MATH-283 (Calculus III) Curriculum

Session 16 - Triple Integrals

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

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

Formulate and evaluate triple iterated integrals over 3D rectangular boxes $B = [a, b] \times [c, d] \times [r, s]$ in Cartesian coordinates, applying Fubini's Theorem across all six differential orderings and leveraging algebraic product separability $\iiint_B g(x)h(y)k(z) \, dV = \left(\int g\right)\left(\int h\right)\left(\int k\right)$ to compute enclosed volumes, total mass, and 3D spatial average values.
Construct rigorous integration limits for general non-rectangular 3D regions $E \subset \mathbb{R}^3$ by executing the 4-step spatial projection protocol: identifying upper and lower bounding surfaces along the inner axis ( $z = u_1(x, y)$ to $z = u_2(x, y)$ ), projecting the 3D boundary onto the 2D coordinate shadow plane $D_{xy}$ , establishing boundary curves, and bounding the outer variable with constants.
Strategically reorder the differentials of a triple integral ( $dz \, dy \, dx \to dx \, dz \, dy$ ) to bypass non-elementary antiderivatives (such as $\int e^{x^2} \, dx$ ), inverting surface functions and analyzing trace intersection curves to determine whether the domain remains unified or partitions into multiple piecewise iterated integrals.

Micro-Lecture

1. Spatial Volume Accumulation & The 3D Riemann Sum

Mathematical Foundation: In single-variable calculus, definite integrals accumulate quantities along a 1D line segment; in multivariable calculus, double integrals accumulate over a 2D planar patch. As established in Multiple Integrals, when physical properties vary continuously across 3D space, we partition a bounded volume $E \subset \mathbb{R}^3$ into an array of infinitesimal rectangular cuboids with side dimensions $\Delta x, \Delta y, \Delta z$ . The volume of each cuboid is $\Delta V = \Delta x \, \Delta y \, \Delta z$ . The triple integral is defined as the infinite refinement limit of the triple Riemann sum: $\iiint_{E} f(x, y, z) \, dV = \lim_{\substack{l \to \infty \\ m \to \infty \\ n \to \infty}} \sum_{i=1}^{l} \sum_{j=1}^{m} \sum_{k=1}^{n} f(x_{ijk}^*, y_{ijk}^*, z_{ijk}^*) \, \Delta x_i \, \Delta y_j \, \Delta z_k$ If $f(x, y, z) = 1$ , the integral yields the spatial volume $V(E) = \iiint_E 1 \, dV$ . If $f(x, y, z) = \rho(x, y, z)$ represents a mass density (in $\text{kg/m}^3$ ) or charge density (in $\text{C/m}^3$ ), the integral calculates the total accumulated mass $M$ or total enclosed electric charge $Q$ .
Engineering Context: Mechanical and aerospace engineers utilize triple integrals to compute mass, centers of gravity, and mass moments of inertia for irregular structural components, ensuring aerodynamic stability and preventing rotational flutter in turbine blades.

2. Fubini's Theorem in Three Dimensions

Mathematical Foundation: When the spatial domain is a rectangular box $B = [a, b] \times [c, d] \times [r, s]$ , the limits of integration for all three coordinate variables are fixed, independent constants. As established in Triple Integrals over a Rectangular Region, Fubini's Theorem guarantees that if $f(x, y, z)$ is continuous across $B$ , the triple integral evaluates as an iterated integral in any of the $3! = 6$ possible differential permutations ( $dz \, dy \, dx$ , $dz \, dx \, dy$ , $dy \, dz \, dx$ , etc.). Furthermore, if the integrand factors into single-variable components $f(x, y, z) = g(x)h(y)k(z)$ , the triple integral decouples into the product of three independent single integrals: $\iiint_{B} g(x)h(y)k(z) \, dV = \left(\int_{a}^{b} g(x) \, dx\right) \left(\int_{c}^{d} h(y) \, dy\right) \left(\int_{r}^{s} k(z) \, dz\right)$
Engineering Context: Electrical and computer engineers evaluate total thermal dissipation and electromagnetic flux in cubic microchip packaging and semiconductor silicon substrates where material properties vary along orthogonal fabrication axes.

3. General Spatial Regions & The Shadow Projection Method

Mathematical Foundation: For general solids with curved faces or inclined planar walls, boundaries depend functionally on neighboring variables. As detailed in Triple Integrals over a General Region, setting up a triple integral requires a systematic "inside-out" perspective: identify the entering floor surface $z = u_1(x, y)$ and exiting ceiling surface $z = u_2(x, y)$ along the inner axis, then project the boundary intersection onto the coordinate shadow plane $D_{xy}$ . The 2D shadow $D_{xy}$ is then treated as a standard double integral region with curve-to-curve middle limits and constant outer limits. When an integrand contains non-elementary single-variable functions, reorienting the spatial sweep order (as documented in Changing the Order of Integration in Triple Integrals) generates algebraic multipliers that make closed-form evaluation possible.
Engineering Context: Civil and environmental engineers use general triple integrals to calculate bulk earthwork excavation volumes and hydraulic reservoir storage capacity, while chemical engineers model pollutant destruction yields across packed catalyst beds.

Skill Block 1

Worked Example A: Evaluating a Monomial Triple Integral over a Rectangular Box

Evaluate the triple integral:

\iiint_{B} 12xyz^2 \, dV

where

B

is the rectangular box defined by

0 \le x \le 1

1 \le y \le 2

, and

0 \le z \le 2

Worked Solution:

Set Up the Iterated Integral:
State Fubini's Theorem for Triple Integrals (defined in Triple Integrals over a Rectangular Region):
$\iiint_{B} f(x, y, z) \, dV = \int_{x=a}^{b} \int_{y=c}^{d} \int_{z=r}^{s} f(x, y, z) \, dz \, dy \, dx$ Substitute the given boundaries: $I = \int_{0}^{1} \int_{1}^{2} \int_{0}^{2} 12xyz^2 \, dz \, dy \, dx$
Evaluate the Innermost Integral ( $dz$ ):
Treat $x$ and $y$ as constants during partial integration:
$\int_{0}^{2} 12xyz^2 \, dz = \left[ 12xy \frac{z^3}{3} \right]_{0}^{2} = \left[ 4xyz^3 \right]_{0}^{2} = 4xy(2)^3 - 0 = 32xy$
Evaluate the Middle Integral ( $dy$ ):
Substitute the inner result into the $y$ -layer, holding $x$ constant:
$\int_{1}^{2} 32xy \, dy = \left[ 16xy^2 \right]_{1}^{2} = 16x(2)^2 - 16x(1)^2 = 64x - 16x = 48x$
Evaluate the Outermost Integral ( $dx$ ):
Substitute the middle result into the $x$ -layer:
$\int_{0}^{1} 48x \, dx = \left[ 24x^2 \right]_{0}^{1} = 24(1)^2 - 0 = 24$ Thus, $\iiint_B 12xyz^2 \, dV = 24$ .

Worked Example B: Average Value of a 3D Linear Potential Field

Find the average value $\overline{f}$ of the scalar field $f(x, y, z) = x + 2y - z$ over the rectangular box $B = [0, 2] \times [0, 3] \times [1, 2]$ .

Worked Solution:

Calculate Geometric Box Volume:
State the Volume of a Rectangular Box Formula:
$V(B) = (b - a)(d - c)(s - r) = (2 - 0)(3 - 0)(2 - 1) = 2 \times 3 \times 1 = 6$
Evaluate the Triple Integral over the Box:
$I = \int_{0}^{2} \int_{0}^{3} \int_{1}^{2} (x + 2y - z) \, dz \, dy \, dx$

Inner $z$ -integral:
$\int_{1}^{2} (x + 2y - z) \, dz = \left[ (x + 2y)z - \frac{z^2}{2} \right]_{1}^{2} = \left(2(x + 2y) - 2\right) - \left(1(x + 2y) - \frac{1}{2}\right) = x + 2y - \frac{3}{2}$

Middle $y$ -integral:
$\int_{0}^{3} \left(x + 2y - \frac{3}{2}\right) \, dy = \left[ xy + y^2 - \frac{3}{2}y \right]_{0}^{3} = 3x + 9 - \frac{9}{2} = 3x + \frac{9}{2}$

Outer $x$ -integral:
$\int_{0}^{2} \left(3x + \frac{9}{2}\right) \, dx = \left[ \frac{3}{2}x^2 + \frac{9}{2}x \right]_{0}^{2} = \frac{3}{2}(4) + \frac{9}{2}(2) = 6 + 9 = 15$
Compute the Spatial Average Value:
State the Average Value Formula for a 3D Function:
$\overline{f} = \frac{1}{V(B)} \iiint_{B} f(x, y, z) \, dV = \frac{15}{6} = 2.5$

Practice Problem 1: Concept & Reading Check

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

(a) State the formal relationship between the differential volume element $dV$ and Cartesian increments $dx, dy, dz$ , and explain why Fubini's Theorem permits all six integration orders to yield identical values on a rectangular box.
(b) In Worked Example A, explain why the variables $x$ and $y$ can be treated as constants during the innermost integration with respect to $z$ .
(c) In the linked engineering application Microprocessor Heat Sink Volumetric Thermal Dissipation (Computer & Mechanical Engineering), how does integrating a 3D volumetric thermal generation density $q(x, y, z)$ differ conceptually from finding the geometric volume of the heat-spreader block?

Solution Key:

(a) The differential volume element $dV = dz \, dy \, dx$ represents the infinitesimal volume of a 3D coordinate prism. Under Fubini's Theorem for continuous functions on rectangular boxes, the boundaries of each variable are independent constants, meaning the order of accumulation does not affect the continuous summation limit of the Riemann sum across the domain.
(b) When performing partial integration with respect to $z$ , the coordinates $x$ and $y$ represent independent orthogonal directions held stationary during the vertical line-slicing pass. Consequently, any expression containing only $x$ and $y$ acts as a constant scaling multiplier with respect to $z$ .
(c) Evaluating $\iiint_B 1 \, dV$ sums unit volume increments, yielding the spatial capacity $V(B)$ in cubic centimeters ( $\text{cm}^3$ ). In contrast, integrating the non-uniform power density $\iiint_B q(x, y, z) \, dV$ weights each spatial differential element $dV$ by its local thermal dissipation rate in $\text{W/cm}^3$ , accumulating the total thermal heat dissipation rate $P_{\text{total}}$ in watts ( $\text{W}$ ).

Show your steps and calculations here.

Practice Problem 2

Evaluate the triple integral over the rectangular box:

I = \int_{-1}^{5} \int_{2}^{4} \int_{0}^{1} (x + yz^2) \, dz \, dy \, dx

Solution Key:

State the Iterated Integral Structure:
State Fubini's Theorem for Triple Integrals (defined in Triple Integrals over a Rectangular Region):
$I = \int_{-1}^{5} \int_{2}^{4} \left[ \int_{0}^{1} (x + yz^2) \, dz \right] dy \, dx$
Evaluate the Innermost Integral ( $dz$ ):
Treat $x$ and $y$ as constants:
$\int_{0}^{1} (x + yz^2) \, dz = \left[ xz + y \frac{z^3}{3} \right]_{0}^{1} = \left( x(1) + \frac{1}{3}y(1)^3 \right) - 0 = x + \frac{1}{3}y$
Evaluate the Middle Integral ( $dy$ ):
Substitute the result into the $y$ -integral over $[2, 4]$ :
$\int_{2}^{4} \left( x + \frac{1}{3}y \right) \, dy = \left[ xy + \frac{1}{6}y^2 \right]_{2}^{4} = \left( 4x + \frac{16}{6} \right) - \left( 2x + \frac{4}{6} \right) = 2x + \frac{12}{6} = 2x + 2$
Evaluate the Outermost Integral ( $dx$ ):
Substitute the linear expression into the $x$ -integral over $[-1, 5]$ :
$\int_{-1}^{5} (2x + 2) \, dx = \left[ x^2 + 2x \right]_{-1}^{5} = \left( 5^2 + 2(5) \right) - \left( (-1)^2 + 2(-1) \right) = (25 + 10) - (1 - 2) = 35 - (-1) = 36$

Show your steps and calculations here.

Practice Problem 3

Evaluate the triple integral over the unit cube $B = [0, 1] \times [0, 1] \times [0, 1]$ :

I = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} e^{x + 2y + 3z} \, dz \, dy \, dx

Solution Key:

Apply the Product Separability Property:
State the Algebraic Separability Theorem (defined in Triple Integrals over a Rectangular Region):
$\iiint_{B} g(x)h(y)k(z) \, dV = \left( \int_{a}^{b} g(x) \, dx \right) \left( \int_{c}^{d} h(y) \, dy \right) \left( \int_{r}^{s} k(z) \, dz \right)$ Decompose the exponential integrand: $e^{x + 2y + 3z} = e^x \cdot e^{2y} \cdot e^{3z}$ Thus: $I = \left( \int_{0}^{1} e^x \, dx \right) \left( \int_{0}^{1} e^{2y} \, dy \right) \left( \int_{0}^{1} e^{3z} \, dz \right)$
Evaluate Each Single-Variable Factor:
- $x$ -factor: $\int_{0}^{1} e^x \, dx = [e^x]_{0}^{1} = e^1 - e^0 = e - 1$
- $y$ -factor: $\int_{0}^{1} e^{2y} \, dy = \left[ \frac{1}{2} e^{2y} \right]_{0}^{1} = \frac{1}{2}(e^2 - e^0) = \frac{e^2 - 1}{2}$
- $z$ -factor: $\int_{0}^{1} e^{3z} \, dz = \left[ \frac{1}{3} e^{3z} \right]_{0}^{1} = \frac{1}{3}(e^3 - e^0) = \frac{e^3 - 1}{3}$
Multiply the Evaluated Terms:
$I = (e - 1) \cdot \left(\frac{e^2 - 1}{2}\right) \cdot \left(\frac{e^3 - 1}{3}\right) = \frac{(e - 1)(e^2 - 1)(e^3 - 1)}{6}$

Show your steps and calculations here.

Practice Problem 4

Evaluate the triple integral over the rectangular region $E$ :

\iiint_{E} (3 - x + 2yz) \, dV \quad \text{where } E = [0, 2] \times [-1, 1] \times [1, 2]

Solution Key:

Formulate the Iterated Triple Integral:
Apply Fubini's Theorem for Triple Integrals:
$I = \int_{0}^{2} \int_{-1}^{1} \int_{1}^{2} (3 - x + 2yz) \, dz \, dy \, dx$
Evaluate the Innermost Integral ( $dz$ ):
Treat $x$ and $y$ as constants:
$\int_{1}^{2} (3 - x + 2yz) \, dz = \left[ (3 - x)z + yz^2 \right]_{1}^{2} = \Big((3 - x)(2) + y(4)\Big) - \Big((3 - x)(1) + y(1)\Big)$ $= (6 - 2x + 4y) - (3 - x + y) = 3 - x + 3y$
Evaluate the Middle Integral ( $dy$ ):
Substitute the expression into the $y$ -layer over $[-1, 1]$ :
$\int_{-1}^{1} (3 - x + 3y) \, dy = \left[ (3 - x)y + \frac{3}{2}y^2 \right]_{-1}^{1}$ Notice that $3y$ is an odd function over the symmetric interval $[-1, 1]$ , so its integral is zero: $= \left( (3 - x)(1) + \frac{3}{2} \right) - \left( (3 - x)(-1) + \frac{3}{2} \right) = (3 - x) - (-(3 - x)) = 2(3 - x) = 6 - 2x$
Evaluate the Outermost Integral ( $dx$ ):
$\int_{0}^{2} (6 - 2x) \, dx = \left[ 6x - x^2 \right]_{0}^{2} = \left( 6(2) - (2)^2 \right) - 0 = 12 - 4 = 8$

Show your steps and calculations here.

Regroup 1

Clarification on Iterated Order: Fubini's Theorem guarantees that on a box with constant boundaries, any of the six differential orders ( $dz\,dy\,dx, dx\,dz\,dy$ , etc.) yields identical results. Choose orders that simplify algebra early (e.g., integrating odd terms over symmetric intervals like $[-1, 1]$ ).
Separability Requirement: Separability applies only when the integrand factors into $g(x)h(y)k(z)$ and all six integration limits are constants. Never attempt to separate when limits contain variables.

Skill Block 2

Worked Example A: First-Octant Tetrahedron with Scaled Boundary Plane

Evaluate the triple integral:

\iiint_{E} z \, dV

where

E

is the solid tetrahedron in the first octant bounded by the planes

x = 0

y = 0

z = 0

, and

2x + y + z = 2

Worked Solution:

Determine Inner $z$ -Limits:
State the Type I Region Definition (defined in Triple Integrals over a General Region):
$u_1(x, y) \le z \le u_2(x, y)$ In the first octant, the floor is the $xy$ -plane ( $z = 0$ ). The top boundary is the slanted plane $2x + y + z = 2 \implies z = 2 - 2x - y$ .
Thus, the inner limits are: $0 \le z \le 2 - 2x - y$
Project onto the $xy$ -Plane to Form 2D Shadow $D$ :
Set $z = 0$ in the plane equation:
$2x + y = 2 \implies y = 2 - 2x$ In the first octant ( $x \ge 0, y \ge 0$ ), the region $D$ is a triangle bounded by $y = 0$ and $y = 2 - 2x$ .
Middle limits: $0 \le y \le 2 - 2x$ .
Outer limits: Set $y = 0 \implies 2x = 2 \implies 0 \le x \le 1$ .
Set Up and Evaluate the Iterated Integral:
$I = \int_{0}^{1} \int_{0}^{2-2x} \int_{0}^{2-2x-y} z \, dz \, dy \, dx$

Inner $z$ -integral:
$\int_{0}^{2-2x-y} z \, dz = \left[ \frac{z^2}{2} \right]_{0}^{2-2x-y} = \frac{1}{2}(2 - 2x - y)^2$

Middle $y$ -integral: Use $u$ -substitution: let $u = 2 - 2x - y \implies dy = -du$ .
When $y = 0, u = 2 - 2x$ ; when $y = 2 - 2x, u = 0$ .
$\int_{2-2x}^{0} \frac{1}{2}u^2 (-du) = \int_{0}^{2-2x} \frac{1}{2}u^2 \, du = \left[ \frac{u^3}{6} \right]_{0}^{2-2x} = \frac{(2 - 2x)^3}{6} = \frac{8(1 - x)^3}{6} = \frac{4}{3}(1 - x)^3$

Outer $x$ -integral:
$\int_{0}^{1} \frac{4}{3}(1 - x)^3 \, dx = \left[ -\frac{4}{9}(1 - x)^4 \right]_{0}^{1} = 0 - \left(-\frac{4}{9}(1)^4\right) = \frac{4}{9}$ Thus, $\iiint_E z \, dV = \frac{4}{9}$ .

Worked Example B: Spatial Volume Enclosed by a Parabolic Cylinder

Find the volume of the solid region $E$ bounded by the parabolic cylinder $y = x^2$ and the planes $z = 0$ and $y + z = 4$ .

Worked Solution:

Determine Inner $z$ -Limits:
The solid extends from the floor $z = 0$ up to the ceiling plane $y + z = 4 \implies z = 4 - y$ .
Thus, $0 \le z \le 4 - y$ .
Determine 2D Projection Shadow $D$ in the $xy$ -Plane:
Setting $z = 0$ in the ceiling plane gives the line $y = 4$ . The lateral wall is the parabolic cylinder $y = x^2$ .
Middle limits ( $y$ ): for any fixed $x$ , $y$ goes from the parabola up to the line: $x^2 \le y \le 4$ .
Outer limits ( $x$ ): the intersection points are $x^2 = 4 \implies -2 \le x \le 2$ .
Formulate and Evaluate the Volume Integral:
State the Volume via Triple Integration Formula:
$V = \iiint_{E} 1 \, dV = \int_{-2}^{2} \int_{x^2}^{4} \int_{0}^{4-y} 1 \, dz \, dy \, dx$

Inner $z$ -integral:
$\int_{0}^{4-y} 1 \, dz = 4 - y$

Middle $y$ -integral:
$\int_{x^2}^{4} (4 - y) \, dy = \left[ 4y - \frac{y^2}{2} \right]_{x^2}^{4} = \left( 16 - 8 \right) - \left( 4x^2 - \frac{x^4}{2} \right) = 8 - 4x^2 + \frac{x^4}{2}$

Outer $x$ -integral: Exploiting the symmetry of the even integrand:
$V = 2 \int_{0}^{2} \left( 8 - 4x^2 + \frac{x^4}{2} \right) \, dx = 2 \left[ 8x - \frac{4}{3}x^3 + \frac{x^5}{10} \right]_{0}^{2}$ $V = 2 \left( 16 - \frac{32}{3} + \frac{16}{5} \right) = 2 \left( \frac{240 - 160 + 48}{15} \right) = 2\left(\frac{128}{15}\right) = \frac{256}{15}$

Practice Problem 5: Concept & Reading Check

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

(a) Define a Type I 3D region, and describe the geometric meaning of the 2D projection shadow $D_{xy}$ in the $xy$ -plane.
(b) In Worked Example A, explain how setting the ceiling surface equation equal to $z = 0$ reveals the boundary line of the 2D shadow region.
(c) In the linked engineering application Aerospace Fuel Tank Center of Mass and Slosh Baffle (Aerospace & Mechanical Engineering), why does determining the vertical center of mass coordinate $\bar{z}$ require evaluating two separate triple integrals ( $M_{xy}$ and $M$ )?

Solution Key:

(a) A Type I region in $\mathbb{R}^3$ is defined as $E = \{(x, y, z) \mid (x, y) \in D_{xy}, \, u_1(x, y) \le z \le u_2(x, y)\}$ . The domain $D_{xy}$ is the orthogonal geometric projection (or shadow) of the 3D solid onto the $xy$ -coordinate plane, which serves as the 2D domain over which the middle and outer single integrals are bounded.
(b) In the first octant, the bottom boundary of the tetrahedron is the horizontal plane $z = 0$ . The boundary where the ceiling surface terminates into the floor occurs along the trace curve where $z = 0$ . Substituting $z = 0$ into $2x + y + z = 2$ gives $2x + y = 2 \implies y = 2 - 2x$ , which forms the hypotenuse boundary of the triangular shadow in the $xy$ -plane.
(c) Center of mass represents the weighted average spatial location of mass throughout the solid. To calculate the vertical coordinate $\bar{z} = \frac{M_{xy}}{M}$ , engineers must evaluate the first moment about the $xy$ -plane $M_{xy} = \iiint_E z \rho(x, y, z) \, dV$ (which weights each parcel of mass by its vertical lever arm $z$ ) and normalize it by the total accumulated mass $M = \iiint_E \rho(x, y, z) \, dV$ .

Show your steps and calculations here.

Practice Problem 6

Evaluate the triple integral over the general region $E$ :

\iiint_{E} (5x - 3y) \, dV

where

E

is the solid tetrahedron in the first octant bounded by

x = 0

y = 0

z = 0

, and

x + y + z = 1

, using the integration order

dz \, dy \, dx

Solution Key:

Set Up the Type I Boundaries:
State the Type I Region Definition (defined in Triple Integrals over a General Region):
$0 \le z \le 1 - x - y, \quad 0 \le y \le 1 - x, \quad 0 \le x \le 1$ Formulate the ordered iterated integral: $I = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} (5x - 3y) \, dz \, dy \, dx$
Evaluate the Innermost Integral ( $dz$ ):
Treat $x$ and $y$ as constants:
$\int_{0}^{1-x-y} (5x - 3y) \, dz = \Big[ (5x - 3y)z \Big]_{0}^{1-x-y} = (5x - 3y)(1 - x - y)$ Expand the product: $= 5x(1 - x) - 5xy - 3y(1 - x) + 3y^2 = (5x - 5x^2) - (5x + 3 - 3x)y + 3y^2 = (5x - 5x^2) - (2x + 3)y + 3y^2$
Evaluate the Middle Integral ( $dy$ ):
$\int_{0}^{1-x} \left( (5x - 5x^2) - (2x + 3)y + 3y^2 \right) \, dy = \left[ (5x - 5x^2)y - \frac{2x+3}{2}y^2 + y^3 \right]_{0}^{1-x}$ Substitute $y = 1 - x$ and factor $(1 - x)^2$ : $= 5x(1 - x)^2 - \frac{2x+3}{2}(1 - x)^2 + (1 - x)^3 = (1 - x)^2 \left[ 5x - \frac{2x+3}{2} + (1 - x) \right]$ $= (1 - x)^2 \left[ 4x + 1 - x - \frac{3}{2} \right] = (1 - x)^2 \left( 3x - \frac{1}{2} \right)$ Expand the polynomial: $= (1 - 2x + x^2)\left(3x - \frac{1}{2}\right) = 3x^3 - \frac{13}{2}x^2 + 4x - \frac{1}{2}$
Evaluate the Outermost Integral ( $dx$ ):
$\int_{0}^{1} \left( 3x^3 - \frac{13}{2}x^2 + 4x - \frac{1}{2} \right) \, dx = \left[ \frac{3}{4}x^4 - \frac{13}{6}x^3 + 2x^2 - \frac{1}{2}x \right]_{0}^{1}$ $= \frac{3}{4} - \frac{13}{6} + 2 - \frac{1}{2} = \frac{9 - 26 + 24 - 6}{12} = \frac{1}{12}$

Show your steps and calculations here.

Practice Problem 7

Find the volume of the first-octant tetrahedron bounded by the intercept plane:

\frac{x}{2} + \frac{y}{1} + \frac{z}{3} = 1

with

x \ge 0, y \ge 0, z \ge 0

, setting up and evaluating a triple integral in rectangular coordinates.

Solution Key:

Determine Integration Limits:
State the Volume via Triple Integration Formula:
$V = \iiint_{E} 1 \, dV$
- Inner $z$ -bounds: Solve for $z \implies z = 3 - \frac{3}{2}x - 3y$ . Floor is $z = 0$ .
- Middle $y$ -bounds: Set $z = 0 \implies \frac{x}{2} + y = 1 \implies 0 \le y \le 1 - \frac{x}{2}$ .
- Outer $x$ -bounds: Set $y = 0 \implies \frac{x}{2} = 1 \implies 0 \le x \le 2$ . $V = \int_{0}^{2} \int_{0}^{1 - x/2} \int_{0}^{3 - 3x/2 - 3y} 1 \, dz \, dy \, dx$
Evaluate Inner $z$ -Integral:
$\int_{0}^{3 - 3x/2 - 3y} 1 \, dz = 3 - \frac{3}{2}x - 3y = 3\left(1 - \frac{x}{2}\right) - 3y$
Evaluate Middle $y$ -Integral:
$\int_{0}^{1 - x/2} \left( 3\left(1 - \frac{x}{2}\right) - 3y \right) \, dy = \left[ 3\left(1 - \frac{x}{2}\right)y - \frac{3}{2}y^2 \right]_{0}^{1 - x/2}$ Substitute $y = 1 - \frac{x}{2}$ : $= 3\left(1 - \frac{x}{2}\right)^2 - \frac{3}{2}\left(1 - \frac{x}{2}\right)^2 = \frac{3}{2}\left(1 - \frac{x}{2}\right)^2 = \frac{3}{2}\left(1 - x + \frac{x^2}{4}\right) = \frac{3}{2} - \frac{3}{2}x + \frac{3}{8}x^2$
Evaluate Outermost $x$ -Integral:
$V = \int_{0}^{2} \left( \frac{3}{2} - \frac{3}{2}x + \frac{3}{8}x^2 \right) \, dx = \left[ \frac{3}{2}x - \frac{3}{4}x^2 + \frac{1}{8}x^3 \right]_{0}^{2}$ $= \left( \frac{3}{2}(2) - \frac{3}{4}(4) + \frac{1}{8}(8) \right) - 0 = 3 - 3 + 1 = 1$ (Note: This matches the geometric pyramid volume formula $V = \frac{1}{6} \cdot a \cdot b \cdot c = \frac{1}{6}(2)(1)(3) = 1$ .)

Show your steps and calculations here.

Practice Problem 8

Let $E = \{(x, y, z) \in \mathbb{R}^3 \mid x \ge 0, y \ge 0, z \ge 0, x + y + z \le 1\}$ be a solid tetrahedron with variable mass density $\rho(x, y, z) = x + y + z$ . Compute the total mass:

M = \iiint_{E} \rho(x, y, z) \, dV

Solution Key:

Formulate the Mass Integral:
State the Total Mass Formula with Variable Density (defined in Triple Integrals over a General Region):
$M = \iiint_{E} \rho(x, y, z) \, dV = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} (x + y + z) \, dz \, dy \, dx$
Evaluate Inner $z$ -Integral:
Treat $x$ and $y$ as constants:
$\int_{0}^{1-x-y} (x + y + z) \, dz = \left[ (x + y)z + \frac{1}{2}z^2 \right]_{0}^{1-x-y} = (x + y)(1 - x - y) + \frac{1}{2}(1 - x - y)^2$ Express $(x + y)$ as $1 - (1 - x - y)$ : $= \Big(1 - (1 - x - y)\Big)(1 - x - y) + \frac{1}{2}(1 - x - y)^2 = (1 - x - y) - \frac{1}{2}(1 - x - y)^2$
Evaluate Middle $y$ -Integral:
Use $u$ -substitution: let $u = 1 - x - y \implies dy = -du$ .
When $y = 0, u = 1 - x$ ; when $y = 1 - x, u = 0$ .
$\int_{1-x}^{0} \left( u - \frac{1}{2}u^2 \right) (-du) = \int_{0}^{1-x} \left( u - \frac{1}{2}u^2 \right) \, du = \left[ \frac{1}{2}u^2 - \frac{1}{6}u^3 \right]_{0}^{1-x} = \frac{1}{2}(1 - x)^2 - \frac{1}{6}(1 - x)^3$
Evaluate Outermost $x$ -Integral:
Use $w$ -substitution: let $w = 1 - x \implies dx = -dw$ .
When $x = 0, w = 1$ ; when $x = 1, w = 0$ .
$M = \int_{1}^{0} \left( \frac{1}{2}w^2 - \frac{1}{6}w^3 \right) (-dw) = \int_{0}^{1} \left( \frac{1}{2}w^2 - \frac{1}{6}w^3 \right) \, dw = \left[ \frac{1}{6}w^3 - \frac{1}{24}w^4 \right]_{0}^{1} = \frac{1}{6} - \frac{1}{24} = \frac{4 - 1}{24} = \frac{3}{24} = \frac{1}{8}$

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Regroup 2

Common Failure Mode: Outer Variable Dependency: Never allow inner variables ( $y$ or $z$ ) to appear in the outer limits ( $dx$ ). The outermost integral must always collapse to pure numerical constants, guaranteeing a single scalar real number output.
Visualizing Shadows: If students struggle with the 2D projection $D$ , instruct them to shine a mental flashlight straight down the $z$ -axis onto the $xy$ -floor. The perimeter where light is blocked defines the curve equations for the middle integral.

Skill Block 3

Worked Example A: Resolving a Non-Elementary Antiderivative via Axis Reordering

Consider the iterated triple integral:

I = \int_{1}^{4} \int_{0}^{1} \int_{2y}^{2} \frac{e^{x^2}}{z} \, dx \, dy \, dz

Rewrite the integral in the order

dy \, dx \, dz

and evaluate its exact value.

Worked Solution:

Analyze Original Bounds:
Extract the inequalities:
$1 \le z \le 4, \quad 0 \le y \le 1, \quad 2y \le x \le 2$ The innermost integral $\int e^{x^2} \, dx$ has no elementary antiderivative. Changing the order of the inner two variables to $dy \, dx$ is required.
Re-slice the 2D Domain in the $xy$ -Plane:
The outer variable $z$ remains on $[1, 4]$ . In the $xy$ -plane, the region is bounded by $x = 2y$ , $x = 2$ , and $y = 0$ .
Solve for $y$ in terms of $x$ :
$x = 2y \implies y = \frac{x}{2}$
For the new order $dy \, dx$ :
- For a fixed $x$ , $y$ goes from $y = 0$ up to $y = \frac{x}{2}$ .
- $x$ spans the constant interval $0 \le x \le 2$ .
Set Up and Evaluate the Rewritten Integral:
$I = \int_{1}^{4} \int_{0}^{2} \int_{0}^{x/2} \frac{e^{x^2}}{z} \, dy \, dx \, dz$

Inner $y$ -integral:
$\int_{0}^{x/2} \frac{e^{x^2}}{z} \, dy = \left[ \frac{e^{x^2}}{z} y \right]_{0}^{x/2} = \frac{x e^{x^2}}{2z}$

Middle $x$ -integral: Use $u$ -substitution: let $u = x^2 \implies du = 2x \, dx \implies x \, dx = \frac{du}{2}$ .
When $x = 0, u = 0$ ; when $x = 2, u = 4$ .
$\int_{0}^{2} \frac{x e^{x^2}}{2z} \, dx = \frac{1}{2z} \int_{0}^{4} \frac{1}{2} e^u \, du = \frac{1}{4z} [e^u]_{0}^{4} = \frac{e^4 - 1}{4z}$

Outer $z$ -integral:
$\int_{1}^{4} \frac{e^4 - 1}{4z} \, dz = \frac{e^4 - 1}{4} \Big[ \ln |z| \Big]_{1}^{4} = \frac{e^4 - 1}{4} (\ln 4 - \ln 1) = \frac{(e^4 - 1)\ln 4}{4}$

Worked Example B: Reordering across Intersecting Cylindrical and Planar Boundaries

Rewrite the triple integral $\int_{0}^{1} \int_{0}^{1-x^2} \int_{0}^{1-x} f(x, y, z) \, dz \, dy \, dx$ into the order $dx \, dy \, dz$ .

Worked Solution:

Extract Original Inequalities:
$0 \le x \le 1, \quad 0 \le y \le 1 - x^2, \quad 0 \le z \le 1 - x$ Bounding surfaces: $x = 0, y = 0, z = 0$ , parabolic cylinder $y = 1 - x^2 \implies x = \sqrt{1 - y}$ , and plane $z = 1 - x \implies x = 1 - z$ .
Determine New Inner $x$ -Limits:
Looking along the $x$ -axis from $x = 0$ , the forward boundary is constrained by two distinct surfaces:
$0 \le x \le \min(1 - z, \sqrt{1 - y})$
Find the Trace Intersection in the $yz$ -Plane:
Equate the two surface equations to determine where the boundary switches:
$1 - z = \sqrt{1 - y} \implies (1 - z)^2 = 1 - y \implies y = 1 - (1 - z)^2 = 2z - z^2$
In the $yz$ -shadow ( $0 \le z \le 1, 0 \le y \le 1$ ), the curve $y = 2z - z^2$ partitions the domain into two regions:
- Region 1 (where $0 \le y \le 2z - z^2$ ): The plane is the active bound: $0 \le x \le 1 - z$ .
- Region 2 (where $2z - z^2 \le y \le 1$ ): The cylinder is the active bound: $0 \le x \le \sqrt{1 - y}$ .
Formulate the Split Iterated Integrals:
$\int_{0}^{1} \int_{0}^{2z-z^2} \int_{0}^{1-z} f(x, y, z) \, dx \, dy \, dz + \int_{0}^{1} \int_{2z-z^2}^{1} \int_{0}^{\sqrt{1-y}} f(x, y, z) \, dx \, dy \, dz$

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 changing the order of integration mandatory when facing an integrand such as $\frac{e^{x^2}}{z}$ with inner differential $dx$ ?
(b) In Worked Example B, explain how the intersection trace $y = 2z - z^2$ indicates that the reordered integral must be split into two pieces.
(c) In the linked engineering application Chemical Reactor Continuous Yield Optimization (Chemical & Environmental Engineering), how does changing the integration order from $dx \, dy \, dz$ to $dy \, dx \, dz$ enable chemical engineers to avoid expensive numerical simulations?

Solution Key:

(a) The function $e^{x^2}$ does not possess an elementary antiderivative. In the order $dx \, dy \, dz$ , the inner integration cannot be completed using standard functions. Reordering to $dy \, dx \, dz$ integrates $y$ first across the interval $[0, x/2]$ , generating the variable factor $x$ , which immediately allows $u$ -substitution with $u = x^2, du = 2x \, dx$ .
(b) The forward extent of the solid along the $x$ -axis is bounded by two competing geometric surfaces: the inclined plane $x = 1 - z$ and the parabolic cylinder $x = \sqrt{1 - y}$ . Setting them equal yields the trace curve $y = 2z - z^2$ . Because one surface cuts in earlier for $y \le 2z - z^2$ and the other for $y \ge 2z - z^2$ , the 2D shadow must be partitioned into two distinct sub-domains.
(c) Analytical solutions eliminate the need to execute 3D finite-difference or finite-volume Navier-Stokes grid calculations for every reactor geometry. Having an explicit closed-form formula allows engineers to optimize catalyst bed length, width, and height in milliseconds within automated design loops.

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Practice Problem 10

Consider the solid tetrahedron $E$ in the first octant bounded by $x = 0, y = 0, z = 0$ , and $x + y + z = 1$ . Starting from the iterated integral in the order $dz \, dy \, dx$ :

I = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} xyz \, dz \, dy \, dx

rewrite the integral in the order

dx \, dz \, dy

and evaluate its exact value.

Solution Key:

Determine New Bounds for Order $dx \, dz \, dy$ :
State the Re-slicing Protocol in the Projection Plane (defined in Changing the Order of Integration in Triple Integrals):
- Inner $x$ -bounds: Isolate $x$ from $x + y + z = 1 \implies 0 \le x \le 1 - y - z$ .
- Shadow in $yz$ -plane: Set $x = 0 \implies y + z = 1$ .
- Middle $z$ -bounds: Solving for $z$ gives $0 \le z \le 1 - y$ .
- Outer $y$ -bounds: Intercept is at $y = 1 \implies 0 \le y \le 1$ . $I = \int_{0}^{1} \int_{0}^{1-y} \int_{0}^{1-y-z} xyz \, dx \, dz \, dy$
Evaluate Inner $x$ -Integral:
Treat $y$ and $z$ as constants:
$\int_{0}^{1-y-z} xyz \, dx = \left[ \frac{1}{2}x^2 yz \right]_{0}^{1-y-z} = \frac{1}{2}yz(1 - y - z)^2$
Evaluate Middle $z$ -Integral:
Expand $(1 - y - z)^2 = (1 - y)^2 - 2(1 - y)z + z^2$ :
$\int_{0}^{1-y} \left( \frac{1}{2}y(1-y)^2 z - y(1-y)z^2 + \frac{1}{2}yz^3 \right) dz$ $= \left[ \frac{1}{4}y(1-y)^2 z^2 - \frac{1}{3}y(1-y)z^3 + \frac{1}{8}yz^4 \right]_{0}^{1-y}$ Substitute $z = 1 - y$ : $= \left( \frac{1}{4} - \frac{1}{3} + \frac{1}{8} \right) y(1 - y)^4 = \left( \frac{6 - 8 + 3}{24} \right) y(1 - y)^4 = \frac{1}{24}y(1 - y)^4$
Evaluate Outermost $y$ -Integral:
Use $u$ -substitution: let $u = 1 - y \implies y = 1 - u$ and $dy = -du$ .
When $y = 0, u = 1$ ; when $y = 1, u = 0$ .
$I = \int_{1}^{0} \frac{1}{24}(1 - u)u^4 (-du) = \int_{0}^{1} \frac{1}{24}(u^4 - u^5) \, du = \frac{1}{24} \left[ \frac{u^5}{5} - \frac{u^6}{6} \right]_{0}^{1} = \frac{1}{24}\left(\frac{1}{30}\right) = \frac{1}{720}$

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Practice Problem 11

Rewrite the volume integral:

V = \int_{0}^{2} \int_{0}^{x} \int_{0}^{x-y} 1 \, dz \, dy \, dx

into the order

dx \, dz \, dy

and evaluate the resulting integral to determine the volume.

Solution Key:

Extract Original Inequalities and Invert for $x$ :
From the original integral:
$0 \le z \le x - y, \quad 0 \le y \le x, \quad 0 \le x \le 2$ The lower bound for $x$ is the slanted plane $z = x - y \implies x = y + z$ .
The upper bound for $x$ is the vertical cut-off plane $x = 2$ .
Thus, the inner limits are: $y + z \le x \le 2$
Determine 2D Projection onto the $yz$ -Plane:
Set the inner limits equal to find the boundary curve:
$y + z = 2 \implies z = 2 - y$
In the first quadrant of the $yz$ -plane ( $y \ge 0, z \ge 0$ ):
- Middle $z$ -bounds: $0 \le z \le 2 - y$ .
- Outer $y$ -bounds: $0 \le y \le 2$ .
Formulate and Evaluate the Rewritten Integral:
$V = \int_{0}^{2} \int_{0}^{2-y} \int_{y+z}^{2} 1 \, dx \, dz \, dy$

Inner $x$ -integral:
$\int_{y+z}^{2} 1 \, dx = [x]_{y+z}^{2} = 2 - (y + z) = (2 - y) - z$

Middle $z$ -integral:
$\int_{0}^{2-y} ((2 - y) - z) \, dz = \left[ (2 - y)z - \frac{z^2}{2} \right]_{0}^{2-y} = (2 - y)^2 - \frac{1}{2}(2 - y)^2 = \frac{1}{2}(2 - y)^2$

Outer $y$ -integral: Use $u$ -substitution: let $u = 2 - y \implies dy = -du$ .
When $y = 0, u = 2$ ; when $y = 2, u = 0$ .
$V = \int_{2}^{0} \frac{1}{2}u^2 (-du) = \int_{0}^{2} \frac{1}{2}u^2 \, du = \left[ \frac{u^3}{6} \right]_{0}^{2} = \frac{8}{6} = \frac{4}{3}$

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Practice Problem 12

Given the triple integral representing the volume of solid $E$ :

V = \int_{0}^{1/2} \int_{0}^{1/2} \int_{0}^{1-2x+2y} 1 \, dz \, dy \, dx

rewrite the integral in the order

dx \, dz \, dy

(splitting the region into two iterated integrals if necessary) and evaluate the volume.

Solution Key:

Analyze Original Boundaries:
Original limits:
$0 \le x \le \frac{1}{2}, \quad 0 \le y \le \frac{1}{2}, \quad 0 \le z \le 1 - 2x + 2y$ Solving the ceiling surface for $x$ : $2x = 1 + 2y - z \implies x = \frac{1 + 2y - z}{2}$ However, $x$ is also bounded by the constant plane $x = 1/2$ . Thus, the upper bound is: $x = \min\left(\frac{1}{2}, \, \frac{1 + 2y - z}{2}\right)$
Locate the Intersection Trace Curve:
Set the two constraints equal:
$\frac{1 + 2y - z}{2} = \frac{1}{2} \implies 1 + 2y - z = 1 \implies z = 2y$
In the $yz$ -plane where $0 \le y \le 1/2$ :
- For $0 \le z \le 2y$ : $\frac{1 + 2y - z}{2} \ge \frac{1}{2}$ , so the constant wall is the active upper bound: $0 \le x \le \frac{1}{2}$ .
- For $2y \le z \le 1 + 2y$ : the slanted plane is the active upper bound: $0 \le x \le \frac{1 + 2y - z}{2}$ .
Formulate the Split Iterated Integrals:
$V = \int_{0}^{1/2} \int_{0}^{2y} \int_{0}^{1/2} 1 \, dx \, dz \, dy + \int_{0}^{1/2} \int_{2y}^{1+2y} \int_{0}^{\frac{1+2y-z}{2}} 1 \, dx \, dz \, dy$
Evaluate the First Integral ( $V_1$ ):
$\int_{0}^{1/2} 1 \, dx = \frac{1}{2}$ $\int_{0}^{2y} \frac{1}{2} \, dz = \left[ \frac{1}{2}z \right]_{0}^{2y} = y$ $\int_{0}^{1/2} y \, dy = \left[ \frac{y^2}{2} \right]_{0}^{1/2} = \frac{1}{2}\left(\frac{1}{4}\right) = \frac{1}{8}$
Evaluate the Second Integral ( $V_2$ ):
$\int_{0}^{\frac{1+2y-z}{2}} 1 \, dx = \frac{1 + 2y - z}{2}$ $\int_{2y}^{1+2y} \frac{1 + 2y - z}{2} \, dz$ Use $u$ -substitution: let $u = 1 + 2y - z \implies dz = -du$ .
When $z = 2y, u = 1$ ; when $z = 1 + 2y, u = 0$ . $\int_{1}^{0} \frac{u}{2} (-du) = \int_{0}^{1} \frac{u}{2} \, du = \left[ \frac{u^2}{4} \right]_{0}^{1} = \frac{1}{4}$ Now perform the outermost integration with respect to $y$ : $\int_{0}^{1/2} \frac{1}{4} \, dy = \left[ \frac{1}{4}y \right]_{0}^{1/2} = \frac{1}{8}$
Combine the Components:
$V = V_1 + V_2 = \frac{1}{8} + \frac{1}{8} = \frac{1}{4}$

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Regroup 3

Trace Intersection Verification: Always check boundary values at the intersection trace (e.g., when $z = 2y$ , both formulas for $x$ yield $1/2$ ). This ensures continuity across the partitioned domains.
Why Split? Splitting occurs because geometry does not care about our coordinate axes; when multiple surfaces clip a volume simultaneously, piecewise integration is mathematically required.

Synthesis Wrap-up

Core Takeaways:
1. Triple integrals over rectangular boxes are computed layer-by-layer; Fubini's Theorem guarantees all 6 order permutations yield identical results.
2. For general regions, follow the 4-step projection protocol: surface ceiling/floor $\to$ 2D projection shadow $D$ $\to$ boundary curves $\to$ outer constants.
3. When an inner antiderivative is non-elementary, reordering the differentials can introduce variable multipliers that unlock standard $u$ -substitution.
Knewton Alta Homework Preview:
- Triple Integrals over a Rectangular Region and a General Region
- Changing the Order of Integration of Triple Integrals