Double Integrals over a General Region

Core Definition

A general region $D \subset \mathbb{R}^2$ is a bounded planar domain whose boundaries are defined by continuous curves rather than fixed rectangular constants. A double integral over a general region is evaluated as an iterated integral where the outer limits are always constants and the inner limits are functions of the outer variable:

Type I Region (Vertical Slices): $D = \{(x, y) \in \mathbb{R}^2 \mid a \le x \le b, \, g_1(x) \le y \le g_2(x)\}$ $\iint_D f(x, y) \, dA = \int_{a}^{b} \left( \int_{g_1(x)}^{g_2(x)} f(x, y) \, dy \right) dx$
Type II Region (Horizontal Slices): $D = \{(x, y) \in \mathbb{R}^2 \mid c \le y \le d, \, h_1(y) \le x \le h_2(y)\}$ $\iint_D f(x, y) \, dA = \int_{c}^{d} \left( \int_{h_1(y)}^{h_2(y)} f(x, y) \, dx \right) dy$

Concept Explanation

When integrating over non-rectangular domains, the geometric profile of the boundary curves dictates how the region must be sliced to formulate clean single integrals:

1. Type I Regions (Vertical Slices: $dy \, dx$ )

In a Type I domain:

The projection of $D$ onto the $x$ -axis forms the constant outer interval $[a, b]$ .
For any fixed $x \in [a, b]$ , a vertical ray entering the region from below enters at the lower boundary curve $y = g_1(x)$ and exits at the upper boundary curve $y = g_2(x)$ .
Crucial Rule: The inner integral is evaluated with respect to $y$ , yielding a function of $x$ only. The outer integral integrates with respect to $x$ across the constant limits $[a, b]$ , ensuring that the final output is a finite numeric value.

2. Type II Regions (Horizontal Slices: $dx \, dy$ )

In a Type II domain:

The projection of $D$ onto the $y$ -axis forms the constant outer interval $[c, d]$ .
For any fixed $y \in [c, d]$ , a horizontal ray passing from left to right enters the region at the left boundary curve $x = h_1(y)$ and exits at the right boundary curve $x = h_2(y)$ .
Crucial Rule: The inner integral is evaluated with respect to $x$ , yielding a function of $y$ only. The outer integral evaluates across constant limits $[c, d]$ .

3. Finding Boundary Limits via Intersection Points

To rigorously establish integration limits:

Find Outer Bounds: Equate boundary curves ( $g_1(x) = g_2(x)$ or $h_1(y) = h_2(y)$ ) and solve for real intersection points. These roots form the endpoints of the outer constant interval.
Determine Top/Bottom or Left/Right Orientation: Test a midpoint value inside the outer interval to determine which curve is strictly greater ( $g_2(x) \ge g_1(x)$ or $h_2(y) \ge h_1(y)$ ).
Compound Regions: If a boundary curve changes character midway through the interval (e.g., a top boundary shifts from a line to a parabola), the region must be partitioned into a union of disjoint sub-regions $D = D_1 \cup D_2$ , where $\iint_D f \, dA = \iint_{D_1} f \, dA + \iint_{D_2} f \, dA$ .

Engineering Applications

Civil & Structural Engineering: Computing resultant hydrostatic force $F_R = \iint_D \rho g (H - y) \, dA$ and center of pressure on submerged parabolic, trapezoidal, or circular dam floodgates.
Construction Management & Mechanics: Calculating planar centroid coordinates $(\bar{x}, \bar{y}) = \left(\frac{1}{A}\iint_D x \, dA, \, \frac{1}{A}\iint_D y \, dA\right)$ for irregular structural steel gusset plates and concrete shear walls.
Mechanical & Fluid Engineering: Evaluating mass flow rate and viscous shear stress distributions across tapered aerodynamic ducts and non-circular pipe conduits.
Aerospace Engineering: Determining geometric area and wing planform centroid for swept-back or delta wing configurations.

Example Problems

Example 1: Type I Region (Parabola and Slanted Line)

Evaluate $\iint_D (x + y) \, dA$ where $D$ is the region bounded by $y = x^2$ and $y = 2x$ .

Solution

Step 1: Locate Intersection Points (Outer Bounds)
Equate boundary equations to find where the curves cross:

x^2 = 2x \implies x^2 - 2x = 0 \implies x(x - 2) = 0

The intersection points occur at

x = 0

and

x = 2

Step 2: Identify Boundary Hierarchy
On the interval $(0, 2)$ , test $x = 1$ : $y = 2(1) = 2$ and $y = (1)^2 = 1$ . Since $2 > 1$ , the line $y = 2x$ is the upper boundary and the parabola $y = x^2$ is the lower boundary. Thus:

D = \{(x, y) \in \mathbb{R}^2 \mid 0 \le x \le 2, \, x^2 \le y \le 2x\}

Step 3: Formulate Iterated Integral
State the Type I Double Integral Formula:

\iint_D f(x, y) \, dA = \int_{a}^{b} \int_{g_1(x)}^{g_2(x)} f(x, y) \, dy \, dx

Substitute limits and integrand:

I = \int_{0}^{2} \int_{x^2}^{2x} (x + y) \, dy \, dx

Step 4: Evaluate Inner Integral ( $dy$ )

\int_{x^2}^{2x} (x + y) \, dy = \left[ xy + \frac{1}{2}y^2 \right]_{y=x^2}^{y=2x} = \Big(x(2x) + \frac{1}{2}(2x)^2\Big) - \Big(x(x^2) + \frac{1}{2}(x^2)^2\Big)

= (2x^2 + 2x^2) - (x^3 + \frac{1}{2}x^4) = 4x^2 - x^3 - \frac{1}{2}x^4

Step 5: Evaluate Outer Integral ( $dx$ )

\int_{0}^{2} \left( 4x^2 - x^3 - \frac{1}{2}x^4 \right) \, dx = \left[ \frac{4}{3}x^3 - \frac{1}{4}x^4 - \frac{1}{10}x^5 \right]_{0}^{2}

= \left( \frac{4}{3}(8) - \frac{1}{4}(16) - \frac{1}{10}(32) \right) - 0 = \frac{32}{3} - 4 - \frac{16}{5}

Find common denominator (

15

= \frac{160}{15} - \frac{60}{15} - \frac{48}{15} = \frac{52}{15}

Example 2: Type II Region (Horizontal Slices)

Evaluate $\iint_D (x + y) \, dA$ where $D$ is bounded on the left by $x = y^2$ and on the right by $x = y$ for $0 \le y \le 1$ .

Solution

Step 1: Formulate Type II Iterated Integral
State the Type II Double Integral Formula:

\iint_D f(x, y) \, dA = \int_{c}^{d} \int_{h_1(y)}^{h_2(y)} f(x, y) \, dx \, dy

Here

c = 0, d = 1

, left bound

h_1(y) = y^2

, and right bound

h_2(y) = y

I = \int_{0}^{1} \int_{y^2}^{y} (x + y) \, dx \, dy

Step 2: Evaluate Inner Integral ( $dx$ )

\int_{y^2}^{y} (x + y) \, dx = \left[ \frac{x^2}{2} + yx \right]_{x=y^2}^{x=y} = \left( \frac{y^2}{2} + y^2 \right) - \left( \frac{y^4}{2} + y^3 \right) = \frac{3}{2}y^2 - y^3 - \frac{1}{2}y^4

Step 3: Evaluate Outer Integral ( $dy$ )

\int_{0}^{1} \left( \frac{3}{2}y^2 - y^3 - \frac{1}{2}y^4 \right) \, dy = \left[ \frac{1}{2}y^3 - \frac{1}{4}y^4 - \frac{1}{10}y^5 \right]_{0}^{1}

= \frac{1}{2} - \frac{1}{4} - \frac{1}{10} = \frac{10}{20} - \frac{5}{20} - \frac{2}{20} = \frac{3}{20}

Connections & References

Parent Concepts: Multiple Integrals, Double Integrals over a Rectangular Region
Sub-concepts & Extensions:
- Switching the Order of Integration
- Double Integrals in Polar Coordinates
Course Links:
- Calc-3 Session 14 - Double Integrals
- Calc-3 Session 15 - Double Integrals in Polar Coordinates

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