
$$

Now, if $F$ is integrable over $R$, then we define the double integral of $f$ over $D$ by
$$
\iint\limits_D f(x, y)\ dA = \iint\limits_R F(x, y)\ dA.
$$

Note that the size of the rectangle doesn't matter, as long as $D \subset R$.

Type I region. If the plane region $D$ lies between two continuous functions of $x$, then $D$ is called a type I region. If $f$ is continuous on a type I region $D$, then:
$$
\iint\limits_D f(x, y)\ dA = \int_a^b\int_{g_1(x)}^{g_2(x)} f(x, y)\ dy\ dx. \tag{1} \label{1}
$$

Type II region. If the plane region $D$ lies between two continuous functions of $y$, then $D$ is called a type II region. If $f$ is continuous on a type II region $D$, then:
$$
\iint\limits_D f(x, y)\ dA = \int_c^d\int_{h_1(y)}^{h_2(y)} f(x, y)\ dx\ dy. \tag{2} \label{2}
$$

Note that $F(x, y) = 0$ when $g_1(x) \leq y \leq g_2(x)$, which enables us to evaluate the integral as an iterated integral. If $D$ is neither a type I or type II region, then we can split $D$ into multiple regions such as $D_1$ and $D_2$. However, $D_1$ and $D_2$ shouldn't overlap (except on the boundary), such that $D = D_1 \cup D_2$. This allows us to split up the integral into multiple integrals, like so:
$$
\iint\limits_D f(x,y)\ dA = \iint\limits_{D_1} f(x,y)\ dA + \iint\limits_{D_2} f(x,y)\ dA. \tag{3} \label{3}
$$

If we integrate the constant function $f(x, y) = 1$ over a region $D$, then we get the area of $D$:
$$
\iint\limits_D 1\ dA = A(D).
$$

This is a useful method to estimate an integral. If $m \leq f(x, y) \leq M$ for all $(x, y)$ in $D$, then
$$
m A(D) \leq \iint\limits_D f(x, y)\ dA \leq MA(D). \tag{4} \label{4}
$$

Example

Suppose we have the following integral
$$
\iint\limits_D \dfrac{y}{x^2 + 1}\ dA,
$$

where $D$ is
$$
D = \{(x,y)\mid 0 \leq x \leq 4,\ 0 \leq y \leq \sqrt{x} \}.
$$

The first thing we will do is to create a plot of $D$, so that we can define $g_1$ and $g_2$. This plot shows the interval on which $D$ is defined.

We can see that it is a type I region because it lies between two continuous functions of $x$. From the graph we can also see that $g_1(x) = 0$ and $g_2(x) = \sqrt{x}$. Using $\eqref{1}$ we get that
$$
\iint\limits_D \dfrac{y}{x^2 + 1}\ dA = \int_0^4\int_0^{\sqrt{x}} \dfrac{y}{x^2 + 1}\ dy\ dx.
$$
Integrating this with respect to $y$ results in a fixed value for $x$, meaning that we can treat $1/(x^2 + 1)$ as a constant, giving:
$$
\int_0^4 \dfrac{x}{x^2 + 1}\ dx.
$$

Using a u-substitution, where we let $u = x^2 + 1$, gives $\frac{1}{2}du = x\ dx$. Substituting this back into the integral and integrating with respect to $x$ results in
$$
\dfrac{1}{4} \ln 14.
$$