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\begin{note*}
    For $\phi = 1$, $\int_D \dd A$ is the area of
    $D$.
\end{note*}

\noindent
To evaluate the area integral, we split the region
$D$ into strips.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/dc7b6ece8a1f11ec.png}
\end{center}
Do $\int \dd x$ for some fixed $y$, and then do
$\int \dd y$.
\[ \int_D \phi \dd A = \int_a^b \dd y
\int_{x_1(y)}^{x_2(y)} \dd x \phi(x, y) \]
($x_1(y)$ and $x_2(y)$ trace the outline of $D$).

\begin{note*}
    This is written as $\int \dd x
    (\text{integrand})$ instead of $\int
    (\text{integrand}) \dd x$. You do $\int \dd x$
    first, and then $\int \dd y$.
\end{note*}

\noindent
Alternatively, we could divide $D$ as
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/8d8ba0948a2011ec.png}
\end{center}
\[ \int_D \phi \dd A = \int_c^d \dd x
\int_{y_1(x)}^{y_2(x)} \dd y \phi(x, y) \]
Now do $\int \dd y$ first and then $\int \dd x$.
\myskip
For suitably well behaved $\phi$ and $D$, any way
of splitting up $\int \dd A$ gives the same
result. (Fubri's theorem).

\begin{example*}
    Let $\phi(x, y) = x^2y$ and $D$ be the
    triangle
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/0af583568a2111ec.png}
    \end{center}
    \begin{align*}
        \int_D \phi \dd A
        &= \int_0^1 \dd y \int_0^{2 - 2y} \dd x
        x^2y \\
        &= \int_0^1 \dd y y
        \left[\frac{x^3}{3}\right]_0^{2 -
        2y} \\
        &= \frac{8}{3} \int_0^1 \dd y y(1 - y)^3
        \\
        &= \frac{2}{15}
    \end{align*}
    or
    \begin{align*}
        \int_D \phi \dd A
        &= \int_0^2 \dd x \int_0^{1 - x^2 / 2}
        \dd y x^2y \\
        &= \int_0^2 \dd x x^2 \left[ \half y^2
        \right]_0^{1 - x^2/2} \\
        &= \half \int_0^2 \dd x x^2 \left( 1 -
        \half x \right)^2 \\
        &= \frac{2}{15}
    \end{align*}
\end{example*}

\noindent
It is often useful to evaluate integrals using
something other than cartesian coordinates.
Consider a change of variables
\[ (x, y) \mapsto (u, v) .\]
We assume that this map is smooth and invertible.
We can then use $(u, v)$ as coordinates on
$\RR^2$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/94eab3428a2111ec.png}
\end{center}
How do we do the integral in $(u, v)$ coordinates?

\begin{claim*}
    The area integral can be written as
    \[ \int_D \phi \dd A = \int_{D'} \dd u \dd v
    |J(u, v)| \phi(u, v) \]
\end{claim*}

\noindent
Here the \emph{Jacobian} is the modulus of the
determinant
\[ |J(u, v)| = \left|
\begin{matrix}
    \pfrac{x}{u} & \pfrac{x}{v} \\
    \pfrac{y}{u} & \pfrac{y}{v}
\end{matrix}
\right| \]
We will also write the matrix
\[ J(u, v) = \pfrac{(x, y)}{(u, v)} \]

\begin{proof}
    We sum over the small parallelograms
    sandwiched between $u, v = \text{constant}$
    lines. Let $x = x(u, v)$ and $y = y(u, v)$.
    \[ \implies \delta x = \pfrac{x}{u} \delta u +
    \pfrac{x}{v} \delta v \]
    and
    \[ \delta y = \pfrac{y}{u} \delta u +
    \pfrac{y}{v} \delta v \]
    \[ \implies
    \begin{pmatrix}
        \delta x \\
        \delta y
    \end{pmatrix}
    =
    \begin{pmatrix}
        \pfrac{x}{u} & \pfrac{x}{v} \\
        \pfrac{y}{u} & \pfrac{y}{v}
    \end{pmatrix}
    \begin{pmatrix}
        \delta u \\
        \delta v
    \end{pmatrix}
    \]
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/61bf3e428a2211ec.png}
    \end{center}
    So
    \[ \bf{a} =
    \begin{pmatrix}
        \pfrac{x}{u} \\
        \pfrac{y}{u}
    \end{pmatrix}
    \delta u \]
    \[ \bf{b} =
    \begin{pmatrix}
        \pfrac{x}{v} \\
        \pfrac{y}{v}
    \end{pmatrix}
    \delta v \]
    The area of parallelogram is
    \[ \delta A = |\bf{a} \times \bf{b}| = \left|
    \pfrac{(x, y)}{(u, v)} \right| \delta u \delta
    v = |J| \delta u \delta v .\]
\end{proof}

\subsubsection*{An Example: 2D Polar Coordinates}

Plane polar coordinates are defined by
\[ x = \rho \cos \phi \qquad y = \rho \sin \phi \]
with $\phi \in [0, \infty)$ and $\phi \in [0,
2\pi)$. Then
\[ J = \pfrac{(x, y)}{(\rho, \phi)} = \left|
\begin{matrix}
    \cos \phi & -\rho \sin \phi \\
    \sin \phi & \rho \cos \phi
\end{matrix}
\right| = \rho \]
The area element is
\[ \delta A = \rho \delta \rho \delta \phi \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/e84a67de8a2211ec.png}
\end{center}
As an example, let $D$ be the region $x, y \ge 0$
and $x^2 + y^2 \le R^2$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/028a57d08a2311ec.png}
\end{center}
This is $0 \le \rho \le R$ and $0 \le \phi \le
\frac{\pi}{2}$. We will integrate $f = e^{-(x^2 +
y^2)/2} = e^{-\rho^2/2}$.
\begin{align*}
    \int_D f \dd A
    &= \int_0^{\pi/2} \dd \phi \int_0^R \dd \rho
    \rho e^{-\rho^2/2} \\
    &= \frac{\pi}{2} [-e^{-\rho^2/2}]_0^R \\
    &= \frac{\pi}{2} (1 - e^{-R^2/2})
\end{align*}