Raffi Hovasapian

Triple Integrals

Slide Duration:

Table of Contents

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

Adding Vectors

14:51

Example 1

16:52

Properties of Vector Addition

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

Multivariable Calculus Triple Integrals

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Section 8: Triple Integrals: Lecture 1 | 27:24 min

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	1 answer Last reply by: Professor Hovasapian Thu May 1, 2014 9:59 PM Post by Andi Bislimi on May 1, 2014 Why 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 Hovasapian Thu Dec 26, 2013 3:05 PM Post by Atreya Mohile on December 26, 2013 Hi 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 Hovasapian Sat Aug 3, 2013 10:37 PM Post by Waleed Junaidy Waleed Junaidy on August 3, 2013 why are we integrating 1 in the first example?
	2 answers Last reply by: Gregory Gutierrez Thu Nov 3, 2016 3:01 AM Post by Josh Winfield on May 19, 2013 Hey 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 Hovasapian Mon Sep 3, 2012 10:54 PM Post by Mohammed Alhumaidi on September 3, 2012 In 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 ∫₀¹ ∫₀² ∫₀³ ( 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 ∫₀¹ ∫₀² ∫₀³ ( x + y + z ) dxdydz = ∫₀¹ ∫₀² ( [1/2]x² + xy + xz ) |₀³ dydz = ∫₀¹ ∫₀² ( [9/2] + 3y + 3z ) dydz. Note that here y and z were treated as constants.

We proceed by computing the double integral ∫₀¹ ∫₀² ( [9/2] + 3y + 3z ) dydz = ∫₀¹ ( [9/2]y + [3/2]y² + 3yz ) |₀² dz = ∫₀¹ ( 15 + 6z ) dz = ( 15z + 3z² ) |₀¹ = 18

Integrate ∫₀^{p \mathord/ \protect phantom p 2 2} ∫_{− 1}² ∫₁³ 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 ∫₀^{p \mathord/ \protect phantom p 2 2} ∫_{− 1}² ∫₁³ x²y²cos(z) dxdydz = ∫₀^{p \mathord/ \protect phantom p 2 2} ∫_{− 1}² ( [1/3]x³y²cos(z) ) |₁³ dydz = ∫₀^{p \mathord/ \protect phantom p 2 2} ∫_{− 1}² ( [26/3]y²cos(z) ) dydz. Note that here y and z were treated as constants.

We proceed by computing the double integral ∫₀^{p \mathord/ \protect phantom p 2 2} ∫_{− 1}² ( [26/3]y²cos(z) ) dydz = ∫₀^{p \mathord/ \protect phantom p 2 2} ( [26/9]y³cos(z) ) |_{− 1}² dz = ∫₀^{p \mathord/ \protect phantom p 2 2} ( 26cos(z) ) dz = ( 26sin(z) ) |₀^{p \mathord/ \protect phantom p 2 2} = 26

Integrate ∫_{− 1 \mathord/ \protect phantom 1 2 2}^{1 \mathord/ \protect phantom 1 2 2} ∫₀¹ ∫₀^y² ye^2z 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 2}^{1 \mathord/ \protect phantom 1 2 2} ∫₀¹ ∫₀^y² ye^2z dzdydx = [1/2]∫_{− 1 \mathord/ \protect phantom 1 2 2}^{1 \mathord/ \protect phantom 1 2 2} ∫₀¹ ( ye^2z ) |₀^y² dydx = [1/2]∫_{− 1 \mathord/ \protect phantom 1 2 2}^{1 \mathord/ \protect phantom 1 2 2} ∫₀¹ ( ye^2y² − 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 2}^{1 \mathord/ \protect phantom 1 2 2} ∫₀¹ ( ye^2y² − y ) dydz = [1/8]∫_{− 1 \mathord/ \protect phantom 1 2 2}^{1 \mathord/ \protect phantom 1 2 2} ( e^2y² − [1/2]y² ) |₀¹ dz = [1/8]∫_{− 1 \mathord/ \protect phantom 1 2 2}^{1 \mathord/ \protect phantom 1 2 2} ( e² − [3/2] ) dz = [1/8] ( e² − [3/2] )( x ) |_{− 1 \mathord/ \protect phantom 1 2 2}^{1 \mathord/ \protect phantom 1 2 2} = [(e²)/8] − [3/16]

Integrate ∫₀¹ ∫_{− z}^z ∫_y²^3y³ 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 ∫_{− z}^z ∫_y²^3y³ z dxdydz = ∫₀¹ ∫_{− z}^z x |_y²^3y³z dydz = ∫₀¹ ∫_{− z}^z ( 3y³ − y² )z dydz. Note that here y and z were treated as constants.

We proceed by computing the double integral ∫₀¹ ∫_{− z}^z ( 3y³ − y² )z dydz = ∫₀¹ ( [3/4]y⁴ − [1/3]y³ ) |_{− z}^zz 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∫₀² ∫₀^{√{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 = ∫_{− 2}² ∫_{− 3}³ ∫_{− √{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 = ∫₀³ ∫₀^{− [4/3]x + 4} ∫₀^{− 4x − 3y + 12} dzdydx .
Integrating in respect to z yields ∫₀³ ∫₀^{− [4/3]x + 4} ∫₀^{− 4x − 3y + 12} dzdydx = ∫₀³ ∫₀^{− [4/3]x + 4} ( z ) |₀^{− 4x − 3y + 12} dydx = ∫₀³ ∫₀^{− [4/3]x + 4} ( − 4x − 3y + 12 ) dydx.

We proceed by solving the double integral ∫₀³ ∫₀^{− [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 2}² ∫_{1 \mathord/ \protect phantom 1 x x}^{− x + [5/2]} ∫₀^{2 − 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

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Triple Integrals

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Intro

Triple Integrals

Intro 0:00
Triple Integrals 0:21

Example 1
Example 2
Example 3
Example 4

Multivariable Calculus

Section 1: Vectors
	Points & Vectors	28:23
	Scalar Product & Norm	30:25
	More on Vectors & Norms	38:18
	Inequalities & Parametric Lines	33:19
	Planes	29:59
	More on Planes	34:18
Section 2: Differentiation of Vectors
	Maps, Curves & Parameterizations	29:48
	Differentiation of Vectors	39:40
Section 3: Functions of Several Variables
	Functions of Several Variable	29:31
	Partial Derivatives	23:31
	Higher and Mixed Partial Derivatives	30:48
Section 4: Chain Rule and The Gradient
	The Chain Rule	28:03
	Tangent Plane	42:25
	Further Examples with Gradients & Tangents	47:11
	Directional Derivative	41:22
	A Unified View of Derivatives for Mappings	39:41
Section 5: Maxima and Minima
	Maxima & Minima	36:41
	Further Examples with Extrema	32:48
	Lagrange Multipliers	32:32
	More Lagrange Multiplier Examples	27:42
	Lagrange Multipliers, Continued	31:47
Section 6: Line Integrals and Potential Functions
	Line Integrals	36:08
	More on Line Integrals	28:04
	Line Integrals, Part 3	29:30
	Potential Functions	40:19
	Potential Functions, Continued	31:45
	Potential Functions, Conclusion & Summary	28:22
Section 7: Double Integrals
	Double Integrals	29:46
	Polar Coordinates	36:17
	Green's Theorem	38:01
	Divergence & Curl of a Vector Field	37:16
	Divergence & Curl, Continued	33:07
	Final Comments on Divergence & Curl	16:49
Section 8: Triple Integrals
	Triple Integrals	27:24
	Cylindrical & Spherical Coordinates	35:33
Section 9: Surface Integrals and Stokes' Theorem
	Parameterizing Surfaces & Cross Product	41:29
	Tangent Plane & Normal Vector to a Surface	37:06
	Surface Area	32:48
	Surface Integrals	46:52
	Divergence & Curl in 3-Space	23:40
	Divergence Theorem in 3-Space	34:12
	Stokes' Theorem, Part 1	22:01
	Stokes' Theorem, Part 2	20:32

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

Related Books

Multivariable Calculus by Stewart, 7th Edition

Author: James Stewart

ISBN: 538497815

Publisher: Brooks Cole

Year: 2011

The book is filled with clear and accurate descriptions of concepts and is also has many relevant, real-world examples.

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