The Development of the Double Integral

Integrals over rectangles are almost the same as integrals over intervals, except that now we are looking at the quantity $f(x,y)$ per unit area instead of the quantity $f(x)$ per unit length.

The simplest examples of such quantities are the volumes that have the surface $f(x,y)$ as height.  Suppose that $R = [a,b] \times [c,d]$ is a rectangle in the $xy$-plane, where $x$ runs from $a$ to $b$ and $y$ runs from $c$ to $d$.  Let's figure out the volume of the solid over the rectangle $R$, between the $xy$-plane and the surface $z=f(x,y)$.

Break the interval $[a,b]$ into $m$ pieces, each of size $\Delta x = \frac{b-a}{m}$.    Break the interval $[c,d]$ into $n$ pieces, each of size $\Delta y = \frac{d-c}{n}$. This breaks $R$ into smaller rectangular boxes, which we call $R_{ij}$, where $i$ indicates the column and $j$ indicates the row.  $R_{ij}$ has area $\Delta A =\Delta x\, \Delta y$.  For each pair $i,j$, pick a sample point $\left(x_{ij}^*, y_{ij}^*\right)$ somewhere in $R_{ij}$.  (For an approximation, a logical choice for the sample point $\left(x_{ij},y_{ij}\right)$ is given by the midpoint rule: $x_{ij}^* = a + \left(i-\frac12\right)\,\Delta x, \qquad y_{ij}^* = c+\left(j-\frac12\right)\Delta y.$) cnx.or	$xy$-plane
Let the height of each tower over $R_{ij}$ be given by $f\left(x_{ij}^*,y_{ij}^*\right)$, i.e. by the height of the surface at our sample point over each rectangle. Find the volume of each tower over each rectangle $R_{ij}$ as $f\left(x_{ij}^*,y_{ij}^*\right) \,\Delta A$. Approximate the total volume by adding up all the tower volumes over all the $R_{ij}$, getting $$V\approx\displaystyle{\sum_{i=1}^m \sum_{j=1}^n f\left(x_{ij}^*, y_{ij}^*\right)\, \Delta A}.$$ www.stetson.edu	3-space, with the "floor" being the $xy$-plane
In this example, you can see both the rectangles $R_{ij}$ on the $xy$-plane, and above them the approximating columns, and the surface under which the volume is to be approximated. www.stetson.edu

With functions of one variable, we take the limit of the approximations of area under a curve $f$ and get an integral:$$A=\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^n f(x_i)\Delta x.$$

Similarly, given our double sum approximating the volume, the limit of this sum is the definition of a (double) integral.