Double Riemann Sums and Double Integrals over Rectangles

This section only requires the material from . This section begins the development of double integrals with rectangular domains and does a mix of approximations and using geometric information to find exact answers. This section can likely be covered in one class and some instructors may want to emphasize more approximations of double integrals and discuss over- versus under-estimation.

In the Preview Activity, we will review a few ideas from integrals in single-variable calculus. The rest of this section will then be used to extend the ideas of integration and its interpretations to functions of two variables over a rectangular region.

In this section, we will use the classic calculus approach to define and understand the definite integral for a function of two variables over a rectangular region of input values. Just as in single-variable calculus, we will develop some algebraic methods to help efficiently evaluate these integrals. Later in this chapter, we will will generalize to regions that are not rectangular and work with functions of three or more variables.

The motivating interpretation of a definite integral for a function of one variable was the area under a curve over a particular interval of input values. We will use the same motivating interpretation for the definite integral of a function of two variables. The definite integral of a nonnegative function of two variables will measure the volume of the region beneath the graph of \(f\) over a region of input points. The definite integral for a function of two variables is called the double integral of the function \(f\). We can show this geometrically as the volume of the solid below the surface given by \(z=f(x,y)\) as shown in . The surface given by \(z=f(x,y)\) is shown in blue and the region of inputs above which we want to find the volume is shown in red in the \(xy\)-plane. The double integral of \(f\) over the red region should measure the volume of the solid shaded in gray.

The region shaded red in is irregular in the coordinates (\(x\) and \(y\)). We will explore this kind of problem in once we have some intuition and tools from examining rectangular regions of integration. To discuss how to approach rectangular regions, we apply the classic calculus approach to the region shown in . In particular, for this rectangular region of input points, we want to

1. \(xy\)
2. \(f(x,y)\)
3. use a limit to find the exact value of the volume under the surface.

In the next activity, we will go through these steps for a particular function and rectangular region of integration as well as introduce some of the notation used in our development of double integrals.

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

Every part of the previous activity can be generalized to work on any bounded rectangular region using any number of subintervals in either the \(x\) or \(y\) directions. This leads us to state these steps in general below and note that using more subintervals corresponds to estimating on a smaller scale, which is step 2 of the classic calculus approach.

If \(f(x,y) \geq 0\) on the rectangle \(R\), we may ask to find the volume of the solid bounded above by \(f\) over \(R\), as illustrated in Figure.

This volume is approximated by a Riemann sum, which sums the volumes of the rectangular prisms shown in Figure. In a double Riemann sum, both \(m\) and \(n\) go to infinity. This means that the number of subrectangles increases without bound, as illustrated in Figure. As this happens, the sum of the volumes of the rectangular boxes approaches the true volume of the solid bounded above by \(z=f(x,y)\) over the region \(R\). Use the sliders at the top of Figure to change the number of subintervals used in the Riemann sum and verify geometrically how the estimated volume will approach true volume under \(f\) as \(n\) and \(m\) become arbitrarily large.

The third step in the classic calculus approach is to take the limit, and when this limit exists, its value is the double integral we have been seeking:

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

We conclude this section with three important interpretations of the double integral that will appear repeatedly throughout the rest of this book. We summarize these three interpretations in the Key Idea below and then discuss each in greater depth.

Suppose that \(f(x,y)\) assumes both positive and negative values on the rectangle \(R\), as shown in Figure. When constructing a Riemann sum, for each \(i\) and \(j\), the product \(f(x_{ij}^*, y_{ij}^*) \enspace \Delta A\) can be interpreted as a signed volume of a box with base area \(\Delta A\) and signed height \(f(x_{ij}^*, y_{ij}^*)\). Since \(f\) can have negative values, this height could be negative. The sum \[\begin{aligned}\end{aligned}\] can then be interpreted as a sum of signed volumes of boxes, with a negative sign attached to those boxes whose heights are below the \(xy\)-plane.

We can then realize the double integral \(\displaystyle\iint_R f(x,y) \, dA\) as a difference in volumes: \(\iint_R f(x,y) \, dA\) tells us the volume of the solids the graph of \(f\) bounds above the \(xy\)-plane over the rectangle \(R\) minus the volume of the solids the graph of \(f\) bounds below the \(xy\)-plane under the rectangle \(R\). This is shown in Figure.

The double integral of \(f\) over the rectangle \(R\) will be positive because there is more volume above the \(xy\)-plane (shown in blue) than volume below the \(xy\)-plane, which is counted as negative and shown in red.

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

We conclude this section with a list of properties of double integrals. Since similar properties are satisfied by single-variable integrals and the arguments for double integrals are essentially the same, we omit their justification.

Let \(f\) and \(g\) be continuous functions on a rectangle \(R = \{(x,y) : a \leq x \leq b, c \leq y \leq d\}\), and let \(k\) be a constant. Then

1. \(\displaystyle\iint_R \left(f(x,y) + g(x,y)\right) \, dA = \iint_R f(x,y) \, dA + \iint_R g(x,y) \, dA\).

2. \(\displaystyle\iint_R kf(x,y) \, dA = k \iint_R f(x,y) \, dA\).

3. If \(f(x,y) \geq g(x,y)\) on \(R\), then \(\displaystyle\iint_R f(x,y) \, dA \geq \iint_R g(x,y) \, dA\).