MATH-283 (Calculus III) Curriculum

Session 14 - Double Integrals

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

By the end of this session, you will be able to:

Formulate, interpret, and evaluate iterated double integrals over rectangular domains using Fubini's Theorem, apply the product separability theorem $\iint_R g(x)h(y)\,dA = \left(\int_a^b g(x)\,dx\right)\left(\int_c^d h(y)\,dy\right)$ , and compute the spatial average value $f_{\text{avg}} = \frac{1}{\text{Area}(R)}\iint_R f(x, y)\,dA$ across planar design envelopes.
Classify, sketch, and evaluate double integrals over general planar regions by distinguishing between Type I (vertical slices: $g_1(x) \le y \le g_2(x)$ ) and Type II (horizontal slices: $h_1(y) \le x \le h_2(y)$ ) geometries, determining curve intersections to establish outer constant bounds.
Execute the 4-step order-switching protocol (Extract $\to$ Sketch $\to$ Invert $\to$ Re-slice) to convert between Type I and Type II iterated integrals, partition compound multi-curve boundaries into disjoint sub-regions, and unlock closed-form solutions for non-elementary integrands such as $e^{x^2}$ , $\frac{\sin y}{y}$ , and $\sqrt{1+y^3}$ .

Micro-Lecture

1. Spatial Accumulation & The Volume Under a Surface

Mathematical Foundation: In single-variable calculus, definite integrals accumulate quantities along a 1-dimensional interval $\int_a^b f(x)\,dx$ . When physical phenomena vary across a 2-dimensional surface, integration generalizes to the double integral $\iint_D f(x, y)\,dA$ , defined in Multiple Integrals as the infinite limit of a double Riemann sum. Geometrically, if $f(x, y) \ge 0$ , the double integral calculates the exact volume of the cylindrical solid lying between the planar region $D$ and the surface $z = f(x, y)$ . If $f(x, y) = 1$ , the integral yields the planar surface area $\text{Area}(D) = \iint_D 1\,dA$ .
Engineering Context: Civil and structural engineers model concrete bridge foundations subject to non-uniform soil reaction pressure; integrating the contact stress distribution $\sigma(x, y)$ over the footing footprint calculates the total structural bearing load supported by the ground.

2. Constant Domains & Fubini's Iterated Slicing

Mathematical Foundation: On a closed rectangular domain $R = [a, b] \times [c, d]$ , all boundary edges are constant lines. As proven in Double Integrals over a Rectangular Region, Fubini's Theorem establishes that double integrals equal iterated single integrals evaluated in either coordinate order: $\iint_R f\,dA = \int_c^d \int_a^b f\,dx\,dy = \int_a^b \int_c^d f\,dy\,dx$ . Choosing the optimal slicing direction eliminates tedious integration by parts when one variable acts as a chain-rule multiplier. Furthermore, dividing the double integral by the rectangular area produces the spatial average value $f_{\text{avg}} = \frac{1}{\text{Area}(R)}\iint_R f\,dA$ .
Engineering Context: Entertainment and electrical engineers evaluate radiant light emissions from flat-panel LED displays, using spatial averages to calculate overall luminaire efficiency and mitigate optical hotspot banding across modular video walls.

3. General Boundaries & Slicing Inversion

Mathematical Foundation: When a domain is bounded by curved or angled profiles, it is classified as a general region (formulated in Double Integrals over a General Region). In Type I domains (vertical slices), outer limits are constants $x \in [a, b]$ and inner limits are functions $y \in [g_1(x), g_2(x)]$ . In Type II domains (horizontal slices), outer limits are constants $y \in [c, d]$ and inner limits are functions $x \in [h_1(y), h_2(y)]$ . When an integrand lacks an elementary antiderivative (e.g., $e^{x^2}$ ), engineers employ Switching the Order of Integration via the 4-step re-slicing protocol to invert boundary equations and evaluate the system.
Engineering Context: Ocean and aerospace engineers evaluate moments of inertia and hydrostatic waterplane stability curves on ship hulls and tapered composite wings, switching integration orders to eliminate algebraic radical singularities in automated finite element routines.

Skill Block 1

Worked Example A: Polynomial Iterated Double Integral over a Rectangle

Evaluate the double integral:

\iint_R (4x^3 + 6xy^2) \, dA \quad \text{where } R = [0, 2] \times [1, 3]

Worked Solution:

Formulate the Iterated Integral:
State Fubini's Theorem for Rectangular Regions (defined in Double Integrals over a Rectangular Region):
$\iint_R f(x, y) \, dA = \int_{a}^{b} \int_{c}^{d} f(x, y) \, dy \, dx$ Substitute the domain limits $x \in [0, 2]$ and $y \in [1, 3]$ : $I = \int_{0}^{2} \int_{1}^{3} (4x^3 + 6xy^2) \, dy \, dx$
Evaluate the Inner Integral ( $dy$ ):
Treat $x$ as a constant and integrate with respect to $y$ :
$\int_{1}^{3} (4x^3 + 6xy^2) \, dy = \left[ 4x^3 y + 2xy^3 \right]_{y=1}^{y=3}$ Substitute the boundary limits: $= \Big(4x^3(3) + 2x(3)^3\Big) - \Big(4x^3(1) + 2x(1)^3\Big)$ $= (12x^3 + 54x) - (4x^3 + 2x) = 8x^3 + 52x$
Evaluate the Outer Integral ( $dx$ ):
Integrate the resulting single-variable function with respect to $x$ :
$I = \int_{0}^{2} (8x^3 + 52x) \, dx = \left[ 2x^4 + 26x^2 \right]_{0}^{2}$ Evaluate at the limits: $I = \Big(2(2)^4 + 26(2)^2\Big) - 0 = 2(16) + 26(4) = 32 + 104 = 136$

Worked Example B: Strategic Order Selection via Fubini's Shortcut

Evaluate the double integral:

\iint_R x \sin(xy) \, dA \quad \text{where } R = [0, 1] \times [0, \pi]

demonstrating why integrating with respect to

y

first is superior to integrating with respect to

x

first.

Worked Solution:

Analyze Slicing Strategy:
State Fubini's Theorem for Rectangular Regions (defined in Double Integrals over a Rectangular Region):
$\iint_R f(x, y) \, dA = \int_{a}^{b} \int_{c}^{d} f(x, y) \, dy \, dx = \int_{c}^{d} \int_{a}^{b} f(x, y) \, dx \, dy$
- If we integrate with respect to $x$ first, $\int x \sin(xy) \, dx$ requires integration by parts because $x$ appears both outside and inside the sine function.
- If we integrate with respect to $y$ first, $x$ acts as a constant multiplier. Since $\frac{\partial}{\partial y}(xy) = x$ , the factor $x$ is the exact derivative of the inner argument, yielding an immediate single-step antiderivative!
Formulate and Evaluate Inner Integral ( $dy$ ):
Set up the iterated integral with $dy$ on the inside:
$I = \int_{0}^{1} \int_{0}^{\pi} x \sin(xy) \, dy \, dx$ Integrate with respect to $y$ , using $\int x \sin(xy) \, dy = -\cos(xy)$ : $\int_{0}^{\pi} x \sin(xy) \, dy = \left[ -\cos(xy) \right]_{y=0}^{y=\pi} = -\cos(\pi x) - (-\cos(0)) = 1 - \cos(\pi x)$
Evaluate Outer Integral ( $dx$ ):
Substitute into the outer integral:
$I = \int_{0}^{1} \Big(1 - \cos(\pi x)\Big) \, dx = \left[ x - \frac{\sin(\pi x)}{\pi} \right]_{0}^{1}$ Evaluate at the limits: $I = \left( 1 - \frac{\sin(\pi)}{\pi} \right) - \left( 0 - \frac{\sin(0)}{\pi} \right) = (1 - 0) - (0 - 0) = 1$

Practice Problem 1: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Under what mathematical conditions does Fubini's Theorem permit swapping the integration order from $dy \, dx$ to $dx \, dy$ , and why must all four limits of integration remain constants for a rectangular region?
(b) In Worked Example B, explain why integrating with respect to $y$ first eliminates the need for integration by parts.
(c) Referencing Optoelectronic LED Display Panel Average Irradiance, explain physically why the spatial average irradiance $E_{\text{avg}}$ divides the total double integral by the rectangular surface area $\text{Area}(R) = (b-a)(d-c)$ .

Solution Key:

(a) State Fubini's Theorem for Rectangular Regions (defined in Double Integrals over a Rectangular Region): $\iint_R f(x, y) \, dA = \int_c^d \int_a^b f(x, y) \, dx \, dy = \int_a^b \int_c^d f(x, y) \, dy \, dx$ This identity is valid for all functions $f(x, y)$ that are continuous on the closed rectangle $R$ . The limits are fixed constants because a rectangle in Cartesian space is explicitly defined by intervals $x \in [a, b]$ and $y \in [c, d]$ .
(b) In the expression $\int x \sin(xy)\,dy$ , $x$ is independent of $y$ . By the chain rule, $\frac{d}{dy}[-\cos(xy)] = x\sin(xy)$ . If integrating with respect to $x$ first, $x$ is the integration variable and would require integration by parts with $u = x$ and $dv = \sin(xy)\,dx$ .
(c) State the Spatial Average Value Formula: $E_{\text{avg}} = \frac{1}{\text{Area}(R)} \iint_R E(x, y) \, dA$ In optoelectronic display engineering, light emission varies across the panel due to heating and circuit resistance. The double integral accumulates total radiant power ( $\text{W}$ ). Dividing by the geometric surface area $\text{Area}(R) = (0.5)(0.4) = 0.2\text{ m}^2$ normalizes the total power to express the mean flux per unit area ( $\text{W/m}^2$ ), which is used to calibrate LED driver brightness and prevent visual hotspot banding.

Show your steps and calculations here.

Practice Problem 2

Evaluate the double integral:

\iint_R (3x^2 - 2y) \, dA \quad \text{where } R = [0, 2] \times [1, 3]

and determine the spatial average value

f_{\text{avg}}

of the function across the region

R

Solution Key:

Formulate the Iterated Integral:
State Fubini's Theorem for Rectangular Regions (defined in Double Integrals over a Rectangular Region):
$\iint_R f(x, y) \, dA = \int_{1}^{3} \int_{0}^{2} (3x^2 - 2y) \, dx \, dy$
Evaluate the Inner Integral ( $dx$ ):
Treat $y$ as a constant:
$\int_{0}^{2} (3x^2 - 2y) \, dx = \left[ x^3 - 2yx \right]_{x=0}^{x=2} = \Big((2)^3 - 2y(2)\Big) - 0 = 8 - 4y$
Evaluate the Outer Integral ( $dy$ ):
$\int_{1}^{3} (8 - 4y) \, dy = \left[ 8y - 2y^2 \right]_{1}^{3} = \Big(8(3) - 2(3)^2\Big) - \Big(8(1) - 2(1)^2\Big)$ $= (24 - 18) - (8 - 2) = 6 - 6 = 0$
Compute the 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 rectangular domain is: $\text{Area}(R) = (2 - 0)(3 - 1) = 2 \times 2 = 4$ $f_{\text{avg}} = \frac{1}{4}(0) = 0$

Show your steps and calculations here.

Practice Problem 3

Evaluate the double integral by selecting the strategically simpler order of integration:

\iint_R y e^{xy} \, dA \quad \text{where } R = [0, 1] \times [0, 2]

Solution Key:

Formulate the Strategically Optimal Iterated Integral:
State Fubini's Theorem for Rectangular Regions (defined in Double Integrals over a Rectangular Region):
$\iint_R y e^{xy} \, dA = \int_{0}^{2} \int_{0}^{1} y e^{xy} \, dx \, dy$
Evaluate the Inner Integral ( $dx$ ):
Holding $y$ constant, $y$ acts as the exact derivative multiplier of $xy$ ( $\frac{\partial}{\partial x}(xy) = y$ ):
$\int_{0}^{1} y e^{xy} \, dx = \left[ e^{xy} \right]_{x=0}^{x=1} = e^{y(1)} - e^{y(0)} = e^y - 1$
Evaluate the Outer Integral ( $dy$ ):
$\int_{0}^{2} (e^y - 1) \, dy = \left[ e^y - y \right]_{0}^{2} = \Big(e^2 - 2\Big) - \Big(e^0 - 0\Big) = e^2 - 2 - 1 = e^2 - 3$

Show your steps and calculations here.

Practice Problem 4

Evaluate the double integral:

\iint_R (2x \cos y - 3y^2) \, dA \quad \text{where } R = [0, 3] \times \left[0, \frac{\pi}{2}\right]

Solution Key:

Formulate the Iterated Integral:
State Fubini's Theorem for Rectangular Regions (defined in Double Integrals over a Rectangular Region):
$\iint_R (2x \cos y - 3y^2) \, dA = \int_{0}^{\pi/2} \int_{0}^{3} (2x \cos y - 3y^2) \, dx \, dy$
Evaluate the Inner Integral ( $dx$ ):
Treat $y$ as a constant:
$\int_{0}^{3} (2x \cos y - 3y^2) \, dx = \left[ x^2 \cos y - 3y^2 x \right]_{x=0}^{x=3} = \Big(9 \cos y - 9y^2\Big) - 0 = 9 \cos y - 9y^2$
Evaluate the Outer Integral ( $dy$ ):
$\int_{0}^{\pi/2} (9 \cos y - 9y^2) \, dy = \left[ 9 \sin y - 3y^3 \right]_{0}^{\pi/2}$ Evaluate at the limits: $= \left( 9 \sin\left(\frac{\pi}{2}\right) - 3\left(\frac{\pi}{2}\right)^3 \right) - 0 = 9(1) - 3\left(\frac{\pi^3}{8}\right) = 9 - \frac{3\pi^3}{8}$ (Note: Multiplying over the interval $x \in [0, 3]$ , $3 \times 3y^2 = 9y^2$ , whose antiderivative is $3y^3 = \frac{3\pi^3}{8}$ .)

Show your steps and calculations here.

Regroup 1

Order Selection Heuristic: How can you determine within 5 seconds whether $dy \, dx$ or $dx \, dy$ is the faster calculation on a rectangle? (Look for cross-variable products where one variable can act as a chain-rule multiplier).
Constant Bounds Integrity: Why is Fubini's Theorem so straightforward over rectangles compared to curved domains? (Because the region is defined entirely by constant endpoints, so switching differentials requires zero change to boundary equations).

Skill Block 2

Worked Example A: Type I General Region (Parabola and Slanted Line)

Evaluate the double integral:

\iint_D (x + y) \, dA

where

D

is the region bounded by the parabola

y = x^2

and the line

y = 2x

Worked Solution:

Find Intersection Points (Outer Bounds):
Set the two boundary curves equal to locate where they intersect:
$x^2 = 2x \implies x^2 - 2x = 0 \implies x(x - 2) = 0$ The outer constant limits are $x = 0$ and $x = 2$ .
Establish the Functional Inner Limits:
Test an intermediate value $x = 1$ : $y = 2(1) = 2$ 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, $D$ is a Type I Region:
$D = \{(x, y) \in \mathbb{R}^2 \mid 0 \le x \le 2, \, x^2 \le y \le 2x\}$
Formulate and Evaluate Inner Integral ( $dy$ ):
State the Type I Double Integral Formula (defined in Double Integrals over a General Region):
$\iint_D f(x, y) \, dA = \int_{a}^{b} \int_{g_1(x)}^{g_2(x)} f(x, y) \, dy \, dx$ $I = \int_{0}^{2} \int_{x^2}^{2x} (x + y) \, dy \, dx$ Evaluate with respect to $y$ , treating $x$ as constant: $\int_{x^2}^{2x} (x + y) \, dy = \left[ xy + \frac{1}{2}y^2 \right]_{y=x^2}^{y=2x} = \Big(x(2x) + \frac{1}{2}(2x)^2\Big) - \Big(x(x^2) + \frac{1}{2}(x^2)^2\Big)$ $= (2x^2 + 2x^2) - (x^3 + \frac{1}{2}x^4) = 4x^2 - x^3 - \frac{1}{2}x^4$
Evaluate Outer Integral ( $dx$ ):
$\int_{0}^{2} \left( 4x^2 - x^3 - \frac{1}{2}x^4 \right) \, dx = \left[ \frac{4}{3}x^3 - \frac{1}{4}x^4 - \frac{1}{10}x^5 \right]_{0}^{2}$ Substitute $x = 2$ : $= \left( \frac{4}{3}(8) - \frac{1}{4}(16) - \frac{1}{10}(32) \right) - 0 = \frac{32}{3} - 4 - \frac{16}{5}$ Find a common denominator ( $15$ ): $= \frac{160}{15} - \frac{60}{15} - \frac{48}{15} = \frac{52}{15}$

Worked Example B: Type II General Region (Horizontal Slices)

Evaluate the double integral:

\iint_D (x + y) \, dA

where

D

is the region bounded by

x = y^2

and

x = y

from

y = 0

y = 2

Worked Solution:

Identify the Type II Boundary Structure:
Because the boundary equations explicitly express $x$ as functions of $y$ , this region is naturally formulated as a Type II Region (defined in Double Integrals over a General Region):
- Outer constant limits for $y$ : $0 \le y \le 2$
- Inner limits for $x$ : Left boundary is $x = y^2$ , right boundary is $x = y$
Formulate the Iterated Integral:
State the Type II Double Integral Formula:
$\iint_D f(x, y) \, dA = \int_{c}^{d} \int_{h_1(y)}^{h_2(y)} f(x, y) \, dx \, dy$ $I = \int_{0}^{2} \int_{y^2}^{y} (x + y) \, dx \, dy$
Evaluate Inner Integral ( $dx$ ):
Treat $y$ as a constant:
$\int_{y^2}^{y} (x + y) \, dx = \left[ \frac{1}{2}x^2 + yx \right]_{x=y^2}^{x=y} = \left( \frac{1}{2}y^2 + y(y) \right) - \left( \frac{1}{2}(y^2)^2 + y(y^2) \right)$ $= \frac{3}{2}y^2 - y^3 - \frac{1}{2}y^4$
Evaluate Outer Integral ( $dy$ ):
$I = \int_{0}^{2} \left( \frac{3}{2}y^2 - y^3 - \frac{1}{2}y^4 \right) \, dy = \left[ \frac{1}{2}y^3 - \frac{1}{4}y^4 - \frac{1}{10}y^5 \right]_{0}^{2}$ Evaluate at $y = 2$ : $= \left( \frac{1}{2}(8) - \frac{1}{4}(16) - \frac{1}{10}(32) \right) - 0 = 4 - 4 - 3.2 = -3.2 = -\frac{16}{5}$

Practice Problem 5: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) What is the structural distinction between a Type I region and a Type II region in terms of which variable has constant outer limits versus functional inner limits?
(b) In Worked Example A, explain how setting $x^2 = 2x$ establishes the outer constant boundaries, and how testing an intermediate point confirms which curve is on top.
(c) Referencing Hydrostatic Pressure Load on a Parabolic Dam Floodgate, explain why fluid pressure depth is integrated along horizontal slices (Type II) rather than vertical slices.

Solution Key:

(a) State the Classification of General Planar Regions (defined in Double Integrals over a General Region):
- Type I (Vertical Slices): $D = \{(x, y) \mid a \le x \le b, \, g_1(x) \le y \le g_2(x)\}$ , where $x$ has constant limits and $y$ has functional limits.
- Type II (Horizontal Slices): $D = \{(x, y) \mid c \le y \le d, \, h_1(y) \le x \le h_2(y)\}$ , where $y$ has constant limits and $x$ has functional limits.
(b) Setting $g_1(x) = g_2(x) \implies x^2 = 2x \implies x(x - 2) = 0$ finds the geometric intersection points $(0, 0)$ and $(2, 4)$ , which bound the domain horizontally. Testing $x = 1 \in (0, 2)$ yields $g_2(1) = 2$ and $g_1(1) = 1$ ; since $2 > 1$ , $y = 2x$ is strictly above $y = x^2$ over the entire interior.
(c) In hydraulic engineering, hydrostatic pressure is governed by fluid depth below the surface. Because the floodgate is bounded on the left by $x = y^2$ and on the right by $x = 4$ , a Type II horizontal slice lies at a uniform water depth $(2 - y)$ , so $P(y) = \rho g(2 - y)$ factors out of the inner $dx$ integral as a pure constant.

Show your steps and calculations here.

Practice Problem 6

Evaluate the double integral over the Type I region:

\iint_D \sqrt{y} \, dA \quad \text{where } D = \{(x, y) \in \mathbb{R}^2 \mid 0 \le x \le 1, \, x^2 \le y \le x\}

Solution Key:

Formulate the Type I Iterated Integral:
State the Type I Double Integral Formula:
$\iint_D f(x, y) \, dA = \int_{a}^{b} \int_{g_1(x)}^{g_2(x)} f(x, y) \, dy \, dx$ $I = \int_{0}^{1} \int_{x^2}^{x} y^{1/2} \, dy \, dx$
Evaluate the Inner Integral ( $dy$ ):
Treat $x$ as a constant:
$\int_{x^2}^{x} y^{1/2} \, dy = \left[ \frac{2}{3} y^{3/2} \right]_{y=x^2}^{y=x} = \frac{2}{3} \left( x^{3/2} - (x^2)^{3/2} \right) = \frac{2}{3} \left( x^{3/2} - x^3 \right)$
Evaluate the Outer Integral ( $dx$ ):
$I = \frac{2}{3} \int_{0}^{1} (x^{3/2} - x^3) \, dx = \frac{2}{3} \left[ \frac{2}{5}x^{5/2} - \frac{1}{4}x^4 \right]_{0}^{1}$ Evaluate at the limits: $= \frac{2}{3} \left( \frac{2}{5} - \frac{1}{4} \right) = \frac{2}{3} \left( \frac{8 - 5}{20} \right) = \frac{2}{3} \left( \frac{3}{20} \right) = \frac{2}{20} = \frac{1}{10}$

Show your steps and calculations here.

Practice Problem 7

Evaluate the double integral over the Type II region:

\iint_D \sqrt{y} \, dA \quad \text{where } D = \{(x, y) \in \mathbb{R}^2 \mid 0 \le y \le 1, \, y \le x \le \sqrt{y}\}

Solution Key:

Formulate the Type II Iterated Integral:
State the Type II Double Integral Formula:
$\iint_D f(x, y) \, dA = \int_{c}^{d} \int_{h_1(y)}^{h_2(y)} f(x, y) \, dx \, dy$ $I = \int_{0}^{1} \int_{y}^{\sqrt{y}} \sqrt{y} \, dx \, dy$
Evaluate the Inner Integral ( $dx$ ):
Since $\sqrt{y}$ contains no $x$ , it is treated as a constant:
$\int_{y}^{\sqrt{y}} \sqrt{y} \, dx = \sqrt{y} [x]_{y}^{\sqrt{y}} = \sqrt{y} (\sqrt{y} - y) = y - y^{3/2}$
Evaluate the Outer Integral ( $dy$ ):
$I = \int_{0}^{1} (y - y^{3/2}) \, dy = \left[ \frac{1}{2}y^2 - \frac{2}{5}y^{5/2} \right]_{0}^{1} = \frac{1}{2} - \frac{2}{5} = \frac{5 - 4}{10} = \frac{1}{10}$ (Notice that Practice Problems 6 and 7 represent the exact same geometric region in the $xy$ -plane, confirming that Type I and Type II evaluations yield identical numeric results).

Show your steps and calculations here.

Practice Problem 8

An optoelectronic display module features a wedge-shaped active zone $R$ bounded by $y = 0$ , $y = x$ , and $x = 2$ (spatial units in meters). Emitted irradiance is given by:

E(x, y) = E_0 (1 - 0.2x) \quad (\text{W/m}^2)

Compute the total radiant power $\iint_R E \, dA$ as a Type I double integral.
Find the spatial average irradiance $E_{\text{avg}}$ across the wedge in terms of $E_0$ .

Solution Key:

Evaluate the Total Irradiance Integral:
State the Type I Double Integral Formula (defined in Double Integrals over a General Region):
$\iint_R E(x, y) \, dA = \int_{0}^{2} \int_{0}^{x} E_0 (1 - 0.2x) \, dy \, dx$ Since the integrand contains no $y$ , it acts as a constant during inner integration: $\int_{0}^{x} E_0 (1 - 0.2x) \, dy = E_0 (1 - 0.2x) [y]_0^x = E_0 (x - 0.2x^2)$ Now integrate with respect to $x$ from $0$ to $2$ : $\iint_R E \, dA = E_0 \int_{0}^{2} \left( x - \frac{1}{5}x^2 \right) \, dx = E_0 \left[ \frac{1}{2}x^2 - \frac{1}{15}x^3 \right]_{0}^{2}$ Evaluate at $x = 2$ : $= E_0 \left( \frac{4}{2} - \frac{8}{15} \right) = E_0 \left( 2 - \frac{8}{15} \right) = E_0 \left( \frac{30 - 8}{15} \right) = \frac{22}{15}E_0$
Compute Spatial Average Irradiance ( $E_{\text{avg}}$ ):
State the Spatial Average Value Formula:
$E_{\text{avg}} = \frac{1}{\text{Area}(R)} \iint_R E \, dA$ The region $R$ is a right triangle with base $2\text{ m}$ and height $2\text{ m}$ : $\text{Area}(R) = \frac{1}{2}(2)(2) = 2\text{ m}^2$ $E_{\text{avg}} = \frac{1}{2} \left( \frac{22}{15}E_0 \right) = \frac{11}{15}E_0 \approx 0.733E_0\text{ W/m}^2$

Show your steps and calculations here.

Regroup 2

Common Slicing Trap: Why does setting limits like $\int_{x^2}^{2x} \int_0^2 f\,dx\,dy$ cause an automatic failure in engineering software? (Because variables appear in the outer limits, leaving a non-constant function instead of a physical scalar number).
Choosing Type I vs. Type II: When is Type II strictly required over Type I? (When solving boundary equations for $y$ introduces piecewise square roots or vertical step discontinuities, whereas solving for $x$ yields single continuous functions).

Skill Block 3

Worked Example A: Dual-Type Geometric Slicing on a Triangular Domain

Let $R$ be the triangular region with vertices $(0, 0), (3, 0),$ and $(3, 2)$ . Evaluate $\iint_R (2x + 3y) \, dA$ by setting it up first as a Type I integral, then verifying the result as a Type II integral.

Worked Solution:

Identify the Boundary Lines:
The triangle has three boundary segments:
- Bottom edge: $y = 0$ along the $x$ -axis from $x = 0$ to $x = 3$ .
- Right edge: $x = 3$ from $y = 0$ to $y = 2$ .
- Hypotenuse edge connecting $(0, 0)$ to $(3, 2)$ : slope $m = \frac{2 - 0}{3 - 0} = \frac{2}{3} \implies y = \frac{2}{3}x \iff x = \frac{3}{2}y$ .
Method 1: Type I Setup (Vertical Slices):
State the Type I Double Integral Formula (defined in Double Integrals over a General Region):
$x \in [0, 3], \quad 0 \le y \le \frac{2}{3}x$ $I = \int_{0}^{3} \int_{0}^{\frac{2}{3}x} (2x + 3y) \, dy \, dx$ Evaluate inner integral ( $dy$ ): $\int_{0}^{\frac{2}{3}x} (2x + 3y) \, dy = \left[ 2xy + \frac{3}{2}y^2 \right]_{0}^{\frac{2}{3}x} = 2x\left(\frac{2}{3}x\right) + \frac{3}{2}\left(\frac{2}{3}x\right)^2 = \frac{4}{3}x^2 + \frac{3}{2}\left(\frac{4}{9}x^2\right) = \frac{4}{3}x^2 + \frac{2}{3}x^2 = 2x^2$ Evaluate outer integral ( $dx$ ): $I = \int_{0}^{3} 2x^2 \, dx = \left[ \frac{2}{3}x^3 \right]_{0}^{3} = \frac{2}{3}(27) = 18$
Method 2: Type II Setup (Horizontal Slices):
State the Type II Double Integral Formula (defined in Double Integrals over a General Region):
- Outer bounds for $y$ : $y \in [0, 2]$
- Horizontal slices enter on the left at the line $x = \frac{3}{2}y$ and exit on the right at the vertical line $x = 3$ : $I = \int_{0}^{2} \int_{\frac{3}{2}y}^{3} (2x + 3y) \, dx \, dy$ Evaluate inner integral ( $dx$ ): $\int_{\frac{3}{2}y}^{3} (2x + 3y) \, dx = \left[ x^2 + 3yx \right]_{x=\frac{3}{2}y}^{x=3} = \Big(9 + 9y\Big) - \left( \left(\frac{3}{2}y\right)^2 + 3y\left(\frac{3}{2}y\right) \right)$ $= 9 + 9y - \left( \frac{9}{4}y^2 + \frac{9}{2}y^2 \right) = 9 + 9y - \frac{27}{4}y^2$ Evaluate outer integral ( $dy$ ): $I = \int_{0}^{2} \left( 9 + 9y - \frac{27}{4}y^2 \right) \, dy = \left[ 9y + \frac{9}{2}y^2 - \frac{9}{4}y^3 \right]_{0}^{2}$ $= 9(2) + \frac{9}{2}(4) - \frac{9}{4}(8) = 18 + 18 - 18 = 18$ Both methods produce the exact same value of $18$ .

Worked Example B: Overcoming a Non-Elementary Antiderivative

Evaluate the double integral by switching the order of integration:

I = \int_{0}^{1} \int_{y}^{1} e^{x^2} \, dx \, dy

Worked Solution:

Execute Step 1 (Extract Original Bounds):
From the integral limits:
$\text{Type II:} \quad 0 \le y \le 1, \quad y \le x \le 1$ Notice that $\int e^{x^2} \, dx$ has no elementary antiderivative.
Execute Step 2 & 3 (Sketch Domain & Invert Curves):
The boundary equations are $y = 0$ , $y = 1$ , $x = 1$ , and $x = y$ .
Inverting $x = y$ gives $y = x$ . The region is a triangle with vertices at $(0, 0), (1, 0),$ and $(1, 1)$ .
Execute Step 4 (Re-slice as Type I):
Apply the 4-Step Re-slicing Protocol (defined in Switching the Order of Integration):
- Outer constant limits for $x$ : $0 \le x \le 1$
- Vertical slices enter at the bottom line $y = 0$ and exit at the slanted line $y = x$ : $0 \le y \le x$ Formulate the new integral: $I = \int_{0}^{1} \int_{0}^{x} e^{x^2} \, dy \, dx$
Evaluate Switched Integral:
The inner integrand $e^{x^2}$ contains no $y$ , so it is treated as a constant:
$\int_{0}^{x} e^{x^2} \, dy = e^{x^2} [y]_0^x = x e^{x^2}$ Now evaluate the outer integral using $u$ -substitution: let $u = x^2 \implies du = 2x \, dx \implies x \, dx = \frac{du}{2}$ : $I = \int_{0}^{1} \frac{1}{2} e^u \, du = \left[ \frac{1}{2}e^u \right]_{0}^{1} = \frac{1}{2}(e^1 - e^0) = \frac{1}{2}(e - 1)$

Practice Problem 9: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Why is it strictly invalid to switch the order of integration over a general region by merely swapping the functional integral signs (e.g., writing $\int_{g_1(x)}^{g_2(x)} \int_a^b f\,dx\,dy$ )?
(b) In Worked Example B, describe how sketching the triangular region $0 \le y \le 1, y \le x \le 1$ allows you to reformulate the limits as $0 \le x \le 1, 0 \le y \le x$ .
(c) In Aircraft Wing Torsional Polar Moment of Inertia, why does re-slicing a tapered airfoil cross-section from vertical chords to horizontal spanwise strips simplify the calculation?

Solution Key:

(a) State the Golden Rule of Iterated Integrals (defined in Switching the Order of Integration): $\iint_D f \, dA \in \mathbb{R}$ The outer integral must always have constant limits $[a, b]$ or $[c, d]$ . Merely moving the functional limits to the outer integral leaves variables in the final evaluated result, violating the fundamental theorem of calculus.
(b) The original limits $y \le x \le 1$ describe horizontal arrows entering at $x = y$ and terminating at the vertical line $x = 1$ , starting from $y = 0$ up to $y = 1$ . When sketched, the domain is the lower right triangle bounded by $y = 0, x = 1,$ and $y = x$ . Slicing vertically, $x$ ranges continuously from $0$ to $1$ , and for each $x$ , the vertical ray runs from the baseline $y = 0$ up to the line $y = x$ .
(c) In aerospace structural analysis, the boundary curve is $y = \sqrt{x}$ . When integrating in Type I chords, intermediate steps involve fractional powers $\int x^2 \sqrt{x}\,dx = \int x^{5/2}\,dx$ . Re-slicing horizontally as Type II inverts the boundary to the clean polynomial $x = y^2$ , so the inner integral integrates pure integer powers $\int_{y^2}^4 x^2\,dx = [\frac{x^3}{3}]_{y^2}^4 = \frac{64 - y^6}{3}$ , eliminating radical fractions.

Show your steps and calculations here.

Practice Problem 10

Let $R$ be bounded by $y = x^3$ , $y = 4$ , and the vertical lines $x = -2$ and $x = 2$ .

Convert the bounds to a Type II region, splitting the domain into two sub-regions $R_1$ and $R_2$ as necessary.
Evaluate the total geometric area $\iint_R 1 \, dA$ .

Solution Key:

Convert to Type II by Inverting Boundaries & Splitting Domain:
State the Boundary Inversion Protocol (defined in Switching the Order of Integration):
$y = x^3 \implies x = y^{1/3}$
The domain spans vertically from $y = (-2)^3 = -8$ to $y = (2)^3 = 8$ . At $y = 4$ , the curve crosses $x = 4^{1/3} \approx 1.587$ :
- Lower Sub-region $R_1$ (from $y = -8$ to $y = 4$ ):
  Left boundary is the vertical line $x = -2$ ; right boundary is the curve $x = y^{1/3}$ . $R_1 = \{(x, y) \in \mathbb{R}^2 \mid -8 \le y \le 4, \, -2 \le x \le y^{1/3}\}$
- Upper Sub-region $R_2$ (from $y = 4$ to $y = 8$ ):
  Left boundary is the curve $x = y^{1/3}$ ; right boundary is the vertical line $x = 2$ . $R_2 = \{(x, y) \in \mathbb{R}^2 \mid 4 \le y \le 8, \, y^{1/3} \le x \le 2\}$
Evaluate the Area Integral as the Sum of Two Type II Integrals:
$\text{Area}(R) = \int_{-8}^{4} \int_{-2}^{y^{1/3}} 1 \, dx \, dy + \int_{4}^{8} \int_{y^{1/3}}^{2} 1 \, dx \, dy$

Evaluate $R_1$ Integral:
$\int_{-8}^{4} [x]_{-2}^{y^{1/3}} \, dy = \int_{-8}^{4} (y^{1/3} + 2) \, dy = \left[ \frac{3}{4}y^{4/3} + 2y \right]_{-8}^{4}$ $= \left( \frac{3}{4}(4)^{4/3} + 2(4) \right) - \left( \frac{3}{4}(-8)^{4/3} + 2(-8) \right)$ Since $4^{4/3} = 4 \times 4^{1/3}$ and $(-8)^{4/3} = ((-8)^{1/3})^4 = (-2)^4 = 16$ : $= \left( 3(4^{1/3}) + 8 \right) - \left( \frac{3}{4}(16) - 16 \right) = 3(4^{1/3}) + 8 - (12 - 16) = 3(4^{1/3}) + 12$

Evaluate $R_2$ Integral:
$\int_{4}^{8} [x]_{y^{1/3}}^{2} \, dy = \int_{4}^{8} (2 - y^{1/3}) \, dy = \left[ 2y - \frac{3}{4}y^{4/3} \right]_{4}^{8}$ $= \left( 2(8) - \frac{3}{4}(8)^{4/3} \right) - \left( 2(4) - \frac{3}{4}(4)^{4/3} \right)$ Since $8^{4/3} = (8^{1/3})^4 = 2^4 = 16$ : $= \left( 16 - \frac{3}{4}(16) \right) - \left( 8 - 3(4^{1/3}) \right) = (16 - 12) - 8 + 3(4^{1/3}) = 3(4^{1/3}) - 4$

Sum the Components:
$\text{Total Area} = \Big(3(4^{1/3}) + 12\Big) + \Big(3(4^{1/3}) - 4\Big) = 6(4^{1/3}) + 8 = 6\sqrt[3]{4} + 8 \approx 17.524$

Show your steps and calculations here.

Practice Problem 11

Evaluate the double integral by switching the order of integration:

\int_{0}^{\pi} \int_{x}^{\pi} \frac{\sin y}{y} \, dy \, dx

Solution Key:

Extract Original Boundaries & Recognize Non-Elementary Integrand:
The original Type I bounds are:
$0 \le x \le \pi, \quad x \le y \le \pi$ Because $\int \frac{\sin y}{y} \, dy$ cannot be solved in elementary terms, we must switch to Type II.
Re-slice as a Type II General Region:
Apply the 4-Step Re-slicing Protocol (defined in Switching the Order of Integration):
- The triangular domain has vertices at $(0, 0), (0, \pi),$ and $(\pi, \pi)$ .
- The vertical extent is $0 \le y \le \pi$ .
- A horizontal slice enters at the left vertical line $x = 0$ and exits at the line $x = y$ : $0 \le x \le y$ Formulate the Type II integral: $I = \int_{0}^{\pi} \int_{0}^{y} \frac{\sin y}{y} \, dx \, dy$
Evaluate the Switched Iterated Integral:
Evaluate inner integral ( $dx$ ):
$\int_{0}^{y} \frac{\sin y}{y} \, dx = \frac{\sin y}{y} [x]_{0}^{y} = \frac{\sin y}{y}(y - 0) = \sin y$ Evaluate outer integral ( $dy$ ): $I = \int_{0}^{\pi} \sin y \, dy = \left[ -\cos y \right]_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2$

Show your steps and calculations here.

Regroup 3

Order Reversal as an Unlocking Key: Notice how in Problems 9 and 11, functions that appeared mathematically "unintegrable" ( $e^{x^2}$ and $\frac{\sin y}{y}$ ) instantly dissolved into high-school $u$ -substitutions the moment the slicing direction was switched.
Region Decomposition Diagnostic: How do you know when a region must be split into multiple integrals during order switching? (Whenever a horizontal or vertical scanning line transitions between different bounding formulas, such as switching from a curve to a flat boundary wall).

Synthesis Wrap-up

Core Takeaways from Session 14:
1. Double integrals extend accumulation into 2D space, calculating volumes, areas, total physical charges, and spatial averages.
2. Over rectangular regions, Fubini's Theorem permits free order swapping; choose the order that creates chain-rule derivatives and avoids integration by parts.
3. Over general regions, outer limits must always be constants; inner limits must be functions. Slicing direction (Type I vertical vs. Type II horizontal) dictates boundary formulation.
4. Switching the order of integration is not just a mathematical curiosity—it is a vital tool for overcoming non-elementary antiderivatives and simplifying multi-component engineering boundary models.
Knewton Alta Homework Alignment:
- Double Integrals over Rectangular Regions Part 1: Basic iterated integration, Fubini's Theorem, and product separability.
- Double Integrals over Rectangular Regions Part 2: Spatial average values and applications to physical mass/charge distributions.
- Double Integrals over General Regions: Setting up and evaluating Type I and Type II regions, and executing the 4-step order-switching protocol.