Integrals over rectangles are almost the same as integrals over intervals, except that now our density $f(x,y)$ is the amount of stuff per unit area instead of stuff per unit length. The simplest example of `stuff' is volume, in which case $f(x,y)$ is height. So suppose that $R = [a,b] \times [c,d]$ is a rectangle, where $x$ runs from $a$ to $b$ and $y$ runs from $c$ to $d$. Let's figure out the volume of the solid between the $x$-$y$ plane and the surface $z=f(x,y)$, and over the rectangle $R$.
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The figure to the right shows this process for the function $f(x,y)=x^2+y^2$, where $R = [-10,10] \times [-10,10]$, and where $n=m=8$. The volume under the surface is approximately the sum of the volumes of a number of towers, and the volume of each tower is $f(x_{ij}^*, y_{ij}^*) \Delta x \Delta y$. As $n \to \infty$ and $m \to \infty$, the approximation gets better and better, and in the limit gives the exact volume under the surface. |
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As with functions of one variable, the limit of this sum is the definition of a (double) integral.
| If $f(x,y)$ is a continuous function of two variables and if $R = [a,b] \times [c,d]$ is a rectangle, then the double integral of $f$ over $R$ is $$\iint_R f(x,y) dA = \iint_R f(x,y) dx dy = \lim_{m \to \infty}\lim_{n\to\infty} \sum_{i=1}^m \sum_{j=1}^n f(x_{ij}^*, y_{ij}^*) \Delta A.$$ A convenient choice for the sample point $(x_{ij}^*,y_{ij}^*)$ is given by the midpoint rule: $x_{ij}^* = a + (i-\frac12)\Delta x$, $y_{ij}^* = c+(j-\frac12)\Delta y$. |
In the following video, we develop the same definition with a slightly different problem, namely computing the total population of the (conveniently rectangular) state of Colorado.