Double Integrals over Rectangular Regions

Learning Objectives

After studying this chapter, you should be able to:

• Recognize when a function of two variables $f(x,y)$ is integrable over a rectangular region $R$.
• Apply the properties of double integrals (linearity, additivity, and monotonicity) to simplify calculations.
• Evaluate double integrals by converting them into iterated integrals using Fubini's Theorem.
• Calculate the volume of a solid bounded by a surface, the area of a plane region, and the average value of a function over a rectangle.

Key Terms & Glossary

• Rectangular Region ($R$): The Cartesian product of two closed intervals $[a, b] \times [c, d]$. It represents the set of all points $(x, y)$ such that $a \leq x \leq b$ and $c \leq y \leq d$.
• Riemann Sum: The sum of the volumes of thin rectangular boxes used to approximate the total volume under a surface: $S_{m,n} = \sum_{i=1}^m \sum_{j=1}^n f(x_{ij}^*, y_{ij}^*) \Delta A$.
• Iterated Integral: An integral performed sequentially with respect to one variable at a time (e.g., first $x$, then $y$), treating the other variable as a constant during each step.
• Double Integral: The limit of the Riemann sum as the dimensions of the sub-rectangles approach zero, representing the signed volume between the surface $z = f(x, y)$ and the $xy$-plane.

The "Big Idea"

In single-variable calculus, the definite integral $\int_a^b f(x)dx$ represents the area under a curve. In multivariable calculus, the double integral extends this logic into the third dimension. Instead of integrating over an interval, we integrate over a region $R$ in the $xy$-plane. The result is the volume of the solid that sits above the rectangle $R$ and below the surface $z = f(x, y)$. This transition from "Length → Area" to "Area → Volume" is the cornerstone of multiple integration.

Formula / Concept Box

[!IMPORTANT] When evaluating an iterated integral, always integrate from the inside out. The inner limits correspond to the inner differential (e.g., $dx$), and the outer limits correspond to the outer differential ($dy$).

Hierarchical Outline

• I. Geometry of the Rectangular Region
  • Definition of $R = [a, b] \times [c, d]$.
  • Partitioning $R$ into sub-rectangles $\Delta A = \Delta x \Delta y$.
• II. The Double Integral as a Limit
  • Choosing sample points $(x_{ij}^*, y_{ij}^*)$.
  • Conditions for integrability (continuity of $f$ on $R$).
• III. Properties of Double Integrals
  • Linearity: $\iint (cf + g) dA = c \iint f dA + \iint g dA$.
  • Additivity: Splitting $R$ into $R_1$ and $R_2$.
• IV. Evaluation via Iterated Integrals
  • Fixing $y$ to integrate with respect to $x$ (and vice versa).
  • Changing the order of integration.
• V. Applications
  • Volume calculation ($z = f(x,y)$ must be $\geq 0$).
  • Finding average values in physical contexts (e.g., average temperature).

Visual Anchors

Conceptual Flow of Integration

Geometric Partitioning

Definition-Example Pairs

Term: Average Value of a Function over a Region

• Definition: The single constant value that, if spread uniformly over the entire area $R$, would result in the same volume as the original function $f(x,y)$.
• Example: If $f(x,y)$ represents the temperature on a metal plate $R$, the average value $\frac{1}{Area} \iint_R f \, dA$ gives the single temperature reading that represents the overall thermal state of the plate.

Term: Iterated Integral

• Definition: A method of calculating a double integral by performing two successive single-variable integrations.
• Example: To find $\iint_R (x+y) \, dA$ where $R=[0,1]\times[0,2]$, you first find $\int_0^1 (x+y) \, dx = [\frac{1}{2}x^2 + xy]_0^1 = \frac{1}{2} + y$, then integrate that result: $\int_0^2 (\frac{1}{2} + y) \, dy$.

Worked Examples

Example 1: Calculating Volume under a Plane

Problem: Find the volume of the solid bounded by the surface $f(x,y) = 4 - x - y$ over the rectangular region $R = [0,1] \times [0,1]$.

Step 1: Set up the iterated integral. $V = \int_0^1 \int_0^1 (4 - x - y) \, dx \, dy$

Step 2: Evaluate the inner integral (with respect to $x$). $\int_0^1 (4 - x - y) \, dx = \left[ 4x - \frac{1}{2}x^2 - xy \right]_0^1 = (4 - \frac{1}{2} - y) - 0 = 3.5 - y$

Step 3: Evaluate the outer integral (with respect to $y$). $\int_0^1 (3.5 - y) \, dy = \left[ 3.5y - \frac{1}{2}y^2 \right]_0^1 = (3.5 - 0.5) = 3$

Final Answer: The volume is 3 cubic units.

Checkpoint Questions

1. If $\iint_R f(x,y) \, dA = 12$ and the area of $R$ is 4, what is the average value of $f$ over $R$?
2. Does the order of integration ($dx \, dy$ vs $dy \, dx$) change the final result of a double integral over a rectangular region for a continuous function?
3. ▶Check Answer for Q1

   The average value is $12 / 4 = 3$.

4. ▶Check Answer for Q2

   No. According to Fubini's Theorem, if the function is continuous, the order does not matter.