For more information, please see full course syllabus of Multivariable Calculus

For more information, please see full course syllabus of Multivariable Calculus

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### Triple Integrals

_{0}

^{1}∫

_{0}

^{2}∫

_{0}

^{3}( 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 ∫
_{0}^{1}∫_{0}^{2}∫_{0}^{3}( x + y + z ) dxdydz = ∫_{0}^{1}∫_{0}^{2}( [1/2]x^{2}+ xy + xz ) |_{0}^{3}dydz = ∫_{0}^{1}∫_{0}^{2}( [9/2] + 3y + 3z ) dydz. Note that here y and z were treated as constants.

_{0}

^{1}∫

_{0}

^{2}( [9/2] + 3y + 3z ) dydz = ∫

_{0}

^{1}( [9/2]y + [3/2]y

^{2}+ 3yz ) |

_{0}

^{2}dz = ∫

_{0}

^{1}( 15 + 6z ) dz = ( 15z + 3z

^{2}) |

_{0}

^{1}= 18

_{0}

^{p \mathord/ \protect phantom p 2 2}∫

_{ − 1}

^{2}∫

_{1}

^{3}x

^{2}y

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

_{0}

^{p \mathord/ \protect phantom p 2 2}∫

_{ − 1}

^{2}( [26/3]y

^{2}cos(z) ) dydz = ∫

_{0}

^{p \mathord/ \protect phantom p 2 2}( [26/9]y

^{3}cos(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

_{ − 1 \mathord/ \protect phantom 1 2 2}

^{1 \mathord/ \protect phantom 1 2 2}∫

_{0}

^{1}∫

_{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}∫_{0}^{1}∫_{0}^{y2}ye^{2z}dzdydx = [1/2]∫_{ − 1 \mathord/ \protect phantom 1 2 2}^{1 \mathord/ \protect phantom 1 2 2}∫_{0}^{1}( ye^{2z}) |_{0}^{y2}dydx = [1/2]∫_{ − 1 \mathord/ \protect phantom 1 2 2}^{1 \mathord/ \protect phantom 1 2 2}∫_{0}^{1}( ye^{2y2}− y ) dydz. Note that here x and y were treated as constants.

_{ − 1 \mathord/ \protect phantom 1 2 2}

^{1 \mathord/ \protect phantom 1 2 2}∫

_{0}

^{1}( ye

^{2y2}− y ) dydz = [1/8]∫

_{ − 1 \mathord/ \protect phantom 1 2 2}

^{1 \mathord/ \protect phantom 1 2 2}( e

^{2y2}− [1/2]y

^{2}) |

_{0}

^{1}dz = [1/8]∫

_{ − 1 \mathord/ \protect phantom 1 2 2}

^{1 \mathord/ \protect phantom 1 2 2}( e

^{2}− [3/2] ) dz = [1/8] ( e

^{2}− [3/2] )( x ) |

_{ − 1 \mathord/ \protect phantom 1 2 2}

^{1 \mathord/ \protect phantom 1 2 2}= [(e

^{2})/8] − [3/16]

_{0}

^{1}∫

_{ − 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 ∫
_{0}^{1}∫_{ − z}^{z}∫_{y2}^{3y3}z dxdydz = ∫_{0}^{1}∫_{ − z}^{z}x |_{y2}^{3y3}z dydz = ∫_{0}^{1}∫_{ − z}^{z}( 3y^{3}− y^{2})z dydz. Note that here y and z were treated as constants.

_{0}

^{1}∫

_{ − z}

^{z}( 3y

^{3}− y

^{2})z dydz = ∫

_{0}

^{1}( [3/4]y

^{4}− [1/3]y

^{3}) |

_{ − z}

^{z}z dz = ∫

_{0}

^{1}− [2/3]z

^{4}dz = ( − [2/15]z

^{5}) |

_{0}

^{1}= − [2/15]

^{2}+ y

^{2}+ (z − 3)

^{2}= 4. Do not integrate.

- The volume of a region is calculated through = where R is the region.
- Note that x
^{2}+ y^{2}+ (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 x
^{2}+ y^{2}= 4 (blue line). Solving for y yields y = ±√{4 − x^{2}} , 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 x
^{2}+ y^{2}+ (z − 3)^{2}= 4. We solve for z and take the negative root for z = 3 ±√{4 − x^{2}− y^{2}} (bottom half of sphere). - Our intervals of integration are therefore x ∈ [0,2], y ∈ [ 0,√{4 − x
^{2}} ] and z ∈ [ 3 − √{4 − x^{2}− y^{2}} ,3 ]. Note that for z, our values range from the bottom of the sphere and the line of intersection.

_{0}

^{2}∫

_{0}

^{√{4 − x2}}∫

_{3 + √{4 − x2 − y2}}

^{3}dzdydx.

^{2}+ z

^{2}= 9. Do not integrate.

- The volume of a region is calculated through = where R is the region.
- Note that y
^{2}+ z^{2}= 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 y
^{2}+ z^{2}= 9 (blue line). Solving for z yields y = ±√{9 − y^{2}} . - 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 − y
^{2}} ,√{9 − y^{2}} ].

_{ − 2}

^{2}∫

_{ − 3}

^{3}∫

_{ − √{9 − y2}}

^{√{9 − y2}}dzdydx.

- 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 = ∫
_{0}^{3}∫_{0}^{ − [4/3]x + 4}∫_{0}^{ − 4x − 3y + 12}dzdydx . - Integrating in respect to z yields ∫
_{0}^{3}∫_{0}^{ − [4/3]x + 4}∫_{0}^{ − 4x − 3y + 12}dzdydx = ∫_{0}^{3}∫_{0}^{ − [4/3]x + 4}( z ) |_{0}^{ − 4x − 3y + 12}dydx = ∫_{0}^{3}∫_{0}^{ − [4/3]x + 4}( − 4x − 3y + 12 ) dydx.

_{0}

^{3}∫

_{0}

^{ − [4/3]x + 4}( − 4x − 3y + 12 ) dydx = ∫

_{0}

^{3}( − 4xy − [3/2]y

^{2}+ 12y ) |

_{0}

^{ − [4/3]x + 4}dx = ∫

_{0}

^{3}( [8/3]x

^{2}− 16x + 24 ) dx = ( [8/9]x

^{3}− 8x

^{2}+ 24x ) |

_{0}

^{3}= 24

^{2}− y

^{2}. 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 − x
^{2}− y^{2}. - Our intervals of integration are therefore x ∈ [ [1/2],2 ], y ∈ [ [1/x], − x + [5/2] ] and z ∈ [ 0,2 − x
^{2}− y^{2}].

_{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.

^{2}+ y − z

^{2}over the region bounded above the xy - plane, inside x

^{2}+ y

^{2}= 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 x
^{2}+ y^{2}= 4 and x > 1. - From our image, we can see that y can be bounded by the cirlce, that is − √{4 − x
^{2}} ≤ y ≤ √{4 − x^{2}} 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 − x
^{2}} ,√{4 − x^{2}} ] and z ∈ [ 0,5 − x − y ].

- 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

*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.

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

### Multivariable Calculus

### 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 x ^{2} + y^{2} = 3^{2}, 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) - x ^{2}, right?*0390

*If I take x ^{2} + y^{2} = 9, and I move the x over, I get 9-x^{2}, 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 - x ^{2}. Now z.*0428

*Well, z, the equation for the sphere is x ^{2} + y^{2} + z^{2} = 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

*z ^{2} = 9 - x^{2} - y^{2}, so z = + or - 9 - x^{2} - y^{2}.*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) - x ^{2} - y^{2}.*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 - x ^{2}, and I am integrating z from 0 to 9 - x^{2} = y^{2} 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) - x ^{2}, that is y.*0541

*z is going to go from 0 to sqrt(9) - x ^{2} - y^{2}, 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 = xy ^{2}z^{3}, over the region bounded by the xy plane, the function y = x^{2}, the function y = x, and z = x^{2} + y^{2}.*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 = x ^{2}, 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 = x ^{2}, 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 = x ^{2}, 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 x^{2} + y^{2}, so we are going to integrate along x from 0 to 1.*0683

*We are going to integrate along y from x ^{2} to x, and then we are going to integrate up z from 0 to x^{2} + y^{2}, 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 x ^{2} to x, right? the x^{2} 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 x^{2} + y^{2}.*0770

*There you go, so our integral is going to equal 0 to 1 for x, x ^{2} to x for y, 0 to x^{2} + y^{2}... we have x, we have y^{2}, we have z^{3}, 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 x ^{2} to x, and it is going to be from 0 to x^{2} + y^{2}.*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 x ^{2} + y^{2} of just plain old dz.*1009

*That is going to equal z going from 0 to x ^{2} + y^{2}, and my answer is x^{2} + y^{2}.*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 x ^{3} + y^{2} -- oops I will write it as xy^{2}, how is that.*1047

*So, x ^{3} + xy^{2}. 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 x ^{2} to x of x^{3} + xy^{2}, 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 x ^{3}y + x × y^{3}/3, and this time I am going from x^{2} 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 x ^{4} + x^{4}/3 - x^{5} + x^{7}/3, which ends up equaling x^{4} + x^{4}/3 - x^{5} - x^{7}/3. That is my new integrand for my last integral.*1126

*Now I am going to integrate from 0 to 1 of, well, x ^{4} + x^{4}/3 - x^{5} - x^{7}/3 dx.*1166

*I end up with the following: I should get, x ^{5}/5 + x^{5}/15 - x^{6}/6 -x^{8}/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), x ^{2} + y^{2} + z^{2} = 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), x ^{2} + y^{2} + z^{2} = 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 x ^{2} + y^{2} + z^{2} = 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 x ^{2} + y^{2} + z^{2} = 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) - x ^{2} - y^{2} 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) - x ^{2} - y^{2} 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

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 ?