Before we introduce double integrals let us recall how the integral of a function of one variable, say \(t\text{,}\) was defined. Geometrically, the definite integral of a function \(f\) over an interval \(I=[a,b]\) represents the area of the region between the graph and the \(t\)-axis. To find an approximation of this area the idea is to use a partition of \(I\) given by \(a=t_0\lt t_1\lt t_2\lt \dots\lt t_n=b\text{.}\) Then we choose a point \(t_i^*\) in every interval \([t_{i-1},t_i]\text{.}\) Now we look at the rectangle of width \(\Delta t_i:=t_i-t_{i-1}\) and height \(f(t_i^*)\text{.}\) Such a rectangle is highlighted in Figure 5.1. The area of this rectangle is \(f(t_i^*)\Delta t_i\text{.}\) Hence the total area of all rectangles as shown in Figure 5.1 is
This total area is an approximation for the area under the graph lightly shaded in Figure 5.1. The above sum is called a Riemann sum of \(f\) over \(I\text{.}\) Taking a finer partition, that is, \(\max\Delta t_i\) smaller, we expect that the corresponding Riemann sum provides a better approximation. Indeed, for continuous functions the limit
exists, and is independent of the intermediate points \(t_i^*\) chosen. The limit is called the Riemann integral or simply the integral of \(f\) over the interval \([a,b]\text{.}\) If \(f(t_i^*)\) is negative then the area is considered negative.
Figure5.1.
The idea to define integrals of functions of two variables (double integrals) is very similar to the above. The area under the graph of a function of one variable corresponds to the volume of the region between the graph of a function of two variables and the \(xy\)-plane. To get an approximation of that volume we replace the rectangles by rectangular boxes, and then pass to the limit by taking smaller and smaller rectangular boxes. Two such approximations are shown in Figure 5.2, together with the limiting case.
Figure5.2.Approximation of the volume under the graph by rectangular boxes
To be more precise let us look at a function, \(f\text{,}\) defined on a rectangle \([a,b]\times[c,d]\) of the plane. By choosing partitions \(a=x_0\lt x_1\lt \dots\lt x_m=b\) and \(c=y_0\lt y_2\lt \dots\lt y_n=d\) of the intervals \([a,b]\) and \([c,d]\) we get a subdivision of the rectangle \([a,b]\times[c,d]\) into small rectangles \([x_{i-1},x_i]\times[y_{j-1},y_j]\text{,}\) one of which is highlighted in Figure 5.3.
Figure5.3.Partition of a rectangle
The lengths of the edges of such a rectangle is \(\Delta x_i\) and \(\Delta y_j\text{,}\) respectively. Now we choose an arbitrary point \(\vect x_{ij}^*:=(x_i^*,y_j^*)\) in that rectangle similarly as done in the one dimensional case with \(t_i^*\text{.}\) Then we look at the rectangular box with basis \([x_{i-1},x_i]\times[y_{j-1},y_j]\) and height \(f(x_i^*,y_j^*)\text{.}\) Such a box is shown in Figure 5.4.
Figure5.4.The graph of a function and the box over a small rectangle.
Its volume is the product of the lengths of its edges, namely \(f(x_i^*,y_j^*)\Delta x_i\Delta y_j\text{.}\) If \(f(x_i^*,y_j^*)\) is negative then the volume is considered negative, as in the case of a function of one variable. Hence the total volume of all boxes is
This sum corresponds to the Riemann sums we looked at before, and is also called a Riemann sum. It turns out that, if \(f\) is continuous on the closed rectangle, then one can pass to the limit as \(\max\Delta x_i\to 0\) and \(\max\Delta y_j\to 0\text{.}\) This leads to the following definition.
Definition5.5.
Suppose that \(f\) is continuous on the rectangle \(R=[a,b]\times[c,d]\text{.}\) Then we call
In this course we will mainly use the second, more concise form.
We would like to integrate over more general regions than just rectangles. We can certainly integrate over a union of finitely many rectangles. If we want to integrate over more general sets then we need to approximate the sets by a union of small rectangles as well. To get upper and lower bounds we approximate from the inside and the outside as shown in Figure 5.7.
Figure5.7.Approximation of a domain from the in- and outside by rectangles.
We then form a sum similar to (5.1) over the relevant rectangles. If the set is reasonably `nice,' that is, its boundary is not too wild, then the limits of the Riemann sums from the inside and the outside are the same, and we can define the double integral over such regions. It is quite difficult to characterise domains for which this is possible, and we omit such a discussion. All domains we encounter in this course have the required properties.
We next collect some properties of the integral. They follow directly from the definition as limits of Riemann sums.
Theorem5.8.
Suppose that \(D,D_1,D_2\subset\mathbb R^2\) are domains on which the integral of every continuous function is well defined. If \(f,g\) are continuous functions on \(D\) and \(\alpha\in\mathbb R\) then
If \(D=D_1\cup D_2\) and \(D_1\cap D_2=\emptyset\) then
\begin{equation*}
\int_Df(\vect x)\,d\vect x =\int_{D_1}f(\vect x)\,d\vect x+\int_{D_2}f(\vect x)\,d\vect x\text{.}
\end{equation*}
Remark5.9.
We introduced the double integral as the volume of the region between the graph of a function and the \(xy\)-plane. Note, however, that this is not the only possible interpretation. If we choose \(f\) to be the constant function with value one, that is, \(f(x,y)=1\text{,}\) then the integral represents the area of the domain:
\begin{equation}
\text{Area of the domain }D=\int_D1\,d\vect x\text{.}\tag{5.2}
\end{equation}
Another physical interpretation is the following. Let \(D\) represent the area occupied by a thin plate made of some material. At every point \((x,y)\) on that plate let \(\varrho(x,y)\) denote the mass density of the material. Then the
\begin{equation*}
\text{Total mass of the plate occupying }D
=\int_D\varrho(\vect x)\,d\vect x\text{.}
\end{equation*}
The theory developed in this section does not allow us to compute the double integral of a given function. We will see how to do this in the next section.
