Double Integrals over General Regions

Summary

To integrate over a nonrectangular plane region $D$ , describe $D$ as a Type I region (functions of $x$ as vertical bounds) or Type II region (functions of $y$ as horizontal bounds), then write an iterated integral with variable limits.

Prerequisites

Double Integrals, Fubini's Theorem

Definition / Procedure

Type I

D = {(x, y) : a \leq x \leq b, g_{1} (x) \leq y \leq g_{2} (x)}, \iint_{D} f d A = \int_{a}^{b} \int_{g_{1} (x)}^{g_{2} (x)} f (x, y) d y d x .

Type II

D = {(x, y) : c \leq y \leq d, h_{1} (y) \leq x \leq h_{2} (y)}, \iint_{D} f d A = \int_{c}^{d} \int_{h_{1} (y)}^{h_{2} (y)} f (x, y) d x d y .

Sketch $D$ , choose the description that yields simple limits, and evaluate the inner integral first.

Conditions / Assumptions

$f$ continuous on the closed bounded region $D$ .
Bounding curves $g_{i}$ , $h_{i}$ continuous.

Worked Example

Let $D$ be the triangle with vertices $(0, 0)$ , $(1, 0)$ , $(1, 1)$ , so $0 \leq x \leq 1$ and $0 \leq y \leq x$ . For $f = 1$ ,

\iint_{D} 1 d A = \int_{0}^{1} \int_{0}^{x} 1 d y d x = \int_{0}^{1} x d x = \frac{1}{2}

(the area of the triangle). As Type II: $0 \leq y \leq 1$ , $y \leq x \leq 1$ gives the same area.

Common Mistakes

Using constant limits on a nonrectangular region.
Reversing order without recomputing bounds from a sketch.

Connections

Fubini's Theorem, Double Integrals in Polar Coordinates for circular regions

References

General plane regions appear in OpenStax Calculus Volume 3.^[1]

OpenStax, Calculus Volume 3, Section 5.2, https://openstax.org/details/books/calculus-volume-3 ↩︎