Now that we know what double integrals are,
we can start to compute them. The key idea is to work with
one variable at a time.

In order to integrate over a rectangle $[a,b] \times [c,d]$, we
first integrate with respect to one variable (say, $y$) for each
fixed value of $x$. That's an ordinary integral, which we
can compute using the fundamental theorem of calculus. We
then integrate the result over the other variable (in this case
$x$), which we can also compute using the fundamental theorem of
calculus. So a 2-dimensional double
integral becomes two ordinary 1-dimensional integrals,
one inside the other. We call this an iterated integral.

There are two ways to see the relation between double integrals
and iterated integrals. In the bottom-up approach, we
evaluate the sum $$\sum_{i=1}^m\sum_{j=1}^n
f\left(x_{i}^*,y_{j}\right)^* \,\Delta x\,\Delta y=\sum_{i=1}^m
\left(\sum_{j=1}^n f\left(x_{i}^*,y_{j}^*\right) \,\Delta
x\,\right) \Delta y,$$ by first summing over all of the boxes with
a fixed $i$ to get the contribution of a column (as indicated with
the parentheses on the second sum), and then adding up the
columns. (We could do this in the other order, by reversing
the summations.)

$f\left(x_{i}^*,y_{j}^*\right)\, \Delta y\, \Delta
x$ is the contribution of the volume
of a tower over a single rectangle to our
double integral, which approximates the volume under the
surface $f(x,y) and over the rectangle.

$\displaystyle\left(\sum_{j=1}^n
f\left(x_{i}^*,y_{j}^*\right) \,\Delta y\,\right) \Delta x$
is the sum of volumes of towers
over all the rectangles in a single column,
approximating the volume under the surface over the
column. Here, $\Delta x$ is the width of the
column. As $n \to \infty$, the sum over $n$ turns into
an integral, and we get $$\displaystyle\left (\int_c^d
f\left(x, y\right)\,dy \right )\, \Delta x.$$

Adding up the columns then gives
$\displaystyle\sum_{i=1}^m \left (\int_c^d f\left(x_i^*,
y\right)\, dy \right )\, \Delta x$. Taking a limit as
$m \to \infty$ turns the sum into an iterated integral,
giving us the actual volume
over the large rectangle, under the surface:$$\int_a^b \left
( \int_c^d f(x,y)\, dy \right ) \,dx.$$

This approach is explained in the following video, and an example
is worked out. (Video Fix? However, there
is a small error. At the beginning it says that we're going to
integrate over the rectangle $[0,1] \times [0,2]$, but for the
rest of the video the region $R$ is actually the rectangle $[0,2]
\times [0,1]$.)