Double Integrals over Rectangular Regions

In 07.02, $\int_a^b f(x)\,dx$ added up thin rectangles to get area under a curve. In two variables, add up thin columns to get volume under a surface.

The definition

Split the rectangle $R = [a, b]\times[c, d]$ into small rectangles of area $\Delta A$. Over each one, build a column whose height is $f$ at a sample point, and add:

$$\iint_R f(x, y)\,dA = \lim\ \sum_{i, j} f(x_{ij}^*, y_{ij}^*)\,\Delta A$$

as the small rectangles shrink. If $f \ge 0$, this is the volume of the solid between the surface $z = f(x, y)$ and the rectangle. Where $f < 0$, the volume counts as negative, just like signed area.

Easy cases

• A constant: $\iint_R c\,dA = c\cdot\text{area}(R)$. Over $[0, 2]\times[0, 5]$, $\iint 3\,dA = 30$: a box $3$ tall.
• $\iint_R 1\,dA$ is the area of $R$.

The midpoint rule

Use the center of each small rectangle as the sample point. For $f = x^2 + y$ on $[0, 2]\times[0, 2]$ with four unit squares, the centers are $(0.5, 0.5)$, $(1.5, 0.5)$, $(0.5, 1.5)$, $(1.5, 1.5)$, where $f = 0.75, 2.75, 1.75, 3.75$. The estimate is $(0.75 + 2.75 + 1.75 + 3.75)\cdot 1 = 9$. The exact value (15.02) is $\frac{28}{3} \approx 9.33$.

Properties

Like single integrals: constants pull out, sums split, and $f \le g$ implies $\iint f \le \iint g$.

Average value

$$f_{\text{avg}} = \frac{1}{\text{area}(R)}\iint_R f\,dA$$

the height of the box with the same base and the same volume (compare 08.06).