Double Integrals over Rectangles
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 5.1: “Double Integrals over Rectangular Regions” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/5-1-double-integrals-over-rectangular-regions |
| Textbook used in class | Stewart, Calculus, Section 15.1: “Double Integrals over Rectangles” |
Quick Reference
Double integral over a rectangle $R = [a,b]\times[c,d]$: $$\iint_R f(x,y)\,dA = \lim_{m,n\to\infty}\sum_{i=1}^m\sum_{j=1}^n f(x_i^*, y_j^*)\,\Delta A.$$
Volume interpretation: If $f \geq 0$, then $\iint_R f(x,y)\,dA$ equals the volume of the solid under $z = f(x,y)$ and above $R$.
Signed volume: If $f$ takes negative values, the integral gives the signed volume (positive above the $xy$-plane, negative below).
Motivation
The single-variable integral $\int_a^b f(x)\,dx$ is the limit of Riemann sums: the area under $y = f(x)$ is approximated by summing thin rectangles. The double integral extends this to two dimensions: the volume under $z = f(x,y)$ above a rectangle $R$ is approximated by summing thin rectangular prisms (columns) of height $f(x_i^*, y_j^*)$ and base area $\Delta A = \Delta x\,\Delta y$.
The definition is conceptually identical to the single-variable case -- only the bookkeeping (two indices, two limits) is new.
Key Concepts
1. The Double Riemann Sum
Partition $[a,b]$ into $m$ subintervals of width $\Delta x = (b-a)/m$ and $[c,d]$ into $n$ subintervals of width $\Delta y = (d-c)/n$. This creates $mn$ subrectangles, each of area $\Delta A = \Delta x\,\Delta y$.
Pick a sample point $(x_i^*, y_j^*)$ in each subrectangle. The double Riemann sum is: $$S_{mn} = \sum_{i=1}^m\sum_{j=1}^n f(x_i^*, y_j^*)\,\Delta A.$$
The double integral is the limit of $S_{mn}$ as $m, n \to \infty$.
2. Midpoint Approximation
For numerical estimates, choosing sample points at midpoints $(\bar x_i, \bar y_j)$ gives the midpoint rule: $$\iint_R f(x,y)\,dA \approx \sum_{i=1}^m\sum_{j=1}^n f(\bar x_i, \bar y_j)\,\Delta A.$$
3. Properties
The double integral shares linearity properties with the single integral:
- $\iint_R cf\,dA = c\iint_R f\,dA$
- $\iint_R (f+g)\,dA = \iint_R f\,dA + \iint_R g\,dA$
- If $f \leq g$ on $R$, then $\iint_R f\,dA \leq \iint_R g\,dA$
Worked Example
Estimate $\iint_R (x - 3y^2)\,dA$ where $R = [0,2]\times[1,2]$, using the midpoint rule with $m = n = 2$. (Stewart 15.1, Example 2.)
Partition: $\Delta x = 1$, $\Delta y = 1/2$. $\Delta A = 1/2$.
Four subrectangles with midpoints: $(1/2, 5/4)$, $(3/2, 5/4)$, $(1/2, 7/4)$, $(3/2, 7/4)$.
Evaluate: $$f(1/2, 5/4) = 1/2 - 3(25/16) = 1/2 - 75/16 = 8/16 - 75/16 = -67/16.$$ $$f(3/2, 5/4) = 3/2 - 75/16 = 24/16 - 75/16 = -51/16.$$ $$f(1/2, 7/4) = 1/2 - 3(49/16) = 8/16 - 147/16 = -139/16.$$ $$f(3/2, 7/4) = 24/16 - 147/16 = -123/16.$$
Sum $= (-67 - 51 - 139 - 123)/16 = -380/16$.
Estimate $= (-380/16)(1/2) = -380/32 = -11.875$.
$\iint_R f\,dA = \left(\int_a^b f\,dx\right)\!\left(\int_c^d f\,dy\right)$. The double integral is NOT the product of two single integrals (in general). The correct computation uses iterated integrals (Section 15.2/Fubini’s theorem): integrate with respect to one variable first, then the other. The product formula holds only in the special case $f(x,y) = g(x)h(y)$ -- a separable function.
Leveled Practice
Problem 1. Using the definition (or the volume interpretation), evaluate $\iint_R 3\,dA$ where $R = [1,4]\times[0,5]$.
Show answer
$\iint_R 3\,dA = 3\cdot\text{Area}(R) = 3\cdot(3)(5) = 45$.
(For a constant function, the double integral is the constant times the area of the rectangle.)