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Triple integrals 1 | Double and triple integrals | Multivariable Calculus | Khan Academy

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IN THIS VIDEO

01Coordinate bounds of the solid
02Why this example is used before triple integrals
03Geometric volume of the rectangular box

FROM THE LEARNING NOTES

0 \le x \le 3,\quad 0 \le y \le 4,\quad 0 \le z \le 2

Coordinate bounds of the solid

SELECTED MOMENT

Problem setup: volume of a rectangular solid

SUGGESTED PREPARATION

Basic geometry of rectangular solids
Cartesian coordinates in three dimensions
Concept of area of a rectangle
Introductory idea of double integrals
basic notion of volume of a rectangular box
elementary idea of an integral as a sum
Basic single-variable definite integrals
Concept of volume as accumulation of small pieces
Notation for iterated integrals
Double integrals as volume under a surface
Antiderivatives of constants
Iterated integration with explicit limits
Rectangular regions in the xy-plane
Rectangular coordinates in 3D
Basic definite integration
Concept of iterated integrals
Physical notion of density
Basic understanding of density functions
Familiarity with rectangular coordinates

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The explanation, unpacked.

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Reviewed learning material · Video analysis · English

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This 120-second clip introduces a rectangular solid with bounds $0 \le x \le 3$ , $0 \le y \le 4$ , and $0 \le z \le 2$ . The instructor writes the bounds, sketches the corresponding box in xyz-space, and notes that although basic geometry already gives the volume, the example is being used to prepare viewers for triple integrals and their relation to double integrals. The volume is computed geometrically as 24 cubic units. No actual triple-integral notation is written within this clip. This 120-second whiteboard clip introduces the meaning of a triple integral through a rectangular box bounded by $0 \le x \le 3$ , $0 \le y \le 4$ , and $0 \le z \le 2$ . A small cube inside the box is used as the basic volume element, with side lengths $dx$ , $dy$ , and $dz$ . The speaker defines the volume differential $dV = dx\,dy\,dz$ , notes that the factors can be reordered because multiplication is commutative, and then begins the integration process conceptually by summing first in the $z$ -direction to form a column. The clip stops before any explicit iterated integral notation is written. This 120-second whiteboard segment introduces a triple integral as a way to compute the volume of a rectangular box. The board shows the bounds $0 \le x \le 3$ , $0 \le y \le 4$ , $0 \le z \le 2$ and the volume element dV = dx dy dz. The instructor builds the iterated integral from the inside out: first integrate in z from 0 to 2 to get the height of a column, then integrate in y from 0 to 4 to form a sheet parallel to the zy-plane, and finally integrate in x from 0 to 3 to obtain the whole solid. The final setup is $\int _{x=0}^{3}\int _{y=0}^{4}\int _{z=0}^{2} dz dy dx$ . The clip ends before the numerical evaluation is carried out. This 120-second whiteboard clip computes the volume of a rectangular box with bounds $0 \le x \le 3$ , $0 \le y \le 4$ , $0 \le z \le 2$ using the iterated triple integral $\int_{0}^{3}\int_{0}^{4}\int_{0}^{2} 1$ \, dz \, dy \, dx. The speaker first explains that the missing integrand is 1, then integrates from the inside out: z gives 2, y gives 8, and x gives 24. Midway through, the clip explicitly connects the result after the z-step to a double integral with constant height $z(x,y)=2$ , emphasizing that for this example the triple integral reduces to the same kind of volume-under-a-surface computation seen earlier. This 120-second whiteboard clip shifts from a completed triple-integral volume calculation for a rectangular box to the setup for mass with variable density. The board shows the box $0\le x\le 3$ , $0\le y\le 4$ , $0\le z\le 2$ and the evaluated integral $\int_0^3\int_0^4\int_0^2 1\, dz\, dy\, dx=24$ . The speaker then introduces a non-uniform density $\rho(x,y,z)=xyz$ , explains that a small mass equals local density times a small volume element, writes $dmass=xyz\,dV$ , and notes that in Cartesian coordinates $dV$ is width times height times depth, though its form changes in other coordinate systems. This video segment introduces the concept of calculating mass using a density function and differential volume elements in rectangular coordinates. The formula for differential mass is presented, along with a visual representation of the region and volume element.

Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.

Chapters

0:00Problem setup: volume of a rectangular solid 0:06Writing the coordinate bounds 0:24Why use this example for triple integrals 0:44Drawing the xyz axes 1:00Sketching the solid from the bounds 1:45Computing the volume geometrically 2:00What does a triple integral mean?2:25Choosing a small cube inside the box 2:58Defining the volume differential

dV

3:23Integrals as infinite sums 3:40Summing first in the

z

-direction 4:00Box bounds and volume element 4:02Writing the inner z-integral 4:25Choosing the order dz, dy, dx 4:35Volume of one vertical column 5:04Summing columns into a zy-plane sheet 5:41Adding sheets in x for total volume 5:58Evaluation announced but not shown 6:00Box bounds and volume element 6:04Implicit integrand is 1 6:12Integrate with respect to z 6:46Connection to a double integral 6:58Constant surface

z(x,y)=2

7:26Finish y- and x-integration 7:53Final volume 24 cubic units 8:00From constant-integrand volume to the goal of mass 8:22Uniform density versus variable density 8:39Writing the density function

\rho(x,y,z)=xyz

8:58Defining the differential mass element

dmass=xyz\,dV

9:43Interpreting

dV

and warning about other coordinates 10:00Introduction to Mass Calculation 10:15Visual Representation of Volume Element

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

The clip opens with a spoken setup: the goal is to find the volume of a solid whose coordinate ranges are about to be specified.

Three inequalities are written in green: $0 \le x \le 3$ , $0 \le y \le 4$ , and $0 \le z \le 2$ . These define an axis-aligned rectangular box in three-dimensional space.

The instructor points out that elementary geometry would already solve the problem by multiplying the three side lengths, but the real purpose is pedagogical: to acclimate the viewer to the form of a triple integral and its connection to a double integral.

A three-axis coordinate frame is drawn and labeled $x$ , $y$ , and $z$ , establishing the spatial setting for the region.

Using the stated bounds, the sketch adds tick marks and dashed edges to form the rectangular solid. The lower face lies in the plane $z=0$ , and the opposite face lies at $z=2$ ; the horizontal extents are $3$ in the $x$ -direction and $4$ in the $y$ -direction.

The volume is then computed directly from the geometry: the base area is $3 \times 4 = 12$ , and multiplying by the height $2$ gives $12 \times 2 = 24$ cubic units.

The clip ends just as the instructor begins to move beyond the purely geometric computation toward the integral viewpoint.

The clip opens by asking what a triple integral means. On screen, a dashed rectangular box sits in a 3D coordinate system, with the bounds $0 \le x \le 3$ , $0 \le y \le 4$ , and $0 \le z \le 2$ written at top left.

To answer the question, the speaker isolates one tiny cube inside the larger box. This small cube is the local volume piece whose size will be measured in the three coordinate directions.

The side lengths are assigned one by one: width $dy$ , height $dz$ , and depth $dx$ . The speaker briefly misspeaks by calling the height $dx$ , then corrects it because the drawn $z$ -axis is the vertical direction.

With those three edge lengths identified, the infinitesimal volume is named $dV$ and written as $dV = dx\,dy\,dz$ . The speaker also remarks that the factors may be reordered because ordinary multiplication is associative and order does not matter.

The explanation then broadens from one cube to the integral idea: an integral is described as an infinite sum of infinitely small pieces such as $dx$ , $dy$ , and $dz$ .

Finally, the first summation direction is chosen. The speaker says to add the small cubes along the $z$ -axis, building a column-like intermediate volume before any further integration steps are taken.

The clip opens on a digital blackboard showing a rectangular box in 3D space. At the top left, the green inequalities $0\le x\le 3$ , $0\le y\le 4$ , and $0\le z\le 2$ define the region. At the top right, the cyan formula $dV = dx\,dy\,dz$ names the infinitesimal volume element. A small cube drawn inside the larger dashed box visually represents one such element.

The instructor begins constructing the integral from the inside outward. He asks for the lowest and highest $z$ -values in the box and identifies them as $z=0$ and $z=2$ . This motivates the innermost integral $\int_{z=0}^{2} dz$ , which corresponds to stacking tiny cubes vertically through the height of the box.

He then fixes the order of integration by writing the differentials as $dz$ , then $dy$ , then $dx$ . The spoken explanation makes clear that this ordering is intentional: $z$ is integrated first, $y$ second, and $x$ last. Thus the emerging expression is an iterated integral whose nesting mirrors successive geometric summations.

With only the inner $z$ -integral in place, the instructor interprets its meaning. For any fixed $x$ and $y$ , $\int_{z=0}^{2} dz$ gives the volume of one vertical column. Because the $z$ -bounds are constant, this column volume simplifies to $2\,dy\,dx$ : height $2$ times base area $dy\,dx$ . This is a key conceptual step from infinitesimal cube to finite-height column.

Next, he adds the middle integral in the $y$ -direction. Since the box satisfies $0\le y\le 4$ , the new layer is $\int_{y=0}^{4}$ . Geometrically, this sums many adjacent columns side by side in the $y$ -direction, producing a sheet parallel to the $zy$ -plane. The written expression becomes $\int_{y=0}^{4}\int_{z=0}^{2} dz\,dy\,dx$ .

Finally, the outer integral is added in the $x$ -direction using the bound $0\le x\le 3$ . This stacks the previously formed sheets along $x$ until the entire rectangular solid is recovered. The completed setup on the board is $\int_{x=0}^{3}\int_{y=0}^{4}\int_{z=0}^{2} dz\,dy\,dx$ , which the speaker identifies as the volume of the whole figure.

The segment ends just as the instructor says that evaluating the integral is straightforward. No numerical evaluation is shown within this clip, so the mathematical content here is the construction and interpretation of the triple integral rather than its final computed value.

The clip opens on a black digital whiteboard with a dashed rectangular box drawn in a 3D coordinate system. At the upper left, the bounds are written in green: $0 \le x \le 3$ , $0 \le y \le 4$ , and $0 \le z \le 2$ . At the upper center, the volume element is written as $dV = dx\,dy\,dz$ . To the right, the iterated integral is shown as $\int_{z=0}^{z=2} dz\,dy\,dx$ .

The speaker points out that no function is written inside the integral, then explains that this means the integrand is implicitly $1$ . A red $1$ is added before $dz$ , making the integral $\int_{0}^{3}\int_{0}^{4}\int_{0}^{2} 1\,dz\,dy\,dx$ .

The first integration is with respect to $z$ . Since the antiderivative of $1$ with respect to $z$ is $z$ , the board shows $[z]_0^2$ . Evaluating at the limits gives $2-0=2$ , so the triple integral reduces to $\int_{0}^{3}\int_{0}^{4} 2\,dy\,dx$ .

The speaker then emphasizes the conceptual link to earlier material: after the $z$ -step, what remains is a double integral. On the right side, the function $z(x,y)=2$ is written, identifying the top of the box as a constant-height surface above the rectangle in the $xy$ -plane.

This connection is explained verbally: $z$ is a function of $x$ and $y$ , but here it is constant and independent of both. Thus the volume under the surface $z=2$ over the base region is exactly the same computation that the reduced double integral represents.

The evaluation continues with the $y$ -integral. The antiderivative of $2$ with respect to $y$ is $2y$ , giving $[2y]_0^4 = 8-0 = 8$ . The remaining single integral is therefore $\int_{0}^{3} 8\,dx$ .

Finally, the antiderivative of $8$ with respect to $x$ is $8x$ , so $[8x]_0^3 = 24-0 = 24$ . The speaker states the result as $24$ units cubed, which is the volume of the rectangular box.

The clip ends by raising the next question—why use triple integrals at all—without answering it within this segment.

The clip opens on a worked volume example for a rectangular solid. On the left is a 3D box with bounds $0\le x\le 3$ , $0\le y\le 4$ , and $0\le z\le 2$ ; on the right is the iterated integral $\int_0^3\int_0^4\int_0^2 1\, dz\, dy\, dx$ and its evaluation to $24$ . The speaker points out that when the integrand is constant inside the volume, a lower-dimensional integral could have sufficed for volume, but the lesson is now changing goals: instead of volume, we want mass.

The next distinction is between uniform and non-uniform density. If density were constant everywhere, total mass would simply be that constant density multiplied by the total volume. But the speaker says to imagine a substance whose density changes from point to point, such as a gas or a mixed material. That motivates treating density as a function of position rather than as a single number.

The board then introduces the standard physics symbol $\rho$ for density and writes a concrete example: $\rho(x,y,z)=xyz$ . This means the density at a point depends on all three coordinates. The earlier box remains the region of interest, so the same rectangular solid is now being reused, not to integrate the constant function $1$ , but to prepare for integrating a variable density over the solid.

From that density function, the speaker derives the local mass element. He argues by units and by interpretation: density has units like mass per volume, so multiplying density at a point by a tiny volume gives a tiny amount of mass. He writes this as $dmass=xyz\,dV$ , or more conceptually $dmass=\rho(x,y,z)\,dV$ . This is the basic building block for a triple-integral mass calculation.

Finally, the clip explains what $dV$ means in the current Cartesian setting. The speaker identifies it with width times height times depth, corresponding to the small side lengths of a tiny rectangular piece inside the box. He also warns that $dV$ is not always written as $dx\,dy\,dz$ ; in other coordinate systems its expression changes. The segment therefore ends after setting up the mass integrand, without yet carrying out the full triple integral for mass.

To find the mass in rectangular coordinates, multiply the density function $\rho(x,y,z)$ by the differential volume element $dV = dx dy dz$ .

The diagram shows a 3D box with dimensions $x=3$ , $y=4$ , $z=2$ , and a small cube representing the differential volume element $dV$ .

Knowledge cards

Coordinate bounds of the solid

The region is defined by three independent intervals: $0 \le x \le 3$ , $0 \le y \le 4$ , and $0 \le z \le 2$ . Together these describe a rectangular box aligned with the coordinate axes.

0 \le x \le 3,\quad 0 \le y \le 4,\quad 0 \le z \le 2

Why this example is used before triple integrals

Although the volume can be found immediately by geometry, the instructor chooses this simple solid to help viewers become familiar with what a triple integral looks like and how it relates to a double integral before tackling more complicated regions.

Geometric volume of the rectangular box

For a rectangular solid, volume equals the product of its three side lengths. Here the side lengths are 3, 4, and 2, so the volume is $3 \times 4 \times 2 = 24$ cubic units.

V = (3)(4)(2) = 24

Wording note: 'cube' versus rectangular box

The spoken introduction refers to a cube, but the displayed bounds give unequal side lengths 3, 4, and 2. Mathematically, the object shown is a rectangular box, not a cube.

Meaning of a triple integral in this clip

The video introduces a triple integral by asking what it means geometrically. The answer given here is that it is built from summing tiny volume pieces inside a three-dimensional region, beginning with one small rectangular cube inside a larger box.

Rectangular box region

The example region is a Cartesian rectangular box with bounds $0 \le x \le 3$ , $0 \le y \le 4$ , and $0 \le z \le 2$ . This simple shape is used before moving to more complicated variable boundaries.

0 \le x \le 3,\; 0 \le y \le 4,\; 0 \le z \le 2

Small cube as the basic volume piece

A tiny cube is drawn somewhere inside the larger box. Its role is to represent one infinitesimal chunk of volume that will later be summed repeatedly to recover the whole region.

Side lengths of the small cube

The cube's width is $dy$ , its height is $dz$ , and its depth is $dx$ . The correction matters because the vertical axis in the drawing is $z$ , so height must correspond to $dz$ , not $dx$ .

\text{width}=dy,\;\text{height}=dz,\;\text{depth}=dx

Volume differential $dV$

The infinitesimal volume of the small cube is called the volume differential and is written as the product of its three side lengths.

dV = dx\,dy\,dz

Reordering the factors in $dV$

Since $dV$ is presented as a product, the speaker notes that $dx$ , $dy$ , and $dz$ can be switched in order without changing the product. This is a statement about multiplication of the differential factors.

dx\,dy\,dz = dy\,dx\,dz = dz\,dy\,dx

Integral as an infinite sum

The clip explains integrals conceptually as tools for taking infinite sums of infinitely small quantities such as $dx$ , $dy$ , and $dz$ . This is the bridge from one tiny cube to the full triple-integral idea.

First summation direction: along $z$

The speaker chooses to begin by adding the small cubes in the $z$ -direction, producing the volume of a column. The clip stops before writing the corresponding iterated integral notation.

Differential volume element dV

The video defines the infinitesimal volume of a small rectangular cube inside a 3D box by $dV = dx\,dy\,dz$ . The accompanying drawing of a tiny cube inside the larger dashed box shows that triple integration is built by summing many such small volume pieces.

dV = dx\,dy\,dz

Rectangular box region

The solid being integrated over is the axis-aligned rectangular box given by $0\le x\le 3$ , $0\le y\le 4$ , and $0\le z\le 2$ . These constant inequalities supply the limits used later in the iterated integral.

0\le x\le 3,\quad 0\le y\le 4,\quad 0\le z\le 2

Order of integration dz, dy, dx

The instructor deliberately writes $dz$ first, then $dy$ , then $dx$ , explaining that this means the integral is taken with respect to $z$ first, then $y$ , then $x$ . The order is tied to a geometric buildup from vertical columns to sheets to the full solid.

dz\,dy\,dx

Inner z-integral as column height

For fixed $x$ and $y$ , the inner integral $\int_{z=0}^{2} dz$ computes the volume of one vertical column. Since the height is constantly 2, the column volume is $2\,dy\,dx$ . This step turns the abstract integral into a concrete geometric piece of the box.

\int_{z=0}^{2} dz = 2,\quad \text{column volume } = 2\,dy\,dx

Middle y-integral as a sheet

After obtaining column volumes, the next integral sums them in the $y$ -direction from 0 to 4. The speaker describes the result as the volume of a sheet parallel to the $zy$ -plane, showing how one-dimensional summation builds a two-dimensional slab inside the box.

\int_{y=0}^{4}

Outer x-integral as total volume

The last integral adds those sheets in the $x$ -direction from 0 to 3. This completes the transition from a single infinitesimal cube to a column, then a sheet, then the entire rectangular solid, yielding the full volume integral.

\int_{x=0}^{3}

Completed triple integral setup

Putting the three stages together, the video arrives at the iterated triple integral $\int_{x=0}^{3}\int_{y=0}^{4}\int_{z=0}^{2} dz\,dy\,dx$ . This expression is presented as the volume of the whole box, although the clip ends before evaluating it numerically.

\int_{x=0}^{3}\int_{y=0}^{4}\int_{z=0}^{2} dz\,dy\,dx

Volume of a rectangular box as a triple integral

The example computes the volume of the box with bounds $0 \le x \le 3$ , $0 \le y \le 4$ , and $0 \le z \le 2$ . Because volume is being measured, the integrand is the constant $1$ , and the volume element is written as $dV = dx\,dy\,dz$ . The displayed iterated integral is $\int_{0}^{3}\int_{0}^{4}\int_{0}^{2} 1\,dz\,dy\,dx$ .

\int_{0}^{3}\int_{0}^{4}\int_{0}^{2} 1 \, dz \, dy \, dx

Why the missing integrand is 1

The speaker explicitly notes that nothing is written under the integral sign, but that this is equivalent to inserting a factor of $1$ . In other words, $dz\,dy\,dx$ is treated as $1\,dz\,dy\,dx$ . This is the standard setup when an integral is being used to compute volume.

dz\,dy\,dx = 1\,dz\,dy\,dx

Inside-out evaluation of the iterated integral

The method shown is to integrate from the innermost differential outward. First, $\int_0^2 1\,dz = [z]_0^2 = 2$ . Then the problem becomes $\int_0^3\int_0^4 2\,dy\,dx$ . Next, $\int_0^4 2\,dy = [2y]_0^4 = 8$ , leaving $\int_0^3 8\,dx = [8x]_0^3 = 24$ .

[z]_0^2=2,\quad [2y]_0^4=8,\quad [8x]_0^3=24

Triple integral reduces to a double integral here

After the $z$ -integration, the remaining expression is $\int_{0}^{3}\int_{0}^{4} 2\,dy\,dx$ . The speaker identifies this as the same type of double integral used previously to compute volume under a surface. For this box, the surface height is constant.

\int_{0}^{3}\int_{0}^{4} 2 \, dy \, dx

Constant surface $z(x,y)=2$

The board writes $z(x,y)=2$ to describe the top face of the box. The speaker explains that this is a constant function, independent of $x$ and $y$ . Therefore the volume under the surface over the rectangular base is computed by integrating the constant height $2$ across the $xy$ -region.

z(x,y)=2

Detailed learning notes

Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.

Symbols · 34

x

Clear evidence

Shown in the video

Evidence

Formula
Observation
Green handwritten inequality ` $0 \le x \le 3$ ` is written in the upper-left area.
Audio
Observation
The speaker says that x is greater than or equal to 0 and less than or equal to 3.

Symbol

Meaning

Coordinate variable for one side length of the rectangular solid; its interval is from 0 to 3.

Domain

$0 \le x \le 3$

y

Clear evidence

Shown in the video

Evidence

Formula
Observation
Green handwritten inequality ` $0 \le y \le 4$ ` is written below the x-bound.
Audio
Observation
The speaker says that y is greater than or equal to 0 and less than or equal to 4.

Symbol

Meaning

Coordinate variable for a second side length of the rectangular solid; its interval is from 0 to 4.

Domain

$0 \le y \le 4$

z

Clear evidence

Shown in the video

Evidence

Formula
Observation
Green handwritten inequality ` $0 \le z \le 2$ ` is written below the y-bound.
Audio
Observation
The speaker says that z is greater than or equal to 0 and less than or equal to 2.

Symbol

Meaning

Coordinate variable for the height of the rectangular solid; its interval is from 0 to 2.

Domain

$0 \le z \le 2$

x,y,z

Clear evidence

Shown in the video

Evidence

Diagram
Observation
Three green axes are drawn and labeled x, y, and z in a three-dimensional coordinate sketch.
Audio
Observation
The speaker identifies the drawn axes as the x-axis, z-axis, and y.

Symbol

x,y,z

Meaning

Labels of the three mutually perpendicular coordinate axes used to draw the rectangular solid.

Domain

Three-dimensional Cartesian coordinate system

x

Clear evidence

Shown in the video

Evidence

Formula
Observation
Top-left handwritten inequalities include $0 \le x \le 3$ .
Audio
Observation
The speaker later labels one edge of the small cube as $dx$ .

Symbol

Meaning

Cartesian coordinate along the depth direction of the large rectangular box in this clip.

Domain

$0 \le x \le 3$

y

Clear evidence

Shown in the video

Evidence

Formula
Observation
Top-left handwritten inequalities include $0 \le y \le 4$ .
Audio
Observation
The speaker labels one edge of the small cube as $dy$ .

Symbol

Meaning

Cartesian coordinate along the width direction of the large rectangular box in this clip.

Domain

$0 \le y \le 4$

z

Clear evidence

Shown in the video

Evidence

Formula
Observation
Top-left handwritten inequalities include $0 \le z \le 2$ .
Audio
Observation
The speaker says the height is $dz$ and that $z$ is up and down.

Symbol

Meaning

Cartesian vertical coordinate, used for the height direction of the large rectangular box.

Domain

$0 \le z \le 2$

dx

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says, "And then its depth is dx."
Formula
Observation
The final written formula is $dV = dx\,dy\,dz$ , containing $dx$ .

Symbol

Meaning

Infinitesimal length increment in the $x$ -direction, corresponding to the depth of the small cube.

dy

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says, "Let's say that its width is dy."
Formula
Observation
The final written formula is $dV = dx\,dy\,dz$ , containing $dy$ .

Symbol

Meaning

Infinitesimal length increment in the $y$ -direction, corresponding to the width of the small cube.

dz

Clear evidence

Shown in the video

Evidence

Audio
Observation
After correcting himself, the speaker says the height is $dz$ .
Formula
Observation
The final written formula is $dV = dx\,dy\,dz$ , containing $dz$ .

Symbol

Meaning

Infinitesimal length increment in the $z$ -direction, corresponding to the height of the small cube.

dV

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says, "you could call that dV, which is kind of the volume differential."
Formula
Observation
The board shows $dV = dx\,dy\,dz$ .

Symbol

Meaning

Volume differential, the infinitesimal volume of a small rectangular box inside the larger region.

x

Clear evidence

Shown in the video

Evidence

Formula
Observation
Top-left green text shows the inequality $0 \le x \le 3$ .
Audio
Observation
The speaker later says to sum in the x direction from $x = 0$ to $x = 3$ .

Symbol

Meaning

Cartesian coordinate along the x-axis; outer integration variable for the rectangular box.

Domain

$0 \le x \le 3$

Knowledge points · 16

Purpose of the example: introducing triple integrals

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says he wants to use the example so viewers get used to what a triple integral looks like and how it relates to a double integral.

Method

Explanation

The video uses a simple rectangular solid not because geometry is insufficient, but to build familiarity with the form of a triple integral and its relation to a double integral before moving to more complicated examples.

Formula

Conditions

The region is a rectangular box with constant coordinate bounds.

Basic geometric volume formula for a rectangular solid

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says that using basic geometry one could multiply width times height times depth to get the volume.

Formula

Explanation

For a rectangular solid, the volume can be found by multiplying the three side lengths corresponding to width, height, and depth.

Formula

V = (\text{width})(\text{height})(\text{depth})

Conditions

The solid is a rectangular box with mutually perpendicular side lengths.

Meaning of a triple integral

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker asks, "So what does a triple integral mean?"

Definition

Explanation

In this clip, a triple integral is introduced conceptually as an object whose meaning is built from summing infinitesimal volumes inside a three-dimensional region.

Conditions

The region is three-dimensional.
The explanation uses a small rectangular box as the basic volume element.

Volume differential $dV$

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker names the small volume $dV$ and calls it the volume differential.
Formula
Observation
The board writes $dV = dx\,dy\,dz$ .

Definition

Explanation

The infinitesimal volume element inside the large box is denoted $dV$ . In Cartesian coordinates it is the product of the three side lengths of a tiny rectangular box.

Formula

dV = dx\,dy\,dz

Conditions

Coordinates are Cartesian.
The small piece is treated as a rectangular box with side increments $dx$ , $dy$ , and $dz$ .

Prerequisites

Dimensions of the small cube

Dimensions of the small cube

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker assigns width $dy$ , corrects height to $dz$ , and assigns depth $dx$ .
Diagram
Observation
A small cube is drawn inside the larger dashed rectangular box.

Definition

Explanation

The small cube inside the larger box has width $dy$ , height $dz$ , and depth $dx$ , matching the three coordinate directions shown on the axes.

Conditions

The cube is aligned with the coordinate axes.
The larger region is a rectangular box.

Commutativity of the factors in $dV$

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says, "you could switch the orders of these ... because multiplication is associative and order doesn't matter".

Method

Explanation

Because $dV$ is written as a product of scalar differentials, the factors $dx$ , $dy$ , and $dz$ may be reordered without changing the value of the product.

Formula

dx\,dy\,dz = dy\,dx\,dz = dz\,dy\,dx

Conditions

This statement concerns multiplication of the differential factors, not yet the full iterated integral setup.

Prerequisites

Volume differential $dV$

Differential volume element dV

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board displays dV = dx dy dz in cyan at the top right.
Audio
Observation
The speaker says to take the sum of these dVs and writes dz, dy, dx in the iterated integral.
Diagram
Observation
A small cyan cube is drawn inside the larger dashed rectangular box, representing one infinitesimal piece of volume.

Definition

Explanation

For a rectangular box aligned with the coordinate axes, an infinitesimal volume element is written as $dV = dx\,dy\,dz$ . In the video this element is represented by a small cube inside the larger box, and the triple integral sums such elements over the whole region.

Formula

dV = dx\,dy\,dz

Conditions

The region is described using Cartesian coordinates.
The small volume element is rectangular, with side lengths dx, dy, and dz.

Rectangular box region in three variables

Clear evidence

Shown in the video

Evidence

Formula
Observation
Green inequalities at top left read $0 \le x \le 3$ , $0 \le y \le 4$ , $0 \le z \le 2$ .
Diagram
Observation
A dashed rectangular box is drawn in a 3D coordinate system with axes labeled x, y, and z.
Audio
Observation
The speaker uses these bounds to choose limits for z, then y, then x.

Definition

Explanation

The solid region being integrated over is the rectangular box $0\le x\le 3$ , $0\le y\le 4$ , $0\le z\le 2$ . The diagram shows this box in 3D space, and the inequalities give the constant limits used for each coordinate direction.

Formula

0\le x\le 3,\quad 0\le y\le 4,\quad 0\le z\le 2

Conditions

The box is axis-aligned.
All three coordinate intervals are constant and independent of the other variables.

Iterated triple integral for volume

Clear evidence

Shown in the video

Evidence

Formula
Observation
The instructor writes an iterated integral with inner z-limits 0 to 2, then adds y-limits 0 to 4, then x-limits 0 to 3, with differentials dz dy dx.
Audio
Observation
The speaker explains that z is integrated first, y next, and x last, and that this computes the volume of the entire figure.

Uncertainties

The final integral is not fully evaluated within the clip; the speaker only says it is straightforward to evaluate.

Method

Explanation

To compute the volume of the rectangular box, the video sets up the iterated integral $\int_{x=0}^{3}\int_{y=0}^{4}\int_{z=0}^{2} dz\,dy\,dx$ . The method is to integrate first in the z-direction to get the height of a column, then sum columns in the y-direction to form a sheet, and finally sum sheets in the x-direction to obtain the total volume.

Formula

\int_{x=0}^{3}\int_{y=0}^{4}\int_{z=0}^{2} dz\,dy\,dx

Conditions

The integrand is 1, so the integral measures volume.
The limits are constant because the region is a rectangular box.
The chosen order is dz first, dy second, dx third.

Prerequisites

Differential volume element dV
Rectangular box region in three variables

Triple integral setup for the volume of a rectangular box

Clear evidence

Shown in the video

Evidence

Formula
Observation
Top-left bounds are $0 \le x \le 3$ , $0 \le y \le 4$ , $0 \le z \le 2$ .
Formula
Observation
Top-center formula is dV = dx dy dz.
Formula
Observation
Iterated integral is shown as $\int_{z=0}^{z=2} 1$ \, dz \, dy \, dx.
Diagram
Observation
A dashed rectangular box is drawn in a 3D coordinate system with axes labeled x, y, z.

Definition

Explanation

The clip sets up the volume of a rectangular box by integrating the constant integrand 1 over the region $0 \le x \le 3$ , $0 \le y \le 4$ , $0 \le z \le 2$ . The differential order shown is dz dy dx, so the integration is performed first with respect to z, then y, then x.

Formula

\int_{0}^{3}\int_{0}^{4}\int_{0}^{2} 1 \, dz \, dy \, dx

Conditions

The region is a rectangular box aligned with the coordinate axes.
The integrand is 1 because the goal is volume.

Evaluating an iterated triple integral from inside out

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker says, "First we're taking the integral with respect to z."
Formula
Observation
Inner evaluation is written as [z]_0^2.
Formula
Observation
Middle evaluation is written as [2y]_0^4.
Formula
Observation
Outer evaluation is written as [8x]_0^3.

Method

Explanation

The method demonstrated is to evaluate the innermost integral first, substitute its limits, carry the resulting expression outward, and repeat until only one variable remains. Each step uses the antiderivative of the current integrand with respect to the current variable.

Formula

\int_{0}^{3}\int_{0}^{4}\int_{0}^{2} 1 \, dz \, dy \, dx

Conditions

The integral is written as an iterated integral with explicit limits.
Each integration is performed with respect to the variable indicated by the differential.

Prerequisites

Triple integral setup for the volume of a rectangular box

Triple integral reduces to a double integral for this box

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker says that after integrating with respect to z, the result is a double integral and that this is the exact integral from previous videos on the double integral.
Formula
Observation
After the inner step, the board shows $\int_{0}^{3}\int_{0}^{4} 2$ \, dy \, dx.
Formula
Observation
The expression $z(x,y)=2$ is written to connect the result to a surface height.

Definition

Explanation

For this rectangular box, integrating 1 with respect to z from 0 to 2 produces the constant 2. The remaining double integral $\int_{0}^{3}\int_{0}^{4} 2$ \, dy \, dx matches the earlier double-integral viewpoint where the height above the xy-region is the constant function $z(x,y)=2$ .

Formula

\int_{0}^{3}\int_{0}^{4} 2 \, dy \, dx

Conditions

The upper surface is $z=2$ and the lower surface is $z=0$ .
The xy-region is $0 \le x \le 3$ and $0 \le y \le 4$ .

Prerequisites

Triple integral setup for the volume of a rectangular box

Claims and conditions · 8

Integral as an infinite sum of infinitesimal pieces

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says integrals help us take infinite sums of infinitely small distances like $dz$ , $dx$ , or $dy$ .

Proposition

Statement

Integrals are described as tools for taking infinite sums of infinitely small quantities such as $dx$ , $dy$ , and $dz$ .

Hypotheses

The quantity being summed is built from infinitesimal pieces.
The setting is the volume computation introduced in this clip.

Quantifiers

infinitely many infinitesimal pieces

Inner z-integral gives column height

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says the integral as written gives the volume of a column for any x and y, and because all quantities are constant it becomes 2 times dy dx.
Formula
Observation
The inner integral shown on the board has limits $z = 0$ to $z = 2$ and differential dz.

Proposition

Statement

For fixed x and y, the inner integral $\int_{z=0}^{2} dz$ equals the height 2, so the contribution of one vertical column is $2\,dy\,dx$ .

Hypotheses

The region has constant z-bounds $0 \le z \le 2$ .
The integrand is 1, so the integral computes volume.
x and y are held fixed while integrating with respect to z.

Quantifiers

For any fixed x and y in the box.

Summing columns in y gives a zy-parallel sheet

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says that after finding the height of a column, one can sum those columns in the y direction from $y = 0$ to $y = 4$ , giving the volume of a sheet parallel to the zy-plane.
Formula
Observation
A second integral sign is added with lower limit 0 and upper limit 4.

Proposition

Statement

Adding the vertical columns over $0\le y\le 4$ produces the volume of a sheet parallel to the $zy$ -plane.

Hypotheses

The previous step has already computed the height of a column in the z-direction.
The y-bound is constant from 0 to 4.

Quantifiers

For the fixed x-slice being summed in the y-direction.

Summing sheets in x gives total volume

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says that all that remains is to add up those sheets in the x direction from $x = 0$ to $x = 3$ to have the volume of the entire figure.
Formula
Observation
A third integral sign is added with lower limit 0 and upper limit 3.

Uncertainties

The numerical value of the total volume is not stated in this clip.

Proposition

Statement

Adding the sheets over $0\le x\le 3$ yields the volume of the entire rectangular box.

Hypotheses

The y-step has produced sheet volumes parallel to the zy-plane.
The x-bound is constant from 0 to 3.

Quantifiers

Over the full x-interval of the box.

An omitted integrand in a volume integral is 1

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker says, "we don't have anything written under here, but we can just assume that there's a 1" and explains that dz dy dx is the same as 1 times dz dy dx.
Formula
Observation
A red 1 is inserted before dz in the integral.

Proposition

Statement

When computing volume with dV = dz dy dx and no other integrand is written, the implicit integrand is 1.

Hypotheses

The integral is being used to compute volume.
The differential element is written without an additional function factor.

Quantifiers

For the displayed volume integral in this clip.

For this example the triple integral is presented as an extension of the double integral

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker says, "what we're doing with the triple integral, it's really, really nothing different" and notes that the remaining double integral is the same kind of integral discussed previously.

Uncertainties

The speaker says he will show why triple integrals are useful "in a second," but that explanation is not contained in this clip.

Proposition

Statement

In this rectangular-box example, the triple integral is presented as doing essentially the same thing as the previously discussed double integral after the z-integration is carried out.

Hypotheses

The region is the displayed box with constant top surface $z=2$ .
The integral is evaluated in the order dz dy dx.

Quantifiers

For the specific example shown in this clip.

Uniform density allows mass = density × volume

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says that if the mass was uniform, you could just multiply its uniform density times its volume and get its mass.

Proposition

Statement

If a body has uniform density, then its total mass is obtained by multiplying that constant density by its volume.

Hypotheses

The density is uniform throughout the body.
The volume of the body is known.

Quantifiers

For a body with constant density throughout its volume.

Constant integrand case can reduce to a lower-dimensional integral for volume

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says that when you have a constant value within the volume, you could have just done a double integral.

Uncertainties

The statement is presented verbally as a comparison to the already-written triple integral; the clip does not fully derive the equivalence.

Proposition

Statement

For computing the volume of the pictured solid, if the integrand is constant within the volume, one could have used a double integral instead of the displayed triple-integral setup.

Hypotheses

The goal is volume rather than mass with variable density.
The integrand over the volume is constant.

Quantifiers

In the volume-computation setting shown on the board.

Derivations and proofs · 6

Geometric computation of the rectangular solid's volume

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says the depth is 3, the width is 4, the base area is 12, and 12 times 2 is 24 cubic units.
Diagram
Observation
The drawn solid has coordinate extents matching x-length 3, y-length 4, and z-height 2.

Numerical verification

Steps

Expression
$3 \times 4 = 12$
Explanation
The base area is computed from the two horizontal side lengths.
Justification
The solid is rectangular, so its base is a rectangle with side lengths 3 and 4.
Shown in the video
Expression
$12 \times 2 = 24$
Explanation
The base area is multiplied by the vertical height to obtain the volume.
Justification
For a rectangular solid, volume equals base area times height.
Shown in the video

Conclusion

The volume of the rectangular solid is 24 cubic units.

Derivation of $dV = dx\,dy\,dz$

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says the small volume equals width times length times height.
Formula
Observation
The board writes $dV = dx\,dy\,dz$ .

Intuitive argument

Steps

Expression
$dV$
Explanation
Start with the infinitesimal volume of the small cube inside the larger box.
Justification
The speaker explicitly introduces $dV$ as the small volume differential.
Shown in the video
Expression
$dx\,dy\,dz$
Explanation
Multiply the three side lengths of the small cube: depth $dx$ , width $dy$ , and height $dz$ .
Justification
The speaker states that the volume is width times length times height and identifies those side lengths.
Shown in the video

Conclusion

The volume differential for the small rectangular box is $dV = dx\,dy\,dz$ .

First summation step: build a column in the $z$ -direction

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says they can first add the cube up in the $z$ direction to get the volume of a column.
Diagram
Observation
The cursor points around the small cube and then toward the vertical $z$ -axis.

Uncertainties

The clip ends before the actual integral notation for the column is written out.

Intuitive argument

Steps

Expression
$dV = dx\,dy\,dz$
Explanation
Begin from the infinitesimal volume element of one small cube.
Justification
This was established earlier on the board.
Shown in the video
Expression
$\text{sum } dV \text{ along } z$
Explanation
Add copies of the small cube upward and downward along the $z$ -axis while keeping the other two dimensions fixed.
Justification
The speaker says to take the cube and add it up in the $z$ direction.
Shown in the video
Expression
$\text{volume of a column}$
Explanation
The result of that first summation is described as the volume of a column.
Justification
The speaker explicitly states that this produces the volume of a column.
Shown in the video

Conclusion

The first conceptual integration direction chosen in the clip is the $z$ -direction, producing a column-like intermediate volume.

Deriving the iterated integral from geometric summation

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker builds the integral step by step: first z from 0 to 2, then y from 0 to 4, then x from 0 to 3.
Formula
Observation
The board progressively shows nested integral signs with the corresponding limits and the differentials dz dy dx.

Intuitive argument

Steps

Expression
$\int_{z=0}^{2} dz$
Explanation
Start with the innermost integration in the z-direction because the smallest cube is stacked vertically from the bottom of the box to the top.
Justification
The displayed bounds include $0 \le z \le 2$ , and the speaker explicitly says to integrate with respect to z first.
Shown in the video
Expression
$2\,dy\,dx$
Explanation
After integrating in z, one vertical column has height 2 and base area dy dx, so its volume is 2 dy dx.
Justification
The speaker states that the column height is 2 and the width and depth are dy and dx.
Shown in the video
Expression
$\int_{y=0}^{4} \int_{z=0}^{2} dz\,dy\,dx$
Explanation
Next, sum those columns in the y-direction to form a sheet parallel to the zy-plane.
Justification
The speaker says to take another integral in the y direction and uses the bounds $y = 0$ to $y = 4$ .
Shown in the video
Expression
$\int_{x=0}^{3} \int_{y=0}^{4} \int_{z=0}^{2} dz\,dy\,dx$
Explanation
Finally, sum the sheets in the x-direction to obtain the volume of the whole box.
Justification
The speaker says to add up the sheets in the x direction from $x = 0$ to $x = 3$ .
Shown in the video

Conclusion

The volume of the rectangular box is represented by the iterated triple integral $\int_{x=0}^{3}\int_{y=0}^{4}\int_{z=0}^{2} dz\,dy\,dx$ .

Evaluation of the triple integral for the box volume

Clear evidence

Shown in the video

Evidence

Formula
Observation
Board shows $\int_{0}^{3}\int_{0}^{4}\int_{0}^{2} 1$ \, dz \, dy \, dx.
Formula
Observation
Board shows [z]_0^2, then $\int_{0}^{3}\int_{0}^{4} 2$ \, dy \, dx, then [2y]_0^4, then $\int_{0}^{3} 8$ \, dx, then [8x]_0^3, then 24.
Audio
Observation
Speaker narrates each antiderivative and limit substitution, ending with "24 units cubed."

Proof

Steps

Expression
$\int_{0}^{3}\int_{0}^{4}\int_{0}^{2} 1 \, dz \, dy \, dx$
Explanation
Start from the displayed iterated integral for the volume of the box.
Justification
The bounds $0 \le x \le 3$ , $0 \le y \le 4$ , $0 \le z \le 2$ and dV = dx dy dz are written on the board.
Shown in the video
Expression
$\int_{0}^{3}\int_{0}^{4} [z]_{0}^{2} \, dy \, dx$
Explanation
Integrate the innermost 1 with respect to z.
Justification
The antiderivative of 1 with respect to z is z, as stated by the speaker.
Shown in the video
Expression
$\int_{0}^{3}\int_{0}^{4} 2 \, dy \, dx$
Explanation
Evaluate [z]_{0}^{2}=2-0.
Justification
Substitute the upper and lower z-limits shown on the integral.
Shown in the video
Expression
$\int_{0}^{3} [2y]_{0}^{4} \, dx$
Explanation
Integrate the middle integrand 2 with respect to y.
Justification
The antiderivative of 2 with respect to y is 2y.
Shown in the video
Expression
$\int_{0}^{3} 8 \, dx$
Explanation
Evaluate [2y]_{0}^{4}=8-0.
Justification
Substitute the y-limits 0 and 4 shown on the board.
Shown in the video
Expression
$[8x]_{0}^{3}$
Explanation
Integrate the outer integrand 8 with respect to x.
Justification
The antiderivative of 8 with respect to x is 8x.
Shown in the video
Expression
$24$
Explanation
Evaluate [8x]_{0}^{3}=24-0.
Justification
Substitute the x-limits 0 and 3 shown on the board.
Shown in the video

Conclusion

The displayed triple integral evaluates to 24 cubic units.

Evaluation of the displayed triple integral for the box volume

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $\int_0^3\int_0^4\int_0^2 1\, dz\, dy\, dx$ and the successive evaluated lines $[z]_0^2$ , $\int_0^3\int_0^4 2\, dy\, dx$ , $[2y]_0^4$ , $\int_0^3 8\, dx$ , $[8x]_0^3$ , and $24$ .
Diagram
Observation
The left diagram labels the rectangular box with $0\le x\le 3$ , $0\le y\le 4$ , and $0\le z\le 2$ .

Numerical verification

Steps

Expression
$\int_0^3\int_0^4\int_0^2 1\, dz\, dy\, dx$
Explanation
Start from the iterated integral written on the board for the rectangular solid.
Justification
Observed directly from the formula on the board.
Shown in the video
Expression
$\int_0^3\int_0^4 [z]_0^2\, dy\, dx$
Explanation
Integrate the innermost integrand $1$ with respect to $z$ .
Justification
Antiderivative of $1$ with respect to $z$ is $z$ .
Shown in the video
Expression
$\int_0^3\int_0^4 2\, dy\, dx$
Explanation
Evaluate $[z]_0^2=2-0=2$ .
Justification
Substitution of the $z$ -limits shown on the board.
Shown in the video
Expression
$\int_0^3 [2y]_0^4\, dx$
Explanation
Integrate the middle integrand $2$ with respect to $y$ .
Justification
Antiderivative of $2$ with respect to $y$ is $2y$ .
Shown in the video
Expression
$\int_0^3 8\, dx$
Explanation
Evaluate $[2y]_0^4=2\cdot 4-2\cdot 0=8$ .
Justification
Substitution of the $y$ -limits shown on the board.
Shown in the video
Expression
$[8x]_0^3$
Explanation
Integrate the outer integrand $8$ with respect to $x$ .
Justification
Antiderivative of $8$ with respect to $x$ is $8x$ .
Shown in the video
Expression
$24$
Explanation
Evaluate $[8x]_0^3=8\cdot 3-8\cdot 0=24$ .
Justification
Substitution of the $x$ -limits shown on the board.
Shown in the video

Conclusion

The displayed triple integral evaluates to $24$ , matching the volume of the rectangular box with side lengths $3$ , $4$ , and $2$ .

Worked examples · 5

Find the volume of the solid with $0 \le x \le 3$ , $0 \le y \le 4$ , $0 \le z \le 2$

Clear evidence

Shown in the video

Evidence

Formula
Observation
Bounds ` $0 \le x \le 3$ `, ` $0 \le y \le 4$ `, and ` $0 \le z \le 2$ ` are written on screen.
Diagram
Observation
A three-dimensional rectangular solid is sketched in the first octant using those bounds.
Audio
Observation
The speaker states the goal is to find the volume of the cube/rectangular solid and later computes it as 24 cubic units.

Uncertainties

The speaker calls the object a cube, but the displayed side lengths are 3, 4, and 2, so the mathematical object shown is a rectangular box rather than a true cube.

Problem

Given a solid bounded by $0 \le x \le 3$ , $0 \le y \le 4$ , and $0 \le z \le 2$ , determine its volume.

Given

$0 \le x \le 3$
$0 \le y \le 4$
$0 \le z \le 2$

Goal

Compute the volume of the solid.

Steps

Expression
$0 \le x \le 3,\quad 0 \le y \le 4,\quad 0 \le z \le 2$
Explanation
Record the coordinate bounds that define the solid.
Justification
These inequalities are explicitly written and spoken in the video.
Shown in the video
Expression
$V = (3)(4)(2)$
Explanation
Use the rectangular-solid volume formula with side lengths 3, 4, and 2.
Justification
The solid is axis-aligned with constant bounds in each coordinate direction.
Shown in the video
Expression
$V = 24$
Explanation
Multiply the side lengths to obtain the numerical volume.
Justification
Arithmetic simplification.
Shown in the video

Answer

24 cubic units

Verification

The result matches the speaker's verbal computation: base area 12 times height 2 gives 24.

Rectangular box region used to motivate triple integrals

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board lists $0 \le x \le 3$ , $0 \le y \le 4$ , and $0 \le z \le 2$ .
Diagram
Observation
A dashed rectangular box is drawn in a 3D coordinate system.

Problem

Use a rectangular box in $xyz$ -space to explain what a triple integral means by summing tiny volume elements.

Given

$0 \le x \le 3$
$0 \le y \le 4$
$0 \le z \le 2$
A small cube is placed somewhere inside the larger box.

Goal

Identify the infinitesimal volume element and describe the first summation direction.

Steps

Expression
$0 \le x \le 3,\; 0 \le y \le 4,\; 0 \le z \le 2$
Explanation
The region is a rectangular box aligned with the coordinate axes.
Justification
These bounds are written directly on the board.
Shown in the video
Expression
$dV = dx\,dy\,dz$
Explanation
Inside the box, choose a tiny rectangular cube and compute its volume as the product of its three side lengths.
Justification
The speaker labels the sides $dx$ , $dy$ , $dz$ and writes the formula on the board.
Shown in the video
Expression
$\text{sum } dV \text{ in the } z\text{-direction}$
Explanation
The next conceptual step is to stack these small cubes vertically to form a column.
Justification
The speaker says they can first add it up in the $z$ direction so that we get the volume of a column.
Shown in the video

Answer

The clip sets up the volume element $dV = dx\,dy\,dz$ and begins the triple-integral interpretation by summing first along $z$ .

Verification

The written bounds, the drawn box, and the spoken explanation all match the same Cartesian volume setup.

Setting up a triple integral for a rectangular box

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board lists $0 \le x \le 3$ , $0 \le y \le 4$ , $0 \le z \le 2$ and constructs the nested integral with those limits.
Diagram
Observation
A dashed rectangular box in 3D coordinates is shown with a small interior cube.
Audio
Observation
The speaker explains that this integral will give the volume of the entire figure.

Uncertainties

The clip stops before the final numerical evaluation is spoken or written.

Problem

Find the volume of the rectangular box defined by $0\le x\le 3$ , $0\le y\le 4$ , and $0\le z\le 2$ using an iterated triple integral.

Given

The region is a rectangular box in 3D Cartesian coordinates.
The coordinate bounds are $0 \le x \le 3$ , $0 \le y \le 4$ , $0 \le z \le 2$ .
The volume element is dV = dx dy dz.

Goal

Write the iterated integral that computes the volume of the box.

Steps

Expression
$dV = dx\,dy\,dz$
Explanation
Use the infinitesimal volume element for a small rectangular cube inside the box.
Justification
This formula is written directly on the board.
Shown in the video
Expression
$\int_{z=0}^{2} dz$
Explanation
Integrate first with respect to z from the bottom face to the top face of the box.
Justification
The speaker identifies the lowest z-value as 0 and the upper bound as 2.
Shown in the video
Expression
$\int_{y=0}^{4} \int_{z=0}^{2} dz\,dy\,dx$
Explanation
Add a y-integral from 0 to 4 to sum the vertical columns into a sheet.
Justification
The speaker says y goes from 0 to 4 and describes this as summing in the y direction.
Shown in the video
Expression
$\int_{x=0}^{3} \int_{y=0}^{4} \int_{z=0}^{2} dz\,dy\,dx$
Explanation
Add the outer x-integral from 0 to 3 to sum all sheets and obtain the full volume.
Justification
The speaker says to go from $x = 0$ to $x = 3$ to get the volume of the entire figure.
Shown in the video

Answer

$\int_{x=0}^{3}\int_{y=0}^{4}\int_{z=0}^{2} dz$ \,dy\,dx

Verification

The setup matches the displayed constant bounds $0 \le x \le 3$ , $0 \le y \le 4$ , $0 \le z \le 2$ and the stated order dz, then dy, then dx.

Volume of the rectangular box $0 \le x \le 3$ , $0 \le y \le 4$ , $0 \le z \le 2$

Clear evidence

Shown in the video

Evidence

Diagram
Observation
A dashed rectangular box is drawn in xyz-space with axis labels x, y, z.
Formula
Observation
Bounds are written as $0 \le x \le 3$ , $0 \le y \le 4$ , $0 \le z \le 2$ .
Formula
Observation
The worked integral is $\int_{0}^{3}\int_{0}^{4}\int_{0}^{2} 1$ \, dz \, dy \, dx and the final written value is 24.

Problem

Compute the volume of the rectangular box shown in the 3D diagram using the displayed triple integral.

Given

$0 \le x \le 3$
$0 \le y \le 4$
$0 \le z \le 2$
dV = dx dy dz
integrand is implicitly 1

Goal

Evaluate the triple integral and interpret the result as volume.

Steps

Expression
$\int_{0}^{3}\int_{0}^{4}\int_{0}^{2} 1 \, dz \, dy \, dx$
Explanation
Set up the volume integral from the box bounds and volume element.
Justification
The board explicitly writes the bounds and dV = dx dy dz.
Shown in the video
Expression
$\int_{0}^{3}\int_{0}^{4} 2 \, dy \, dx$
Explanation
After integrating with respect to z, the inner result is 2.
Justification
[z]_0^ $2 = 2 - 0 = 2$ .
Shown in the video
Expression
$\int_{0}^{3} 8 \, dx$
Explanation
After integrating with respect to y, the middle result is 8.
Justification
[2y]_0^ $4 = 8 - 0 = 8$ .
Shown in the video
Expression
$24$
Explanation
After integrating with respect to x, the final numerical value is 24.
Justification
[8x]_0^ $3 = 24 - 0 = 24$ .
Shown in the video

Answer

24 cubic units

Verification

The speaker states the final result as 24 units cubed, and the board shows the final written value 24.

Rectangular solid: from volume integral to variable-density mass setup

Clear evidence

Shown in the video

Evidence

Diagram
Observation
The board shows a rectangular solid in 3D with labeled bounds $0\le x\le 3$ , $0\le y\le 4$ , and $0\le z\le 2$ .
Formula
Observation
The board shows the volume integral $\int_0^3\int_0^4\int_0^2 1\, dz\, dy\, dx=24$ and later the density example $\rho(x,y,z)=xyz$ with $dmass=xyz\,dV$ .
Audio
Observation
The speaker contrasts finding volume with finding mass when density is not uniform.

Uncertainties

The clip sets up the mass problem but does not carry out the full triple integral for mass before it ends.

Problem

For the rectangular solid shown on the board, first compute its volume using a triple integral, then set up the corresponding mass calculation when the density is non-uniform and given by $\rho(x,y,z)=xyz$ .

Given

The solid is a rectangular box with $0\le x\le 3$ , $0\le y\le 4$ , and $0\le z\le 2$ .
The volume computation on the board uses integrand $1$ .
The example density is $\rho(x,y,z)=xyz$ .
The mass element is written as $dmass=xyz\,dV$ .

Goal

Compute the box volume and express the infinitesimal mass element for the variable-density version of the same solid.

Steps

Expression
$\int_0^3\int_0^4\int_0^2 1\, dz\, dy\, dx$
Explanation
Write the triple integral for the volume of the rectangular box.
Justification
Volume of a 3D region can be computed by integrating $1$ over the region.
Shown in the video
Expression
$24$
Explanation
Evaluate the displayed iterated integral step by step to obtain the volume.
Justification
The board explicitly shows the intermediate evaluations leading to $24$ .
Shown in the video
Expression
$\rho(x,y,z)=xyz$
Explanation
Replace the constant-density/volume situation with a variable density example.
Justification
The speaker introduces a non-uniform density and writes this specific function on the board.
Shown in the video
Expression
$dmass = xyz\,dV$
Explanation
Express the infinitesimal mass as density times infinitesimal volume.
Justification
The speaker states that the mass of a small volume is that volume times the density at that point.
Shown in the video

Answer

The volume of the box is $24$ . For the variable-density example, the differential mass element is $dmass=xyz\,dV$ .

Verification

The volume result $24$ agrees with the geometric side lengths $3\times 4\times 2$ . The mass setup follows directly from the stated rule 'small mass = density at that point times small volume.'

Visual events · 13

Writing the coordinate bounds

Clear evidence

Shown in the video

Evidence

Animation
Observation
Green inequalities appear sequentially in the upper-left corner: first ` $0 \le x \le 3$ `, then ` $0 \le y \le 4$ `, then ` $0 \le z \le 2$ `.

Objects

Handwritten inequalities in green

Changes

The x-bound appears first.
The y-bound is added below it.
The z-bound is added below the y-bound.

Invariants

All three bounds start at 0.
Each bound is an interval along one coordinate axis.

Interpretation

The visual sequence defines a rectangular region in three-dimensional space by specifying independent intervals for x, y, and z.

Drawing the three-dimensional coordinate frame

Clear evidence

Shown in the video

Evidence

Animation
Observation
Three green axes are drawn from a common origin and labeled x, y, and z.

Objects

x-axis
y-axis
z-axis
origin

Changes

Axes are drawn one after another.
Axis labels x, y, and z are added.

Invariants

The axes share a common origin.
The sketch represents a three-dimensional coordinate system.

Interpretation

The coordinate frame provides the spatial setting for the solid defined by the earlier inequalities.

Sketching the rectangular solid

Clear evidence

Shown in the video

Evidence

Animation
Observation
Tick marks are added along the axes, then dashed green edges are drawn to outline a rectangular solid in the first octant.

Uncertainties

Exact tick placement is schematic rather than measured precisely on screen.

Objects

Rectangular solid
dashed edges
axis tick marks

Changes

Tick marks indicate the extents along each axis.
Dashed lines connect corresponding endpoints to form the box.
The top face and vertical edges are added last.

Invariants

The solid remains aligned with the coordinate axes.
The lower face lies in the plane $z = 0$ .
The side lengths correspond to the intervals 0 to 3, 0 to 4, and 0 to 2.

Interpretation

The animation converts the algebraic bounds into a geometric picture of the region whose volume is being discussed.

Large rectangular box in Cartesian space

Clear evidence

Shown in the video

Evidence

Diagram
Observation
A 3D coordinate system with axes labeled $x$ , $y$ , and $z$ is shown, with a dashed rectangular box inside it.
Formula
Observation
The inequalities $0 \le x \le 3$ , $0 \le y \le 4$ , and $0 \le z \le 2$ remain visible at top left.

Objects

coordinate axes $x$ , $y$ , $z$
dashed rectangular box
handwritten inequalities $0 \le x \le 3$ , $0 \le y \le 4$ , $0 \le z \le 2$

Changes

The cursor moves around the box and axes while the drawing itself stays fixed.

Invariants

The outer region remains a rectangular box.
The displayed coordinate bounds do not change.

Interpretation

The picture establishes a simple Cartesian solid whose volume will be built from infinitesimal sub-volumes.

Insertion of a small representative cube

Clear evidence

Shown in the video

Evidence

Diagram
Observation
A smaller solid cube is drawn inside the larger dashed box.
Audio
Observation
The speaker says, "That's my cube. It's someplace in this larger cube".

Objects

large dashed rectangular box
small solid cube inside it

Changes

A new small cube appears inside the larger region.
The speaker then labels its edges as $dy$ , $dz$ , and $dx$ .

Invariants

The small cube lies entirely within the larger box.
The coordinate axes and outer bounds remain unchanged.

Interpretation

The small cube represents one infinitesimal volume element used to build the triple integral.

Writing the volume differential formula

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board writes $dV = dx\,dy\,dz$ in the upper right.
Audio
Observation
The speaker calls $dV$ the volume differential.

Objects

formula $dV = dx\,dy\,dz$
small cube diagram

Changes

The symbolic expression for the infinitesimal volume is added to the board.

Invariants

The underlying geometry remains the same rectangular box with a small interior cube.

Interpretation

The visual formula converts the geometric side lengths of the tiny cube into the standard notation for a volume element.

Pointing out the first integration direction

Clear evidence

Shown in the video

Evidence

Diagram
Observation
The cursor points near the small cube and then toward the vertical $z$ -axis.
Audio
Observation
The speaker says they can add the cube up in the $z$ direction to get the volume of a column.

Objects

small cube
vertical $z$ -axis
cursor

Changes

Attention shifts from the small cube to the vertical direction.

Invariants

No new formula is written during this interval.
The outer box and bounds remain visible.

Interpretation

The gesture indicates that the first summation in the triple-integral construction is taken along the $z$ -axis.

Overall whiteboard layout and writing sequence

Clear evidence

Shown in the video

Evidence

Diagram
Observation
The screen shows a black digital blackboard with green inequalities at top left, cyan formula dV = dx dy dz at top right, and a 3D coordinate drawing below.
Animation
Observation
A cursor moves around the board while new integral signs and limits are written in sequence.

Objects

Top-left green inequalities for x, y, and z bounds.
Top-right cyan formula dV = dx dy dz.
Lower-left 3D coordinate axes labeled x, y, z.
Dashed rectangular box in the first octant.
Small cyan cube inside the larger box.
Nested integral signs added in cyan and pink/magenta.

Changes

First the inner z-limits 0 and 2 are written beside the first integral sign.
Then dz, dy, dx are placed to indicate the integration order.
Next a second integral sign is added with y-limits 0 and 4.
Finally a third integral sign is added with x-limits 0 and 3.

Invariants

The region remains the same rectangular box throughout the clip.
The coordinate bounds listed at top left do not change.
The volume element formula dV = dx dy dz remains visible.

Interpretation

The visual progression turns a geometric picture of a box made of tiny cubes into an algebraic iterated integral whose nesting order mirrors successive summation in z, then y, then x.

Geometric buildup from column to sheet to full box

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker describes first a column, then a sheet parallel to the zy-plane, then the entire figure.
Diagram
Observation
The small cube inside the dashed box visually supports the idea of building larger pieces from infinitesimal volumes.

Uncertainties

The animation does not explicitly shade a full column, sheet, or solid; these are described verbally while the static box drawing remains on screen.

Objects

One infinitesimal cube.
A vertical column obtained by summing in z.
A sheet obtained by summing in y.
The full rectangular solid obtained by summing in x.

Changes

The explanation moves from a single small cube to a vertical stack in the z-direction.
It then extends that stack across the y-interval to make a sheet.
It finally repeats that sheet across the x-interval to recover the whole box.

Invariants

The box dimensions remain 3 by 4 by 2 in the x, y, z directions.
The integration order stays dz, dy, dx.

Interpretation

The mathematical relation being illustrated is that triple integration decomposes volume into successive one-dimensional summations: height first, then width, then depth.

Rectangular box diagram linked to coordinate bounds

Clear evidence

Shown in the video

Evidence

Diagram
Observation
A dashed rectangular box is drawn in a 3D coordinate system with axes labeled x, y, z.
Formula
Observation
Green inequalities at top left list the coordinate ranges.
Formula
Observation
Red arrows connect the written bounds to the corresponding axis directions on the box.

Objects

3D coordinate axes labeled x, y, z
dashed rectangular box
green bound inequalities
red arrows

Changes

The red 1 is added inside the integral early in the clip.
Later, intermediate antiderivatives and evaluated constants are written beneath the original integral.
Near the end, the view scrolls downward to make room for the remaining steps.

Invariants

The box dimensions remain 3 by 4 by 2 throughout the clip.
The integration order remains dz dy dx once established.

Interpretation

The diagram visually identifies the region whose volume is being computed by the iterated integral.

Constant top surface $z(x,y)=2$

Clear evidence

Shown in the video

Evidence

Formula
Observation
Green text $z(x,y)=2$ is written on the right side.
Audio
Observation
Speaker says the surface is $z=2$ and that it is a constant function independent of x and y.
Diagram
Observation
The top face of the box corresponds to the plane $z=2$ .

Objects

box top face
expression $z(x,y)=2$
xy-base region

Changes

The written function $z(x,y)=2$ appears after the inner z-integration has produced the constant 2.

Invariants

The height above the xy-region stays constant at 2.

Interpretation

This visual connects the triple-integral computation to the earlier double-integral idea of volume under a surface.

Static box diagram plus progressive handwritten mass setup

Clear evidence

Shown in the video

Evidence

Diagram
Observation
The left half of the screen contains a green 3D coordinate sketch of a dashed rectangular box with axis labels and bound annotations; the right half contains the iterated integral and its evaluation lines.
Animation
Observation
As the clip progresses, new green handwriting is added beneath the earlier work: first $\rho(x,y,z)=xyz$ , then $dmass=xyz\,dV$ .

Objects

3D coordinate axes
dashed rectangular box
bound labels $0\le x\le 3$ , $0\le y\le 4$ , $0\le z\le 2$
iterated integral $\int_0^3\int_0^4\int_0^2 1\, dz\, dy\, dx$
evaluation lines ending at $24$
handwritten density formula $\rho(x,y,z)=xyz$
handwritten mass-element formula $dmass=xyz\,dV$

Changes

New formulas are written under the existing volume work during the second half of the clip.
The focus shifts from the already-evaluated volume integral to the newly introduced density and mass-element expressions.

Invariants

The rectangular box diagram remains on screen throughout.
The previously written volume integral and its final value $24$ remain visible while the new mass setup is added.

Interpretation

The visual progression shows that the same solid region is being reused: first as a volume example, then as the domain for a variable-density mass calculation.

Misconceptions · 8

Calling the solid a cube despite unequal side lengths

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says 'find the volume of a cube' while the displayed bounds are $0 \le x \le 3$ , $0 \le y \le 4$ , and $0 \le z \le 2$ .
Formula
Observation
The three side lengths shown on screen are unequal.

Misconception

The spoken wording suggests the object is a cube.

Clarification

Mathematically, the displayed region is a rectangular box with side lengths 3, 4, and 2, not a cube, because a cube requires all side lengths to be equal.

Confusing which coordinate gives the height

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker first says the height is $dx$ , then immediately corrects himself: "Sorry, no, its height is dz".

Misconception

One might mistakenly assign the vertical height of the small cube to $dx$ instead of $dz$ .

Clarification

In the drawing, $z$ is the up-and-down axis, so the height of the small cube is $dz$ ; $dx$ is the depth direction.

Confusing the written order of differentials with the geometric order of summation

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says he will write dz first just so it reminds us that we are going to take the integral with respect to z first, then y, then x.

Misconception

One might think the differentials can be placed in any order without affecting which variable is integrated first.

Clarification

In the video’s notation, writing dz first signals that the innermost integration is with respect to z, followed by y and then x.

Assuming every inner integral must produce a nonconstant function

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says the result will be a function of x and y, but since all quantities here are constant it is actually a constant value.

Misconception

Because an inner integral is performed while holding other variables fixed, one may expect the result always to depend on those outer variables.

Clarification

Here the z-limits are constant, so after integrating in z the column volume is just the constant height 2 times dy dx, not a varying function of x and y.

Assuming a blank integrand means zero

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker explicitly addresses that nothing is written under the integral and says to assume there is a 1.

Misconception

A reader might think that if no function is written inside a volume integral, there is no integrand or the integrand is unspecified in a way that prevents evaluation.

Clarification

The clip states that dz dy dx is the same as 1 \, dz dy dx, so the missing integrand is 1.

Thinking a triple integral is a completely different operation here

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker says the triple integral is "really, really nothing different" and anticipates the question why do it at all.

Uncertainties

The promised justification for why triple integrals are useful is not given within this clip.

Misconception

A learner may think the triple integral introduces a wholly new computational process unrelated to double integrals.

Clarification

In this example, after integrating with respect to z, the remaining computation is exactly a double integral with constant height $z(x,y)=2$ .

Confusing $dV$ with only the Cartesian product $dx\,dy\,dz$

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says $dV$ doesn't always have to be $dx\,dy\,dz$ if we're doing other coordinates, mentioning polar coordinates as an example where it could be slightly different.

Misconception

One might think the volume element in a triple integral is always $dx\,dy\,dz$ .

Clarification

In Cartesian coordinates for this box, $dV$ corresponds to width times height times depth, but in other coordinate systems the expression for $dV$ changes.

Using total volume times a single density when density varies

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker contrasts the uniform-density case, where mass is just density times volume, with the non-uniform case, where density changes from point to point.

Misconception

One might try to compute mass by multiplying a single density value by the whole volume even when density is not uniform.

Clarification

If density varies with position, mass must be built from local contributions $\rho(x,y,z)\,dV$ rather than one global product.

Concept relations · 16

Basic geometric volume formula for a rectangular solid → Purpose of the example: introducing triple integrals

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker contrasts the easy geometric method with the goal of getting used to what a triple integral looks like.

Contrast

Explanation

The video explicitly notes that basic geometry already solves this example, but the pedagogical purpose is to transition toward triple integrals.

Basic geometric volume formula for a rectangular solid → Find the volume of the solid with $0 \le x \le 3$ , $0 \le y \le 4$ , $0 \le z \le 2$

Clear evidence

Derived from the video

Evidence

Formula
Observation
The inequalities ` $0 \le x \le 3$ `, ` $0 \le y \le 4$ `, and ` $0 \le z \le 2$ ` are written first.
Animation
Observation
The later sketch uses those same extents to draw the rectangular solid.

Application

Explanation

The geometric volume formula is applied to the specific rectangular solid defined by the displayed coordinate bounds.

Dimensions of the small cube → Volume differential $dV$

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker labels the cube sides and then writes the product formula.
Formula
Observation
$dV = dx\,dy\,dz$ appears after the side labels are discussed.

Proof dependency

Explanation

The definition of $dV$ depends on first identifying the three side lengths $dx$ , $dy$ , and $dz$ of the small cube.

Volume differential $dV$ → Commutativity of the factors in $dV$

Clear evidence

Shown in the video

Evidence

Audio
Observation
After writing the product, the speaker remarks that the order can be switched because multiplication is associative and order does not matter.

Application

Explanation

The commutativity remark applies directly to the product expression defining $dV$ .

Meaning of a triple integral → Volume differential $dV$

Clear evidence

Shown in the video

Evidence

Audio
Observation
The clip opens by asking what a triple integral means and later explains integrals as infinite sums of infinitesimal pieces.
Formula
Observation
The infinitesimal piece used here is $dV = dx\,dy\,dz$ .

Contains

Explanation

The conceptual meaning of the triple integral in this clip contains the construction of the volume element $dV$ as the basic summed piece.

Volume differential $dV$ → First summation step: build a column in the $z$ -direction

Clear evidence

Shown in the video

Evidence

Audio
Observation
After defining $dV$ , the speaker says they can first add it up in the $z$ direction to get a column.

Application

Explanation

The volume element $dV$ is the object being summed in the first integration step along $z$ .

Differential volume element dV → Iterated triple integral for volume

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows dV = dx dy dz and then builds an integral summing such elements.
Audio
Observation
The speaker refers to taking the sum of these dVs when setting up the integral.

Application

Explanation

The differential volume element dV = dx dy dz is the basic piece being summed by the triple integral setup.

Rectangular box region in three variables → Iterated triple integral for volume

Clear evidence

Shown in the video

Evidence

Formula
Observation
The inequalities $0 \le x \le 3$ , $0 \le y \le 4$ , $0 \le z \le 2$ are used directly as the limits of the nested integrals.
Audio
Observation
The speaker chooses the z, y, and x limits from these displayed inequalities.

Application

Explanation

The constant coordinate bounds of the rectangular box determine the limits of each iterated integral.

Iterated triple integral for volume → Inner z-integral gives column height

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says the inner integral gives the volume of a column and equals 2 dy dx.
Formula
Observation
The innermost integral shown is from $z = 0$ to $z = 2$ with dz.

Proof dependency

Explanation

The geometric interpretation of a vertical column depends on evaluating the inner z-integral first.

Summing columns in y gives a zy-parallel sheet → Summing sheets in x gives total volume

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker first forms a sheet by summing in y, then obtains the whole figure by summing in x.
Formula
Observation
Additional integral signs are nested outside the earlier ones with y-limits 0 to 4 and x-limits 0 to 3.

Generalizes

Explanation

The full solid volume is obtained by generalizing the sheet construction across the remaining x-direction.

Triple integral setup for the volume of a rectangular box → Evaluating an iterated triple integral from inside out

Clear evidence

Shown in the video

Evidence

Formula
Observation
The bounds and dV are written first, then the iterated evaluation proceeds step by step.

Application

Explanation

The rectangular-box setup is evaluated using the inside-out iterated integration method.

Evaluating an iterated triple integral from inside out → Triple integral reduces to a double integral for this box

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker says the result after the z-step is a double integral and compares it to previous double-integral videos.
Formula
Observation
The board shows $\int_{0}^{3}\int_{0}^{4} 2$ \, dy \, dx after the inner step.

Contains

Explanation

The triple-integral evaluation contains a stage that is precisely a double integral over the xy-region with constant height 2.

Find an answer · 23

What region is defined by $0 \le x \le 3$ , $0 \le y \le 4$ , and $0 \le z \le 2$ ?

Clear evidence

Shown in the video

Evidence

Formula
Observation
The three inequalities defining the region are visible on screen.

Knowledge points

Basic geometric volume formula for a rectangular solid
Find the volume of the solid with $0 \le x \le 3$ , $0 \le y \le 4$ , $0 \le z \le 2$

Why does the instructor use a simple box if geometry already gives the volume?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker explains that the example is meant to introduce triple integrals and their relation to double integrals.

Knowledge points

Purpose of the example: introducing triple integrals

What is the volume of the rectangular solid shown in this clip?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker computes 12 times 2 to get 24 cubic units.

Knowledge points

Find the volume of the solid with $0 \le x \le 3$ , $0 \le y \le 4$ , $0 \le z \le 2$
Geometric computation of the rectangular solid's volume

How does this clip begin explaining the meaning of a triple integral?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker asks, "So what does a triple integral mean?"

Knowledge points

Meaning of a triple integral

Why is the infinitesimal volume written as $dV = dx\,dy\,dz$ ?

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $dV = dx\,dy\,dz$ .
Audio
Observation
The speaker describes it as width times length times height.

Knowledge points

Volume differential $dV$
Dimensions of the small cube
Derivation of $dV = dx\,dy\,dz$

Can the factors in $dV$ be reordered, and why?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says the order can be switched because multiplication is associative and order does not matter.

Knowledge points

Commutativity of the factors in $dV$

Which direction is summed first in this introduction to the triple integral?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says they can first add the cube up in the $z$ direction to get the volume of a column.

Knowledge points

First summation step: build a column in the $z$ -direction

Why is the height of the small cube $dz$ rather than $dx$ in this drawing?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker corrects the height label from $dx$ to $dz$ .

Knowledge points

Confusing which coordinate gives the height
Dimensions of the small cube

What does dV mean in a triple integral for volume?

Clear evidence

Shown in the video

Evidence

Formula
Observation
Top-right formula reads dV = dx dy dz.
Audio
Observation
The speaker talks about summing these dVs.

Knowledge points

Differential volume element dV

Why does the video integrate with respect to z first?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says he writes dz first to remind us that the integral is taken with respect to z first.
Formula
Observation
The innermost integral has limits $z = 0$ to $z = 2$ .

Knowledge points

Iterated triple integral for volume
Inner z-integral gives column height

What are the limits of integration for the rectangular box shown?

Clear evidence

Shown in the video

Evidence

Formula
Observation
The displayed bounds are $0 \le x \le 3$ , $0 \le y \le 4$ , $0 \le z \le 2$ .
Audio
Observation
The speaker uses these exact numbers as the limits of the nested integrals.

Knowledge points

Rectangular box region in three variables
Iterated triple integral for volume

How does the triple integral build volume from columns to sheets to the whole solid?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker describes a column, then a sheet parallel to the zy-plane, then the entire figure.
Diagram
Observation
A small cube inside a larger dashed box illustrates the buildup.

Knowledge points

Inner z-integral gives column height
Summing columns in y gives a zy-parallel sheet
Summing sheets in x gives total volume
Geometric buildup from column to sheet to full box

Coverage and review notes

Covered · Opening black screen transitions into the spoken setup of the volume problem.

Covered · The coordinate bounds $0 \le x \le 3$ , $0 \le y \le 4$ , and $0 \le z \le 2$ are written and read aloud.

Covered · The speaker notes that geometry would suffice and states the pedagogical aim of introducing triple integrals.

Covered · A three-dimensional coordinate system is drawn and the rectangular solid is sketched from the given bounds.

Covered · The speaker computes the volume geometrically as 24 cubic units and begins transitioning toward the integral approach.

Covered · Opening question about the meaning of a triple integral.

Covered · Speaker motivates taking the volume of a very small volume inside the region.

Covered · A small cube is drawn and its side lengths are identified as $dy$ , $dz$ , and $dx$ .

Covered · The formula $dV = dx\,dy\,dz$ is written and discussed.

Covered · The speaker explains integrals as infinite sums of infinitesimal distances.

Covered · The first summation direction is chosen as the $z$ -direction to form a column.

Covered · Initial board state and construction of the inner z-integral with dz dy dx order.

Covered · Explanation that the inner integral gives the volume of one vertical column, equal to 2 dy dx.

Covered · Addition of the y-integral from 0 to 4 and interpretation as forming a sheet parallel to the zy-plane.

Covered · Addition of the x-integral from 0 to 3 to obtain the full volume setup; evaluation is announced but not performed.

Covered · The box, bounds, dV = dx dy dz, and implicit integrand 1 are introduced.

Covered · The inner z-integration is evaluated to 2 and rewritten as a double integral.

Covered · The speaker connects the remaining double integral to $z(x,y)=2$ and says the triple integral is not fundamentally different here.

Covered · The y- and x-integrations are completed, yielding 24 cubic units.

Covered · Audio contrasts constant-integrand volume computation with the new goal of finding mass; the board already shows the box and the volume integral.

Covered · The speaker explains uniform versus non-uniform density and motivates a variable density function.

Covered · The density symbol $\rho$ is introduced and the example $\rho(x,y,z)=xyz$ is written on the board.

Covered · The speaker defines the small-mass element as density times small volume and writes $dmass=xyz\,dV$ .

Covered · The speaker interprets $dV$ as width times height times depth in Cartesian coordinates and warns that other coordinate systems change its form.

Covered · No additional distinct mathematical content is introduced in the final fraction of the clip beyond the ongoing explanation of $dV$ .

Covered · All mathematical content in the video has been covered.

Explore the knowledge in this video

Reviewed subject paths

Double and triple integrals

Triple integrals 1 | Double and triple integrals | Multivariable Calculus | Khan Academy

The explanation, unpacked.

Chapters

Learning script

Knowledge cards

Coordinate bounds of the solid

Why this example is used before triple integrals

Geometric volume of the rectangular box

Wording note: 'cube' versus rectangular box

Meaning of a triple integral in this clip

Rectangular box region

Small cube as the basic volume piece

Side lengths of the small cube

Volume differential dVdVdV

Reordering the factors in dVdVdV

Integral as an infinite sum

First summation direction: along zzz

Differential volume element dV

Rectangular box region

Order of integration dz, dy, dx

Inner z-integral as column height

Middle y-integral as a sheet

Outer x-integral as total volume

Completed triple integral setup

Volume of a rectangular box as a triple integral

Why the missing integrand is 1

Inside-out evaluation of the iterated integral

Triple integral reduces to a double integral here

Constant surface z(x,y)=2z(x,y)=2z(x,y)=2

Detailed learning notes

x

y

z

x,y,z

x

y

z

dx

dy

dz

dV

x

Purpose of the example: introducing triple integrals

Basic geometric volume formula for a rectangular solid

Meaning of a triple integral

Volume differential dVdVdV

Dimensions of the small cube

Commutativity of the factors in dVdVdV

Differential volume element dV

Rectangular box region in three variables

Iterated triple integral for volume

Triple integral setup for the volume of a rectangular box

Evaluating an iterated triple integral from inside out

Triple integral reduces to a double integral for this box

Integral as an infinite sum of infinitesimal pieces

Inner z-integral gives column height

Summing columns in y gives a zy-parallel sheet

Summing sheets in x gives total volume

An omitted integrand in a volume integral is 1

For this example the triple integral is presented as an extension of the double integral

Uniform density allows mass = density × volume

Constant integrand case can reduce to a lower-dimensional integral for volume

Geometric computation of the rectangular solid's volume

Derivation of dV=dx dy dzdV = dx\,dy\,dzdV=dxdydz

First summation step: build a column in the zzz-direction

Deriving the iterated integral from geometric summation

Evaluation of the triple integral for the box volume

Evaluation of the displayed triple integral for the box volume

Find the volume of the solid with 0≤x≤30 \le x \le 30≤x≤3, 0≤y≤40 \le y \le 40≤y≤4, 0≤z≤20 \le z \le 20≤z≤2

Rectangular box region used to motivate triple integrals

Setting up a triple integral for a rectangular box

Volume of the rectangular box 0≤x≤30 \le x \le 30≤x≤3, 0≤y≤40 \le y \le 40≤y≤4, 0≤z≤20 \le z \le 20≤z≤2

Rectangular solid: from volume integral to variable-density mass setup

Writing the coordinate bounds

Drawing the three-dimensional coordinate frame

Sketching the rectangular solid

Large rectangular box in Cartesian space

Insertion of a small representative cube

Writing the volume differential formula

Pointing out the first integration direction

Volume differential $dV$

Reordering the factors in $dV$

First summation direction: along $z$

Constant surface $z(x,y)=2$

Volume differential $dV$

Commutativity of the factors in $dV$

Derivation of $dV = dx\,dy\,dz$

First summation step: build a column in the $z$ -direction

Find the volume of the solid with $0 \le x \le 3$ , $0 \le y \le 4$ , $0 \le z \le 2$

Volume of the rectangular box $0 \le x \le 3$ , $0 \le y \le 4$ , $0 \le z \le 2$

Constant top surface $z(x,y)=2$

Confusing $dV$ with only the Cartesian product $dx\,dy\,dz$

Basic geometric volume formula for a rectangular solid → Find the volume of the solid with $0 \le x \le 3$ , $0 \le y \le 4$ , $0 \le z \le 2$

Dimensions of the small cube → Volume differential $dV$

Volume differential $dV$ → Commutativity of the factors in $dV$

Meaning of a triple integral → Volume differential $dV$

Volume differential $dV$ → First summation step: build a column in the $z$ -direction

What region is defined by $0 \le x \le 3$ , $0 \le y \le 4$ , and $0 \le z \le 2$ ?

Why is the infinitesimal volume written as $dV = dx\,dy\,dz$ ?

Can the factors in $dV$ be reordered, and why?

Why is the height of the small cube $dz$ rather than $dx$ in this drawing?