Double Integrals over a Rectangular Region

Concept Explanation

When integrating over a rectangle in the $xy$-plane, all four boundaries are fixed geometric constants ($x = a, x = b, y = c, y = d$). This geometric symmetry provides unique analytical simplifications:

1. Fubini's Theorem (Iterated Integrals)

If $f(x, y)$ is continuous on the rectangle $R = [a, b] \times [c, d]$, the double integral can be evaluated as two successive single-variable definite integrals:

• Integrating $dy \, dx$ (Vertical Slices First): Hold $x$ constant and evaluate the inner integral $\int_c^d f(x, y) \, dy$ to produce a single-variable function $A(x)$ representing the cross-sectional area of the solid at position $x$. Then evaluate $\int_a^b A(x) \, dx$.
• Integrating $dx \, dy$ (Horizontal Slices First): Hold $y$ constant and evaluate the inner integral $\int_a^b f(x, y) \, dx$ to produce $A(y)$, then evaluate $\int_c^d A(y) \, dy$.

Because the boundaries are constants, reversing the order of integration simply requires exchanging the differential order ($dx \, dy \leftrightarrow dy \, dx$) and swapping the corresponding constant limits of integration without altering boundary equations.

2. Product Separability Property

If the multivariable integrand factors cleanly into the product of a single-variable function of $x$ and a single-variable function of $y$, such that $f(x, y) = g(x) \cdot h(y)$, then over a rectangular domain $R = [a, b] \times [c, d]$:

This decouples the multivariable calculation into the product of two independent single-variable integrals, significantly accelerating evaluation.

3. Spatial Average Value Formula

In single-variable calculus, the average value of $f(x)$ over $[a, b]$ is $f_{\text{avg}} = \frac{1}{b - a} \int_a^b f(x) \, dx$. For a two-variable function $f(x, y)$ over a region $R$, the spatial average value is defined by dividing the total accumulated double integral by the planar surface area of the domain:

For a rectangular domain $R = [a, b] \times [c, d]$, the area is simply $\text{Area}(R) = (b - a)(d - c)$.

Engineering Applications

• Optoelectronics & Display Engineering: Evaluating average surface irradiance $E_{\text{avg}}$ and total emitted radiant luminous flux across flat-panel LED displays and solar photovoltaic panels.
• Electrical & Computer Engineering: Calculating total electrostatic charge accumulation $Q = \iint_R \sigma(x, y) \, dA$ across rectangular microchip dies and capacitive sensor pads.
• Mechanical & Thermal Engineering: Determining total heat transfer rate $q = \iint_R h(x, y)(T_s - T_\infty) \, dA$ across rectangular radiator cooling fins and heat sinks.
• Civil & Structural Engineering: Computing total contact load and average soil bearing pressure beneath rectangular concrete foundation footings.

Example Problems

Example 1: Basic Iterated Integration & Spatial Average

Evaluate $\iint_R (3x^2 - 2y) \, dA$ over the rectangular region $R = [0, 2] \times [1, 3]$, and determine the spatial average value $f_{\text{avg}}$.

Solution

Step 1: State Governing Formula
State Fubini's Theorem for Rectangular Regions:

$$\iint_R f(x, y) \, dA = \int_{c}^{d} \int_{a}^{b} f(x, y) \, dx \, dy$$

Here $a = 0, b = 2$ and $c = 1, d = 3$. Set up the iterated integral:

$$I = \int_{1}^{3} \int_{0}^{2} (3x^2 - 2y) \, dx \, dy$$

Step 2: Evaluate the Inner Integral ($dx$)
Treat $y$ as a constant:

$$\int_{0}^{2} (3x^2 - 2y) \, dx = \left[ x^3 - 2yx \right]_{x=0}^{x=2} = \Big((2)^3 - 2y(2)\Big) - 0 = 8 - 4y$$

Step 3: Evaluate the Outer Integral ($dy$)

$$\int_{1}^{3} (8 - 4y) \, dy = \left[ 8y - 2y^2 \right]_{1}^{3} = \Big(8(3) - 2(3)^2\Big) - \Big(8(1) - 2(1)^2\Big)$$

$$= (24 - 18) - (8 - 2) = 6 - 6 = 0$$

Step 4: Compute Spatial Average Value
State the Spatial Average Value Formula:

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

The area of the rectangle is $\text{Area}(R) = (2 - 0)(3 - 1) = 2 \times 2 = 4$.

$$f_{\text{avg}} = \frac{1}{4} (0) = 0$$

Example 2: Strategic Order Selection via Fubini's Theorem

Evaluate the double integral:

Solution

Step 1: Analyze Integration Order Strategy
If we integrate with respect to $x$ first, the factor $x$ in $x e^{xy}$ forces integration by parts. However, if we integrate with respect to $y$ first, $x$ is treated as a constant multiplier that perfectly matches the internal derivative of the exponent $u = xy \implies du = x \, dy$. Therefore, we choose order $dy \, dx$.

Step 2: Set up the Iterated Integral
State Fubini's Theorem for Rectangular Regions:

$$\iint_R x e^{xy} \, dA = \int_{0}^{1} \int_{0}^{2} x e^{xy} \, dy \, dx$$

Step 3: Evaluate Inner Integral ($dy$)
Treating $x$ as constant:

$$\int_{0}^{2} x e^{xy} \, dy = \left[ e^{xy} \right]_{y=0}^{y=2} = e^{2x} - e^0 = e^{2x} - 1$$

Step 4: Evaluate Outer Integral ($dx$)

$$\int_{0}^{1} (e^{2x} - 1) \, dx = \left[ \frac{1}{2} e^{2x} - x \right]_{0}^{1} = \left( \frac{1}{2}e^2 - 1 \right) - \left( \frac{1}{2}e^0 - 0 \right)$$

$$= \frac{1}{2}e^2 - 1 - \frac{1}{2} = \frac{1}{2}e^2 - \frac{3}{2}$$

---

Connections & References

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