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Double Integrals over a Rectangular Region

Formulas & Definitions
Core Definition

For a continuous function of two variables f(x,y)f(x, y)f(x,y) defined over a closed rectangular domain R=[a,b]Γ—[c,d]={(x,y)∈R2∣a≀x≀b, c≀y≀d}R = [a, b] \times [c, d] = \{(x, y) \in \mathbb{R}^2 \mid a \le x \le b, \, c \le y \le d\}R=[a,b]Γ—[c,d]={(x,y)∈R2∣a≀x≀b,c≀y≀d}, the double integral ∬Rf(x,y) dA\iint_R f(x, y) \, dA∬R​f(x,y)dA evaluates the limit of Riemann sums over planar grid partitions:

∬Rf(x,y) dA=lim⁑Δxβ†’0Ξ”yβ†’0βˆ‘i=1mβˆ‘j=1nf(xijβˆ—,yijβˆ—) Δx Δy\iint_R f(x, y) \, dA = \lim_{\substack{\Delta x \to 0 \\ \Delta y \to 0}} \sum_{i=1}^m \sum_{j=1}^n f(x_{ij}^*, y_{ij}^*) \, \Delta x \, \Delta y∬R​f(x,y)dA=Ξ”xβ†’0Ξ”yβ†’0​lim​i=1βˆ‘m​j=1βˆ‘n​f(xijβˆ—β€‹,yijβˆ—β€‹)Ξ”xΞ”y
By Fubini's Theorem, the double integral equals the iterated single integrals evaluated in either order:
∬Rf(x,y) dA=∫cd∫abf(x,y) dx dy=∫ab∫cdf(x,y) dy dx\iint_R f(x, y) \, dA = \int_{c}^{d} \int_{a}^{b} f(x, y) \, dx \, dy = \int_{a}^{b} \int_{c}^{d} f(x, y) \, dy \, dx∬R​f(x,y)dA=∫cdβ€‹βˆ«ab​f(x,y)dxdy=∫abβ€‹βˆ«cd​f(x,y)dydx

Concept Explanation

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

1. Fubini's Theorem (Iterated Integrals)

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

  • Integrating dy dxdy \, dxdydx (Vertical Slices First): Hold xxx constant and evaluate the inner integral ∫cdf(x,y) dy\int_c^d f(x, y) \, dy∫cd​f(x,y)dy to produce a single-variable function A(x)A(x)A(x) representing the cross-sectional area of the solid at position xxx. Then evaluate ∫abA(x) dx\int_a^b A(x) \, dx∫ab​A(x)dx.
  • Integrating dx dydx \, dydxdy (Horizontal Slices First): Hold yyy constant and evaluate the inner integral ∫abf(x,y) dx\int_a^b f(x, y) \, dx∫ab​f(x,y)dx to produce A(y)A(y)A(y), then evaluate ∫cdA(y) dy\int_c^d A(y) \, dy∫cd​A(y)dy.

Because the boundaries are constants, reversing the order of integration simply requires exchanging the differential order (dx dy↔dy dxdx \, dy \leftrightarrow dy \, dxdxdy↔dydx) 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 xxx and a single-variable function of yyy, such that f(x,y)=g(x)β‹…h(y)f(x, y) = g(x) \cdot h(y)f(x,y)=g(x)β‹…h(y), then over a rectangular domain R=[a,b]Γ—[c,d]R = [a, b] \times [c, d]R=[a,b]Γ—[c,d]:

∬Rg(x)h(y) dA=(∫abg(x) dx)(∫cdh(y) dy)\iint_R g(x)h(y) \, dA = \left( \int_{a}^{b} g(x) \, dx \right) \left( \int_{c}^{d} h(y) \, dy \right)∬R​g(x)h(y)dA=(∫ab​g(x)dx)(∫cd​h(y)dy)
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)f(x)f(x) over [a,b][a, b][a,b] is favg=1bβˆ’a∫abf(x) dxf_{\text{avg}} = \frac{1}{b - a} \int_a^b f(x) \, dxfavg​=bβˆ’a1β€‹βˆ«ab​f(x)dx. For a two-variable function f(x,y)f(x, y)f(x,y) over a region RRR, the spatial average value is defined by dividing the total accumulated double integral by the planar surface area of the domain:

favg=1Area(R)∬Rf(x,y) dAf_{\text{avg}} = \frac{1}{\text{Area}(R)} \iint_R f(x, y) \, dAfavg​=Area(R)1β€‹βˆ¬R​f(x,y)dA
For a rectangular domain R=[a,b]Γ—[c,d]R = [a, b] \times [c, d]R=[a,b]Γ—[c,d], the area is simply Area(R)=(bβˆ’a)(dβˆ’c)\text{Area}(R) = (b - a)(d - c)Area(R)=(bβˆ’a)(dβˆ’c).

Engineering Applications

  • Optoelectronics & Display Engineering: Evaluating average surface irradiance EavgE_{\text{avg}}Eavg​ and total emitted radiant luminous flux across flat-panel LED displays and solar photovoltaic panels.
  • Electrical & Computer Engineering: Calculating total electrostatic charge accumulation Q=∬RΟƒ(x,y) dAQ = \iint_R \sigma(x, y) \, dAQ=∬R​σ(x,y)dA across rectangular microchip dies and capacitive sensor pads.
  • Mechanical & Thermal Engineering: Determining total heat transfer rate q=∬Rh(x,y)(Tsβˆ’T∞) dAq = \iint_R h(x, y)(T_s - T_\infty) \, dAq=∬R​h(x,y)(Tsβ€‹βˆ’Tβˆžβ€‹)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 ∬R(3x2βˆ’2y) dA\iint_R (3x^2 - 2y) \, dA∬R​(3x2βˆ’2y)dA over the rectangular region R=[0,2]Γ—[1,3]R = [0, 2] \times [1, 3]R=[0,2]Γ—[1,3], and determine the spatial average value favgf_{\text{avg}}favg​.

Solution

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

∬Rf(x,y) dA=∫cd∫abf(x,y) dx dy\iint_R f(x, y) \, dA = \int_{c}^{d} \int_{a}^{b} f(x, y) \, dx \, dy∬R​f(x,y)dA=∫cdβ€‹βˆ«ab​f(x,y)dxdy
Here a=0,b=2a = 0, b = 2a=0,b=2 and c=1,d=3c = 1, d = 3c=1,d=3. Set up the iterated integral:
I=∫13∫02(3x2βˆ’2y) dx dyI = \int_{1}^{3} \int_{0}^{2} (3x^2 - 2y) \, dx \, dyI=∫13β€‹βˆ«02​(3x2βˆ’2y)dxdy

Step 2: Evaluate the Inner Integral (dxdxdx)
Treat yyy as a constant:

∫02(3x2βˆ’2y) dx=[x3βˆ’2yx]x=0x=2=((2)3βˆ’2y(2))βˆ’0=8βˆ’4y\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∫02​(3x2βˆ’2y)dx=[x3βˆ’2yx]x=0x=2​=((2)3βˆ’2y(2))βˆ’0=8βˆ’4y

Step 3: Evaluate the Outer Integral (dydydy)

∫13(8βˆ’4y) dy=[8yβˆ’2y2]13=(8(3)βˆ’2(3)2)βˆ’(8(1)βˆ’2(1)2)\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)∫13​(8βˆ’4y)dy=[8yβˆ’2y2]13​=(8(3)βˆ’2(3)2)βˆ’(8(1)βˆ’2(1)2)
=(24βˆ’18)βˆ’(8βˆ’2)=6βˆ’6=0= (24 - 18) - (8 - 2) = 6 - 6 = 0=(24βˆ’18)βˆ’(8βˆ’2)=6βˆ’6=0

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

favg=1Area(R)∬Rf(x,y) dAf_{\text{avg}} = \frac{1}{\text{Area}(R)} \iint_R f(x, y) \, dAfavg​=Area(R)1β€‹βˆ¬R​f(x,y)dA
The area of the rectangle is Area(R)=(2βˆ’0)(3βˆ’1)=2Γ—2=4\text{Area}(R) = (2 - 0)(3 - 1) = 2 \times 2 = 4Area(R)=(2βˆ’0)(3βˆ’1)=2Γ—2=4.
favg=14(0)=0f_{\text{avg}} = \frac{1}{4} (0) = 0favg​=41​(0)=0

Example 2: Strategic Order Selection via Fubini's Theorem

Evaluate the double integral:

∬Rxexy dAwhereΒ R=[0,1]Γ—[0,2]\iint_R x e^{xy} \, dA \quad \text{where } R = [0, 1] \times [0, 2]∬R​xexydAwhereΒ R=[0,1]Γ—[0,2]

Solution

Step 1: Analyze Integration Order Strategy
If we integrate with respect to xxx first, the factor xxx in xexyx e^{xy}xexy forces integration by parts. However, if we integrate with respect to yyy first, xxx is treated as a constant multiplier that perfectly matches the internal derivative of the exponent u=xyβ€…β€ŠβŸΉβ€…β€Šdu=x dyu = xy \implies du = x \, dyu=xy⟹du=xdy. Therefore, we choose order dy dxdy \, dxdydx.

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

∬Rxexy dA=∫01∫02xexy dy dx\iint_R x e^{xy} \, dA = \int_{0}^{1} \int_{0}^{2} x e^{xy} \, dy \, dx∬R​xexydA=∫01β€‹βˆ«02​xexydydx

Step 3: Evaluate Inner Integral (dydydy)
Treating xxx as constant:

∫02xexy dy=[exy]y=0y=2=e2xβˆ’e0=e2xβˆ’1\int_{0}^{2} x e^{xy} \, dy = \left[ e^{xy} \right]_{y=0}^{y=2} = e^{2x} - e^0 = e^{2x} - 1∫02​xexydy=[exy]y=0y=2​=e2xβˆ’e0=e2xβˆ’1

Step 4: Evaluate Outer Integral (dxdxdx)

∫01(e2xβˆ’1) dx=[12e2xβˆ’x]01=(12e2βˆ’1)βˆ’(12e0βˆ’0)\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)∫01​(e2xβˆ’1)dx=[21​e2xβˆ’x]01​=(21​e2βˆ’1)βˆ’(21​e0βˆ’0)
=12e2βˆ’1βˆ’12=12e2βˆ’32= \frac{1}{2}e^2 - 1 - \frac{1}{2} = \frac{1}{2}e^2 - \frac{3}{2}=21​e2βˆ’1βˆ’21​=21​e2βˆ’23​


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