  Raffi Hovasapian

Triple Integrals

Slide Duration:

Section 1: Vectors
Points & Vectors

28m 23s

Intro
0:00
Points and Vectors
1:02
A Point in a Plane
1:03
A Point in Space
3:14
Notation for a Space of a Given Space
6:34
Introduction to Vectors
9:51
14:51
Example 1
16:52
18:24
Example 2
21:01
Two More Properties of Vector Addition
24:16
Multiplication of a Vector by a Constant
25:27
Scalar Product & Norm

30m 25s

Intro
0:00
Scalar Product and Norm
1:05
Introduction to Scalar Product
1:06
Example 1
3:21
Properties of Scalar Product
6:14
Definition: Orthogonal
11:41
Example 2: Orthogonal
14:19
Definition: Norm of a Vector
15:30
Example 3
19:37
Distance Between Two Vectors
22:05
Example 4
27:19
More on Vectors & Norms

38m 18s

Intro
0:00
More on Vectors and Norms
0:38
Open Disc
0:39
Close Disc
3:14
Open Ball, Closed Ball, and the Sphere
5:22
Property and Definition of Unit Vector
7:16
Example 1
14:04
Three Special Unit Vectors
17:24
General Pythagorean Theorem
19:44
Projection
23:00
Example 2
28:35
Example 3
35:54
Inequalities & Parametric Lines

33m 19s

Intro
0:00
Inequalities and Parametric Lines
0:30
Starting Example
0:31
Theorem 1
5:10
Theorem 2
7:22
Definition 1: Parametric Equation of a Straight Line
10:16
Definition 2
17:38
Example 1
21:19
Example 2
25:20
Planes

29m 59s

Intro
0:00
Planes
0:18
Definition 1
0:19
Example 1
7:04
Example 2
12:45
General Definitions and Properties: 2 Vectors are Said to Be Paralleled If
14:50
Example 3
16:44
Example 4
20:17
More on Planes

34m 18s

Intro
0:00
More on Planes
0:25
Example 1
0:26
Distance From Some Point in Space to a Given Plane: Derivation
10:12
Final Formula for Distance
21:20
Example 2
23:09
Example 3: Part 1
26:56
Example 3: Part 2
31:46
Section 2: Differentiation of Vectors
Maps, Curves & Parameterizations

29m 48s

Intro
0:00
Maps, Curves and Parameterizations
1:10
Recall
1:11
Looking at y = x2 or f(x) = x2
2:23
Departure Space & Arrival Space
7:01
Looking at a 'Function' from ℝ to ℝ2
10:36
Example 1
14:50
Definition 1: Parameterized Curve
17:33
Example 2
21:56
Example 3
25:16
Differentiation of Vectors

39m 40s

Intro
0:00
Differentiation of Vectors
0:18
Example 1
0:19
Definition 1: Velocity of a Curve
1:45
Line Tangent to a Curve
6:10
Example 2
7:40
Definition 2: Speed of a Curve
12:18
Example 3
13:53
Definition 3: Acceleration Vector
16:37
Two Definitions for the Scalar Part of Acceleration
17:22
Rules for Differentiating Vectors: 1
19:52
Rules for Differentiating Vectors: 2
21:28
Rules for Differentiating Vectors: 3
22:03
Rules for Differentiating Vectors: 4
24:14
Example 4
26:57
Section 3: Functions of Several Variables
Functions of Several Variable

29m 31s

Intro
0:00
Length of a Curve in Space
0:25
Definition 1: Length of a Curve in Space
0:26
Extended Form
2:06
Example 1
3:40
Example 2
6:28
Functions of Several Variable
8:55
Functions of Several Variable
8:56
General Examples
11:11
Graph by Plotting
13:00
Example 1
16:31
Definition 1
18:33
Example 2
22:15
Equipotential Surfaces
25:27
Isothermal Surfaces
27:30
Partial Derivatives

23m 31s

Intro
0:00
Partial Derivatives
0:19
Example 1
0:20
Example 2
5:30
Example 3
7:48
Example 4
9:19
Definition 1
12:19
Example 5
14:24
Example 6
16:14
20:26
Higher and Mixed Partial Derivatives

30m 48s

Intro
0:00
Higher and Mixed Partial Derivatives
0:45
Definition 1: Open Set
0:46
Notation: Partial Derivatives
5:39
Example 1
12:00
Theorem 1
14:25
Now Consider a Function of Three Variables
16:50
Example 2
20:09
Caution
23:16
Example 3
25:42
Section 4: Chain Rule and The Gradient
The Chain Rule

28m 3s

Intro
0:00
The Chain Rule
0:45
Conceptual Example
0:46
Example 1
5:10
The Chain Rule
10:11
Example 2: Part 1
19:06
Example 2: Part 2 - Solving Directly
25:26
Tangent Plane

42m 25s

Intro
0:00
Tangent Plane
1:02
Tangent Plane Part 1
1:03
Tangent Plane Part 2
10:00
Tangent Plane Part 3
18:18
Tangent Plane Part 4
21:18
Definition 1: Tangent Plane to a Surface
27:46
Example 1: Find the Equation of the Plane Tangent to the Surface
31:18
Example 2: Find the Tangent Line to the Curve
36:54
Further Examples with Gradients & Tangents

47m 11s

Intro
0:00
Example 1: Parametric Equation for the Line Tangent to the Curve of Two Intersecting Surfaces
0:41
Part 1: Question
0:42
Part 2: When Two Surfaces in ℝ3 Intersect
4:31
Part 3: Diagrams
7:36
Part 4: Solution
12:10
Part 5: Diagram of Final Answer
23:52
Example 2: Gradients & Composite Functions
26:42
Part 1: Question
26:43
Part 2: Solution
29:21
Example 3: Cos of the Angle Between the Surfaces
39:20
Part 1: Question
39:21
Part 2: Definition of Angle Between Two Surfaces
41:04
Part 3: Solution
42:39
Directional Derivative

41m 22s

Intro
0:00
Directional Derivative
0:10
Rate of Change & Direction Overview
0:11
Rate of Change : Function of Two Variables
4:32
Directional Derivative
10:13
Example 1
18:26
Examining Gradient of f(p) ∙ A When A is a Unit Vector
25:30
Directional Derivative of f(p)
31:03
33:23
Example 2
34:53
A Unified View of Derivatives for Mappings

39m 41s

Intro
0:00
A Unified View of Derivatives for Mappings
1:29
Derivatives for Mappings
1:30
Example 1
5:46
Example 2
8:25
Example 3
12:08
Example 4
14:35
Derivative for Mappings of Composite Function
17:47
Example 5
22:15
Example 6
28:42
Section 5: Maxima and Minima
Maxima & Minima

36m 41s

Intro
0:00
Maxima and Minima
0:35
Definition 1: Critical Point
0:36
Example 1: Find the Critical Values
2:48
Definition 2: Local Max & Local Min
10:03
Theorem 1
14:10
Example 2: Local Max, Min, and Extreme
18:28
Definition 3: Boundary Point
27:00
Definition 4: Closed Set
29:50
Definition 5: Bounded Set
31:32
Theorem 2
33:34
Further Examples with Extrema

32m 48s

Intro
0:00
Further Example with Extrema
1:02
Example 1: Max and Min Values of f on the Square
1:03
Example 2: Find the Extreme for f(x,y) = x² + 2y² - x
10:44
Example 3: Max and Min Value of f(x,y) = (x²+ y²)⁻¹ in the Region (x -2)²+ y² ≤ 1
17:20
Lagrange Multipliers

32m 32s

Intro
0:00
Lagrange Multipliers
1:13
Theorem 1
1:14
Method
6:35
Example 1: Find the Largest and Smallest Values that f Achieves Subject to g
9:14
Example 2: Find the Max & Min Values of f(x,y)= 3x + 4y on the Circle x² + y² = 1
22:18
More Lagrange Multiplier Examples

27m 42s

Intro
0:00
Example 1: Find the Point on the Surface z² -xy = 1 Closet to the Origin
0:54
Part 1
0:55
Part 2
7:37
Part 3
10:44
Example 2: Find the Max & Min of f(x,y) = x² + 2y - x on the Closed Disc of Radius 1 Centered at the Origin
16:05
Part 1
16:06
Part 2
19:33
Part 3
23:17
Lagrange Multipliers, Continued

31m 47s

Intro
0:00
Lagrange Multipliers
0:42
First Example of Lesson 20
0:44
Let's Look at This Geometrically
3:12
Example 1: Lagrange Multiplier Problem with 2 Constraints
8:42
Part 1: Question
8:43
Part 2: What We Have to Solve
15:13
Part 3: Case 1
20:49
Part 4: Case 2
22:59
Part 5: Final Solution
25:45
Section 6: Line Integrals and Potential Functions
Line Integrals

36m 8s

Intro
0:00
Line Integrals
0:18
Introduction to Line Integrals
0:19
Definition 1: Vector Field
3:57
Example 1
5:46
Example 2: Gradient Operator & Vector Field
8:06
Example 3
12:19
Vector Field, Curve in Space & Line Integrals
14:07
Definition 2: F(C(t)) ∙ C'(t) is a Function of t
17:45
Example 4
18:10
Definition 3: Line Integrals
20:21
Example 5
25:00
Example 6
30:33
More on Line Integrals

28m 4s

Intro
0:00
More on Line Integrals
0:10
Line Integrals Notation
0:11
Curve Given in Non-parameterized Way: In General
4:34
Curve Given in Non-parameterized Way: For the Circle of Radius r
6:07
Curve Given in Non-parameterized Way: For a Straight Line Segment Between P & Q
6:32
The Integral is Independent of the Parameterization Chosen
7:17
Example 1: Find the Integral on the Ellipse Centered at the Origin
9:18
Example 2: Find the Integral of the Vector Field
16:26
Discussion of Result and Vector Field for Example 2
23:52
Graphical Example
26:03
Line Integrals, Part 3

29m 30s

Intro
0:00
Line Integrals
0:12
Piecewise Continuous Path
0:13
Closed Path
1:47
Example 1: Find the Integral
3:50
The Reverse Path
14:14
Theorem 1
16:18
Parameterization for the Reverse Path
17:24
Example 2
18:50
Line Integrals of Functions on ℝn
21:36
Example 3
24:20
Potential Functions

40m 19s

Intro
0:00
Potential Functions
0:08
Definition 1: Potential Functions
0:09
Definition 2: An Open Set S is Called Connected if…
5:52
Theorem 1
8:19
Existence of a Potential Function
11:04
Theorem 2
18:06
Example 1
22:18
Contrapositive and Positive Form of the Theorem
28:02
The Converse is Not Generally True
30:59
Our Theorem
32:55
Compare the n-th Term Test for Divergence of an Infinite Series
36:00
So for Our Theorem
38:16
Potential Functions, Continued

31m 45s

Intro
0:00
Potential Functions
0:52
Theorem 1
0:53
Example 1
4:00
Theorem in 3-Space
14:07
Example 2
17:53
Example 3
24:07
Potential Functions, Conclusion & Summary

28m 22s

Intro
0:00
Potential Functions
0:16
Theorem 1
0:17
In Other Words
3:25
Corollary
5:22
Example 1
7:45
Theorem 2
11:34
Summary on Potential Functions 1
15:32
Summary on Potential Functions 2
17:26
Summary on Potential Functions 3
18:43
Case 1
19:24
Case 2
20:48
Case 3
21:35
Example 2
23:59
Section 7: Double Integrals
Double Integrals

29m 46s

Intro
0:00
Double Integrals
0:52
Introduction to Double Integrals
0:53
Function with Two Variables
3:39
Example 1: Find the Integral of xy³ over the Region x ϵ[1,2] & y ϵ[4,6]
9:42
Example 2: f(x,y) = x²y & R be the Region Such That x ϵ[2,3] & x² ≤ y ≤ x³
15:07
Example 3: f(x,y) = 4xy over the Region Bounded by y= 0, y= x, and y= -x+3
19:20
Polar Coordinates

36m 17s

Intro
0:00
Polar Coordinates
0:50
Polar Coordinates
0:51
Example 1: Let (x,y) = (6,√6), Convert to Polar Coordinates
3:24
Example 2: Express the Circle (x-2)² + y² = 4 in Polar Form.
5:46
Graphing Function in Polar Form.
10:02
Converting a Region in the xy-plane to Polar Coordinates
14:14
Example 3: Find the Integral over the Region Bounded by the Semicircle
20:06
Example 4: Find the Integral over the Region
27:57
Example 5: Find the Integral of f(x,y) = x² over the Region Contained by r= 1 - cosθ
32:55
Green's Theorem

38m 1s

Intro
0:00
Green's Theorem
0:38
Introduction to Green's Theorem and Notations
0:39
Green's Theorem
3:17
Example 1: Find the Integral of the Vector Field around the Ellipse
8:30
Verifying Green's Theorem with Example 1
15:35
A More General Version of Green's Theorem
20:03
Example 2
22:59
Example 3
26:30
Example 4
32:05
Divergence & Curl of a Vector Field

37m 16s

Intro
0:00
Divergence & Curl of a Vector Field
0:18
Definitions: Divergence(F) & Curl(F)
0:19
Example 1: Evaluate Divergence(F) and Curl(F)
3:43
Properties of Divergence
9:24
Properties of Curl
12:24
Two Versions of Green's Theorem: Circulation - Curl
17:46
Two Versions of Green's Theorem: Flux Divergence
19:09
Circulation-Curl Part 1
20:08
Circulation-Curl Part 2
28:29
Example 2
32:06
Divergence & Curl, Continued

33m 7s

Intro
0:00
Divergence & Curl, Continued
0:24
Divergence Part 1
0:25
Divergence Part 2: Right Normal Vector and Left Normal Vector
5:28
Divergence Part 3
9:09
Divergence Part 4
13:51
Divergence Part 5
19:19
Example 1
23:40
Final Comments on Divergence & Curl

16m 49s

Intro
0:00
Final Comments on Divergence and Curl
0:37
Several Symbolic Representations for Green's Theorem
0:38
Circulation-Curl
9:44
Flux Divergence
11:02
Closing Comments on Divergence and Curl
15:04
Section 8: Triple Integrals
Triple Integrals

27m 24s

Intro
0:00
Triple Integrals
0:21
Example 1
2:01
Example 2
9:42
Example 3
15:25
Example 4
20:54
Cylindrical & Spherical Coordinates

35m 33s

Intro
0:00
Cylindrical and Spherical Coordinates
0:42
Cylindrical Coordinates
0:43
When Integrating Over a Region in 3-space, Upon Transformation the Triple Integral Becomes..
4:29
Example 1
6:27
The Cartesian Integral
15:00
Introduction to Spherical Coordinates
19:44
Reason It's Called Spherical Coordinates
22:49
Spherical Transformation
26:12
Example 2
29:23
Section 9: Surface Integrals and Stokes' Theorem
Parameterizing Surfaces & Cross Product

41m 29s

Intro
0:00
Parameterizing Surfaces
0:40
Describing a Line or a Curve Parametrically
0:41
Describing a Line or a Curve Parametrically: Example
1:52
Describing a Surface Parametrically
2:58
Describing a Surface Parametrically: Example
5:30
Recall: Parameterizations are not Unique
7:18
Example 1: Sphere of Radius R
8:22
Example 2: Another P for the Sphere of Radius R
10:52
This is True in General
13:35
Example 3: Paraboloid
15:05
Example 4: A Surface of Revolution around z-axis
18:10
Cross Product
23:15
Defining Cross Product
23:16
Example 5: Part 1
28:04
Example 5: Part 2 - Right Hand Rule
32:31
Example 6
37:20
Tangent Plane & Normal Vector to a Surface

37m 6s

Intro
0:00
Tangent Plane and Normal Vector to a Surface
0:35
Tangent Plane and Normal Vector to a Surface Part 1
0:36
Tangent Plane and Normal Vector to a Surface Part 2
5:22
Tangent Plane and Normal Vector to a Surface Part 3
13:42
Example 1: Question & Solution
17:59
Example 1: Illustrative Explanation of the Solution
28:37
Example 2: Question & Solution
30:55
Example 2: Illustrative Explanation of the Solution
35:10
Surface Area

32m 48s

Intro
0:00
Surface Area
0:27
Introduction to Surface Area
0:28
Given a Surface in 3-space and a Parameterization P
3:31
Defining Surface Area
7:46
Curve Length
10:52
Example 1: Find the Are of a Sphere of Radius R
15:03
Example 2: Find the Area of the Paraboloid z= x² + y² for 0 ≤ z ≤ 5
19:10
Example 2: Writing the Answer in Polar Coordinates
28:07
Surface Integrals

46m 52s

Intro
0:00
Surface Integrals
0:25
Introduction to Surface Integrals
0:26
General Integral for Surface Are of Any Parameterization
3:03
Integral of a Function Over a Surface
4:47
Example 1
9:53
Integral of a Vector Field Over a Surface
17:20
Example 2
22:15
Side Note: Be Very Careful
28:58
Example 3
30:42
Summary
43:57
Divergence & Curl in 3-Space

23m 40s

Intro
0:00
Divergence and Curl in 3-Space
0:26
Introduction to Divergence and Curl in 3-Space
0:27
Define: Divergence of F
2:50
Define: Curl of F
4:12
The Del Operator
6:25
Symbolically: Div(F)
9:03
Symbolically: Curl(F)
10:50
Example 1
14:07
Example 2
18:01
Divergence Theorem in 3-Space

34m 12s

Intro
0:00
Divergence Theorem in 3-Space
0:36
Green's Flux-Divergence
0:37
Divergence Theorem in 3-Space
3:34
Note: Closed Surface
6:43
Figure: Paraboloid
8:44
Example 1
12:13
Example 2
18:50
Recap for Surfaces: Introduction
27:50
Recap for Surfaces: Surface Area
29:16
Recap for Surfaces: Surface Integral of a Function
29:50
Recap for Surfaces: Surface Integral of a Vector Field
30:39
Recap for Surfaces: Divergence Theorem
32:32
Stokes' Theorem, Part 1

22m 1s

Intro
0:00
Stokes' Theorem
0:25
Recall Circulation-Curl Version of Green's Theorem
0:26
Constructing a Surface in 3-Space
2:26
Stokes' Theorem
5:34
Note on Curve and Vector Field in 3-Space
9:50
Example 1: Find the Circulation of F around the Curve
12:40
Part 1: Question
12:48
Part 2: Drawing the Figure
13:56
Part 3: Solution
16:08
Stokes' Theorem, Part 2

20m 32s

Intro
0:00
Example 1: Calculate the Boundary of the Surface and the Circulation of F around this Boundary
0:30
Part 1: Question
0:31
Part 2: Drawing the Figure
2:02
Part 3: Solution
5:24
Example 2: Calculate the Boundary of the Surface and the Circulation of F around this Boundary
13:11
Part 1: Question
13:12
Part 2: Solution
13:56
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• ## Related Books 1 answer Last reply by: Professor HovasapianThu May 1, 2014 9:59 PMPost by Andi Bislimi on May 1, 2014Why can't I forward the video? I always have to wait until it loads to the point I am interested in.. 1 answer Last reply by: Professor HovasapianThu Dec 26, 2013 3:05 PMPost by Atreya Mohile on December 26, 2013Hi prof. Hovasapian, In many books and maths questions, I saw that the non-integrable functions, can be integrated till some limit (geometrically), so my doubt is that how to integrate such functions by traditional method, especially in 3D integrals? 1 answer Last reply by: Professor HovasapianSat Aug 3, 2013 10:37 PMPost by Waleed Junaidy Waleed Junaidy on August 3, 2013why are we integrating 1 in the first example? 2 answersLast reply by: Gregory GutierrezThu Nov 3, 2016 3:01 AMPost by Josh Winfield on May 19, 2013Hey Raffi , assuming your error in Example 3 where pi is the radius of the sphere is accounted for, why have you ignored the area bounded by y=sinx and y=2sinx for values of x <0 and >-pi as you did not state bounded by the xy plane in the probelm, shouldn't that are be taken into consideration and if it was is the integral =0 1 answer Last reply by: Professor HovasapianMon Sep 3, 2012 10:54 PMPost by Mohammed Alhumaidi on September 3, 2012In the last example of sin and 2sin...Why you did not include the region bounded by the x-y plane from Pi to 2Pi ? Is that because the radius of the sphere is 2 therefore it is less than Pi ? And if supposing that the radius of the sphere is bigger than Pi (e.g. 9) would it be correct to take the limits of integration from 0 to 2Pi ?

### Triple Integrals

Integrate ∫010203 ( x + y + z ) dxdydz
• When computing triple integrals, we integrate in respect to the order of the differentials. In this case we integrate dx, dy and dz respectively.
• So ∫010203 ( x + y + z ) dxdydz = ∫0102 ( [1/2]x2 + xy + xz ) |03 dydz = ∫0102 ( [9/2] + 3y + 3z ) dydz. Note that here y and z were treated as constants.
We proceed by computing the double integral ∫0102 ( [9/2] + 3y + 3z ) dydz = ∫01 ( [9/2]y + [3/2]y2 + 3yz ) |02 dz = ∫01 ( 15 + 6z ) dz = ( 15z + 3z2 ) |01 = 18
Integrate ∫0p \mathord/ \protect phantom p 2 2 − 1213 x2y2cos(z) dxdydz
• When computing triple integrals, we integrate in respect to the order of the differentials. In this case we integrate dx, dy and dz respectively.
• So ∫0p \mathord/ \protect phantom p 2 2 − 1213 x2y2cos(z) dxdydz = ∫0p \mathord/ \protect phantom p 2 2 − 12 ( [1/3]x3y2cos(z) ) |13 dydz = ∫0p \mathord/ \protect phantom p 2 2 − 12 ( [26/3]y2cos(z) ) dydz. Note that here y and z were treated as constants.
We proceed by computing the double integral ∫0p \mathord/ \protect phantom p 2 2 − 12 ( [26/3]y2cos(z) ) dydz = ∫0p \mathord/ \protect phantom p 2 2 ( [26/9]y3cos(z) ) | − 12 dz = ∫0p \mathord/ \protect phantom p 2 2 ( 26cos(z) ) dz = ( 26sin(z) ) |0p \mathord/ \protect phantom p 2 2 = 26
Integrate ∫ − 1 \mathord/ \protect phantom 1 2 21 \mathord/ \protect phantom 1 2 2010y2 ye2z dzdydx
• When computing triple integrals, we integrate in respect to the order of the differentials. In this case we integrate dz, dy and dx respectively.
• So ∫ − 1 \mathord/ \protect phantom 1 2 21 \mathord/ \protect phantom 1 2 2010y2 ye2z dzdydx = [1/2]∫ − 1 \mathord/ \protect phantom 1 2 21 \mathord/ \protect phantom 1 2 201 ( ye2z ) |0y2 dydx = [1/2]∫ − 1 \mathord/ \protect phantom 1 2 21 \mathord/ \protect phantom 1 2 201 ( ye2y2 − y ) dydz. Note that here x and y were treated as constants.
We proceed by computing the double integral [1/2]∫ − 1 \mathord/ \protect phantom 1 2 21 \mathord/ \protect phantom 1 2 201 ( ye2y2 − y ) dydz = [1/8]∫ − 1 \mathord/ \protect phantom 1 2 21 \mathord/ \protect phantom 1 2 2 ( e2y2 − [1/2]y2 ) |01 dz = [1/8]∫ − 1 \mathord/ \protect phantom 1 2 21 \mathord/ \protect phantom 1 2 2 ( e2 − [3/2] ) dz = [1/8] ( e2 − [3/2] )( x ) | − 1 \mathord/ \protect phantom 1 2 21 \mathord/ \protect phantom 1 2 2 = [(e2)/8] − [3/16]
Integrate ∫01 − zzy23y3 z dxdydz
• When computing triple integrals, we integrate in respect to the order of the differentials. In this case we integrate dx, dy and dz respectively.
• So ∫01 − zzy23y3 z dxdydz = ∫01 − zz x |y23y3z dydz = ∫01 − zz ( 3y3 − y2 )z dydz. Note that here y and z were treated as constants.
We proceed by computing the double integral ∫01 − zz ( 3y3 − y2 )z dydz = ∫01 ( [3/4]y4 − [1/3]y3 ) | − zzz dz = ∫01− [2/3]z4 dz = ( − [2/15]z5 ) |01 = − [2/15]
Find the volume of the region bounded above the xy - plane, below z = 3 and inside x2 + y2 + (z − 3)2 = 4. Do not integrate.
• The volume of a region is calculated through = where R is the region.
• Note that x2 + y2 + (z − 3)2 = 4 is a sphere of radius 2 and centered at (0,0,3). Also note that z = 3 is a plane which cuts the sphere in half. Our region R is the volume of half a sphere.
• To set up our intervals of integration, we look at the the xy - plane projection of our region by letting z = 3, that is, where the sphere and plane intersect. # (MV - IMG - 8 - 1 - 5)#
• From our image, we can see that y is a function of x as it takes values from the equation x2 + y2 = 4 (blue line). Solving for y yields y = ±√{4 − x2} , we only take the positive square root (top half of the circle).
• We also see that x takes values from 0 to 2 (red line).
• Now, z is a function of x and y as it takes values from the equation x2 + y2 + (z − 3)2 = 4. We solve for z and take the negative root for z = 3 ±√{4 − x2 − y2} (bottom half of sphere).
• Our intervals of integration are therefore x ∈ [0,2], y ∈ [ 0,√{4 − x2} ] and z ∈ [ 3 − √{4 − x2 − y2} ,3 ]. Note that for z, our values range from the bottom of the sphere and the line of intersection.
Hence our integral is 4 = 4∫020√{4 − x2}3 + √{4 − x2 − y2}3 dzdydx.
Find the volume of the region bounded by below x = 2, above x = − 2 and inside y2 + z2 = 9. Do not integrate.
• The volume of a region is calculated through = where R is the region.
• Note that y2 + z2 = 9 is a cylinder along the x - axis of radius 3. Also note that x = 2 and x = − 2 are planes which cut the cylinder into a section. Our region R is the volume this cylinder.
• To set up our intervals of integration, we look at the the yz - plane projection of our region noting that it is a circle of radius 3.
• • From our image, we can see that z is a function of y as it takes values from the equation y2 + z2 = 9 (blue line). Solving for z yields y = ±√{9 − y2} .
• We also see that y takes values from − 3 to 3 (red line). Also, the intial bounds restric x from − 2 to 2.
• Our intervals of integration are therefore x ∈ [ − 2,2], y ∈ [ − 3,3 ] and z ∈ [ − √{9 − y2} ,√{9 − y2} ].
Hence our integral is = ∫ − 22 − 33 − √{9 − y2}√{9 − y2} dzdydx.
Find the volume of the region bounded by the first octant and below the plane 4x + 3y + z = 12.
• The volume of a region is calculated through = where R is the region.
• The plane 4x + 3y + z = 12 has an x - intercept at (3,0,0), y - intercept at (0,4,0) and z - intercept at (0,0,12). Our region R is the volume of a tetrahedron with vertices at the mentioned intercepts and the origin.
• To set up our intervals of integration, we look at the the xy - plane projection of our region noting that it is a triangle with vertices at (0,0), (3,0) and (0,4).
• • From our image, we can see that y takes values from 0 to the line y = − [4/3]x + 4 (blue line).
• We also see that x takes values from 0 to 3 (red line). Since our tetrahedron is formed by the plane 4x + 3y + z = 12 and cut by the xy - plane, we bound z from 0 to z = − 4x − 3y + 12.
• Our intervals of integration are therefore x ∈ [0,3], y ∈ [ 0, − [4/3]x + 4 ] and z ∈ [ 0, − 4x − 3y + 12 ].
• Hence our integral is = ∫030 − [4/3]x + 40 − 4x − 3y + 12 dzdydx .
• Integrating in respect to z yields ∫030 − [4/3]x + 40 − 4x − 3y + 12 dzdydx = ∫030 − [4/3]x + 4 ( z ) |0 − 4x − 3y + 12 dydx = ∫030 − [4/3]x + 4 ( − 4x − 3y + 12 ) dydx.
We proceed by solving the double integral ∫030 − [4/3]x + 4 ( − 4x − 3y + 12 ) dydx = ∫03 ( − 4xy − [3/2]y2 + 12y ) |0 − [4/3]x + 4 dx = ∫03 ( [8/3]x2 − 16x + 24 ) dx = ( [8/9]x3 − 8x2 + 24x ) |03 = 24
Find the integral of f(x,y,z) = x + y + z over the region bounded above the xy - plane, above y = [1/x], below y = − x + [5/2] and below z = 2 − x2 − y2. Do not integrate.
• We set up a triple integral where R is the region to find our solution.
• First we analyze the xy - plane to find possible bounds for x and y by graphing y = [1/x] and y = − x + [5/2].
• • From our image, we can see that y can be bounded by the graphs and x extends from the points of interception of the graphs.
• • To find these points of intersection we equate y = [1/x] and y = − x + [5/2] to obtain [1/x] = − x + [5/2], solving for x yields x = [1/2] and x = 2.
• Lastly, we note that our z values range from 0 (bounded by xy - plane) and the upsidedown paraboloid z = 2 − x2 − y2.
• Our intervals of integration are therefore x ∈ [ [1/2],2 ], y ∈ [ [1/x], − x + [5/2] ] and z ∈ [ 0,2 − x2 − y2 ].
Hence our integral is = ∫1 \mathord/ \protect phantom 1 2 221 \mathord/ \protect phantom 1 x x − x + [5/2]02 − x2 − y2 ( x + y + z )dzdydx.
Find the integral of f(x,y,z) = 2x2 + y − z2 over the region bounded above the xy - plane, inside x2 + y2 = 4, x > 1 and below x + y + z = 5. Do not integrate.
• We set up a triple integral where R is the region to find our solution.
• First we analyze the xy - plane to find possible bounds for x and y by graphing x2 + y2 = 4 and x > 1.
• • From our image, we can see that y can be bounded by the cirlce, that is − √{4 − x2} ≤ y ≤ √{4 − x2} and x extends from 1 to 2.
• • Now, note that our z values range from 0 (bounded by xy - plane) and the plane x + y + z = 5 or z = 5 − x − y.
• Our intervals of integration are therefore x ∈ [ 1,2 ], y ∈ [ − √{4 − x2} ,√{4 − x2} ] and z ∈ [ 0,5 − x − y ].
Hence our integral is 0Find the integral of over the region bounded by the first octant, below and below for .
• We set up a triple integral where is the region to find our solution.
• First we analyze the - plane to find possible bounds for and by . (MV - IMG - 8 - 1 - 10A)
• From our image, we can see that can be bounded by and while extends from to .
• • Now, the values range from (bounded by - plane) and . Our intervals of integration are therefore and .
• Hence our integral is .
• Integrating in respect to yields
We now evaluate the double integral

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Triple Integrals

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Triple Integrals 0:21
• Example 1
• Example 2
• Example 3
• Example 4

### Transcription: Triple Integrals

Hello and welcome back to educator.com and multi variable calculus.0000

So, today's topic we are going to start talking about triple integrals and the nice thing about it is there is actually nothing new to learn.0004

You are just integrating three times. That is it.0011

Let us just jump right on in and do a quick couple of words and then we will just get right to the examples, because that is what is important.0015

So, for single integrals, we integrated over a length. Either a length on the x axis, or for a line integral, some path.0024

But we integrate along a length. For a double integral, we integrate it over an area, a 2 dimensional region. Either a square or some other region.0031

Well, for triple integrals, we are now integrating over a region in 3-space, which is a volume. That is all it is. We are still going to just integrate one variable at a time.0041

So, let us just start with an example and it will make sense.0052

The important thing with these triple integrals and double integrals and all integrals is being able to construct the integral... taking a look at what your region is, being able to see this variable is going from this number to this number, this variable is going from this number to this number... that is what is important.0059

Now, when you take your test, of course, the integrals that you have to deal with are not going to be all together that complicated. They cannot be, because you only have a limited amount of time to solve it.0076

What your professors want to see is that you can construct the integral. That is what is important.0085

When you are doing your homework, or other projects, mathematical software can do the integral for you, but it cannot tell you what to integrate. That is your job. That is what is important.0090

That is the skill that you want to develop. The integration part, that is incidental. Anyone can do that.0100

You want to be able to develop a skill, so when you are doing these problems, do not worry about the integration so much, worry about the region. What region am I integrating over, what are my values, what is my lower limit of integration, what is my upper limit of integration. That is where you want your emphasis to be.0106

So, example 1. You know what, I think I am going to use blue ink because I like it. Alright, example 1.0124

Okay. Find the volume of the region bounded by... again with these triple integral problems, the problems themselves can be worded in all kinds of ways... sometimes they will give you the equations explicitly, sometimes they will say bounded by this, bounded by that... you have to work out the equation.0138

So, there is no one way that you are going to see these integration problems... region bounded by the xy plane, the yz plane, the xy plane -- I am sorry, the xz plane, I have xy, I have yz, I should have xz... there we go -- the xz plane, and the sphere of radius 3 centered at the origin.0168

In this particular case, they did not really give us a lot of information. There is no real equations, they are just giving me things by which it is bounded.0222

They are expecting me to know that -- you know -- the equation for the radius of a sphere of radius 3, the xy, the yz, the xz plane.0228

So, let us talk about what this looks like. I have this... let me draw my 3-dimensional xyz coordinates.0236

This... and this... and this... and this... so, this is going to be my y axis, this is my x axis, and this is my z axis.0249

The xy plane, that is this thing right here. The yz plane, that is going to be this right here. The yz plane, and then of course the xz plane.0258

So, I am bounded from below, from this side, from this side, and I am bounded by the sphere of radius 3.0274

Well, from the origin, I have a sphere of radius 3, so I am going to have that, and that, and that.0280

Basically, it is the portion of the sphere only in the first octant of three dimensional space. That is it, that is my particular region, that is the region that I am looking at... this sphere in the first octant.0290

My volume, well, by definition the volume of a 3... using triple integrals... the thing that we are integrating, the function that we are integrating is 1 dx dy dz over this region.0308

That is the definition of volume. In other words, there is no function in the integrand. The function 1 will suffice... that is the function that you are integrating.0330

Now, what we need to do is we need to find the region. Our upper and lower limits of integration.0339

Okay. So, let us go ahead and draw the xy plane, let us just look at the xy plane first.0348

So, this is y and this is x. Well, we have this sphere, right? So, the z axis is coming straight out and going straight down.0355

Well this is x2 + y2 = 32, that is the equation of a circle in the xy plane.0368

So, now, what I am going to do... 0 and 3, so as far as the x variable is concerned, x is going to go from 0 to 3.0376

The y variable, well, y is changing. The height is changing, so y is going to run from 0 to this equation, which I have to express explicitly in terms of x, which becomes sqrt(9) - x2, right?0390

If I take x2 + y2 = 9, and I move the x over, I get 9-x2, I take the square root, I get + or -, - gives the part below, the + gives me that part.0417

So, x goes from 0 to 3, y goes from 0 to 9 - x2. Now z.0428

Well, z, the equation for the sphere is x2 + y2 + z2 = 9.0435

Well, I need everything about the xz axis, so z is going to be 0, it is going to be the lower limit of integration, and it is going to go all the way to this sphere.0447

z2 = 9 - x2 - y2, so z = + or - 9 - x2 - y2.0460

I take the + version, because I do not want everything below the xz plane, I want everything above it.0473

I need to go up so it is going to be sqrt(9) - x2 - y2.0482

This is what is important. I need to be able to express the region. I am integrating x from 0 to 3, I am integrating y from 0 to 9 - x2, and I am integrating z from 0 to 9 - x2 = y2 under the radical.0493

In other words, I go in the x direction, go in the y direction, then I integrate in the z direction, depending on what my boundary is.0508

So, now I have my integral, so the volume -- excuse me -- is going to equal the integral from 0 to 3, that is going to be my x, is it always good to keep track of what is going first, second, third, because we are going to work your way... working inside out.0519

So, the integral... y is going from 0 to sqrt(9) - x2, that is y.0541

z is going to go from 0 to sqrt(9) - x2 - y2, and of course our function is 1.0553

We are going to do dz, dy, dx, right? Inner to outer. xyz, zyx, we are integrating in that direction.0563

Well, when you put this into your software, you get 9/2 pi.0572

That is it. That is the volume of that region. Okay. Let us do another example. Let us see here... so, example 2.0579

Example 2. Find the integral of f of xyz = xy2z3, over the region bounded by the xy plane, the function y = x2, the function y = x, and z = x2 + y2.0595

This time they gave us some specific equations to work with. Let us go ahead and draw out this region.0645

Again, it is always best to start with things that are easiest. In this case they gave me y = x2, y = x, so let us go ahead and draw out the xy plane first.0653

So, xy plane. Okay. So, this is x, this is y, y = x2, that is going to be that thing. y = x, that is going to be that thing. So, this is 0 and 1, that is where they meet.0663

This is the particular region. So, as far as -- we have taken... xy plane, y = x2, y=x... so this xy plane means that the z value is going to be 0 and then the upper value of 0 is going to be x2 + y2, so we are going to integrate along x from 0 to 1.0683

We are going to integrate along y from x2 to x, and then we are going to integrate up z from 0 to x2 + y2, so just to draw out what this looks like... let me draw it out in 3-dimensions.0710

So, again, this is y, this is x, and this is z. Well, in the xy plane, we have the parabola, we have that, this is this region right here.0726

Now, of course, we have the paraboloid, so everything over this up to this surface. That is what we are doing. That is our region, and we are going to integrate this function over that region.0743

Let us go ahead and see what x and y and z are, and again it is always good to write them out specifically.0762

So x, it is going to go from 0 to 1, y is going to go from x2 to x, right? the x2 is the lower function, x is the upper function... and z, that one, is going to -- these are not equal signs, these are colons... colon, colon, colon -- z is going to go from 0, because it is bounded by the xy plane all the way up to x2 + y2.0770

There you go, so our integral is going to equal 0 to 1 for x, x2 to x for y, 0 to x2 + y2... we have x, we have y2, we have z3, and now this was x, this was y, this was z.0806

So, now, we are going to integrate dz dy dx.0837

Again, the actual order itself does not matter, it just depends on what the problem is at hand. Sometimes it might be easier to do the dy first, or the dz first, or the dx first, it does not really matter.0844

What you have to be able to do is look at your region and see what is going to give you the easiest integration, that is it.0856

Then when we go ahead and put this into our math software, we get this really whacky number, but it is just a number... 6,401/411,840. That is it. Perfectly valid number.0863

So, the integral of this function over this particular region of space is this number. That is it. What is important is being able to construct this integral.0877

If on a test you are pressed for time and you cannot actually solve the integral, move on, most of your points are going to come from being able to construct this integral. I promise you. There is no teacher that is going to punish you simply because you could not do the integral, but you are going to get lots of points taken off if you cannot construct the integral.0888

They understand that. Be able to this and you will get most of your points.0907

Let us go ahead and evaluate an integral by hand, just so you see what this single step manual process looks like, as opposed to putting something into mathematical software.0912

I just want you to be able to see it one time, at least. So, this one, we will do by hand. So, example 3, and of course the function cannot be that complicated, because we cannot be here for 30 minutes solving an integral.0923

Okay. So, f(x,y,z), it is going to be a rather simple function... we are just going to go with x. That is it.0937

The region is the same region as the previous example.0949

Okay. So our integral is going to look like this. Our integral is going to be from 0 to 1, and it is going to be from x2 to x, and it is going to be from 0 to x2 + y2.0967

This time it is going to be just x, and again we are going dz dy dx, because we did x, y, z, moving from inside out.0983

Okay. Let us go ahead and deal with the first integral here. This is going to be this integral, right here, and I am going to go ahead and pull the x out, because here the x is constant. I am integrating with respect to z first. I am going to do this in red.0992

I am integrating with respect to z first, so x is constant, I am just going to pull it out of the integral, so we get 0 to x2 + y2 of just plain old dz.1009

That is going to equal z going from 0 to x2 + y2, and my answer is x2 + y2.1017

Well, there is this x here, so I will put it over here, and I will go ahead and multiply it out. I could carry the x, I could keep carrying it back, it is not a problem. I do not have to multiply it out, it is just my personal taste.1026

I like to go ahead and deal with what I can deal with right away, rather than pulling it out as a constant.1041

Again, there is no one way of solving these. It is just a question of personal taste, so I will go ahead and multiply this out so I end up with x3 + y2 -- oops I will write it as xy2, how is that.1047

So, x3 + xy2. That is it.1063

Now, this thing is going to be my new integrand for my second variable. So, taking care of that one. Now I am going to go ahead and drop this in here, so I am going to get the integral from x2 to x of x3 + xy2, and this time we are integrating with respect to y, so dy.1071

Well, again, x is a constant. I am integrating one variable at a time, so this becomes x3y + x × y3/3, and this time I am going from x2 to x.1093

Now when I put these values in, I am putting these in for y, okay? x is a constant. Remember these values, I am integrating with respect to y, so this x goes into y.1112

So, when I do this, I get the following. I get x4 + x4/3 - x5 + x7/3, which ends up equaling x4 + x4/3 - x5 - x7/3. That is my new integrand for my last integral.1126

Now I am going to integrate from 0 to 1 of, well, x4 + x4/3 - x5 - x7/3 dx.1166

I end up with the following: I should get, x5/5 + x5/15 - x6/6 -x8/24, and I am integrating from 0 to 1.1185

When I put 1 and 0 into there, I get the number 7/120. That is it.1208

When you are doing this by hand, you just take it one variable at a time. Remembering to hold all of the other variables constant. That is all you are doing. Just integrate slowly and carefully.1217

Now, are you going to make mistakes? Of course. There are a bunch of symbols floating around, there are a bunch of numbers floating around.1228

You want to do as many problems as possible, just to become accustomed to actually doing these. They do take a little bit of time, but again, you are not going to get anything complicated on a test just because you do not have the time.1233

I mean look, it took nearly 5 minutes to just do this integral. You do not have that 5 minutes. Constructing the integral, that is what is important.1247

Okay. So, let us do a final example here. So, this is going to be example 4.1255

We will let our function, f(x,y,z) = some trigonometric function this time... xz × cos(xy). That is going to be the function we integrate.1265

We want to find the integral of f over the region bounded by y = sin(x), y = 2sin(x), x2 + y2 + z2 = 4, and z greater than -- oops, sorry, not greater than or equal to -- and z < or = 0.1280

Okay. So, these are our boundaries. y = sin(x), y = 2sin(x), x2 + y2 + z2 = 4, and z > or = 0.1330

Okay, we are going to have everything below the xy plane, because z is going to be negative, so let us go ahead and deal with the xy plane first. Let us just see what this looks like.1342

So, we have y = sin(x), so that is going to equal that graph right there... y = 2sin(x), that is going to be that graph right there -- sorry, my hand is a little shaky right there but you know what the graph looks like.1357

It looks like x is going to run from, so x is going to run from 0 to pi, from here to here, because we are looking at this region in the xy plane.1375

z is less than 0, so it is going to be everything below this plane and x2 + y2 + z2 = 4... that is the sphere of radius 2.1390

that is what you have got. Now, let me draw this in 3-dimensions as well as I can. It is kind of difficult.1403

Again, you do not necessarily have to see it. If you have all of this information, if you have the kind of mind that can sort of pick out what is going where, you do not need to draw it out.1410

Remember, this is -- oops, crazy lines again... I love those things -- so this is x, this is y, this is z, so y = x so we have that is sin(x), that is sin(2x), so this is the region... then of course we have the...everything below that.1420

So we have the sphere -- it is a little weird, I know... you know what, I am not doing a very good job of drawing it but you know what is happening.1450

It is basically this region, everything below the xy plane touching the sphere x2 + y2 + z2 = 4.1471

Okay. Now let us go ahead and recall what is going where.1482

x, we said is moving from 0 all the way to pi. Well, y is moving from sin(x) to sin(x), so sin(x) to 2sin(x), that is lower and upper, and z... well z is going from -sqrt(4) - x2 - y2 up to 0.1488

The lower limit of integration is the actual sphere itself, and it is going up to 0, because it is below the xy plane.1527

So, this is what is important. Now our integral is going to equal the integral from 0 to pi, the integral from -- oops, not 0 -- the integral from sin(x) to sin(2x), and the integral from -sqrt(4) - x2 - y2 up to 0.1537

Of course we have our function which is xz cos(xy), and again this is x, this is y, this is z... so we are working out... so we are going to do dz dy and dx.1571

When we put this in our software we get the following number. 13.657. Again, Perfectly valid number.1590

So, when you are doing these integrals, this is what you want. We want to be able to construct the integral.1601

We want to be able to integrate along x, you are going to integrate along y, then you are going to integrate along z... x, y, z, three things.1606

You need to be able to choose these things. That is what is important. The rest is just integration.1619

Okay. So, that is our introduction to triple integrals. In the next couple of lessons, we are going to be talking about cylindrical coordinates and spherical coordinates, and triple integrals in those particular coordinate systems.1627

Thank you so much for joining us here at educator.com, we will see you next time. Bye-bye.1641

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