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

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

### Double Integrals

- To set up the integral pay close attendtion to the order of the differentials, dy and dx.
- Since dy is prior to dx we set up the integral so that we first cover the intervals for y and then those for x. That is, the intervals of y are a function of x.
- Our shaded region is the area between the curves y = x
^{2}and y = 1. Hence x^{2}≤ y ≤ 1. Note that this area runs from - 1 ≤ x ≤ 1.

_{ − 1}

^{1}∫

_{x2}

^{1}f dydx. Note the order of integration corresponds to the opposite order of the differentials.

- To set up the integral pay close attendtion to the order of the differentials, dy and dx.
- Since dx is prior to dy we set up the integral so that we first cover the intervals for x and then those for y. That is, the intervals of x are a function of y.
- Our shaded region is the left side of the circle x
^{2}+ y^{2}= 1. Solving for x yields x = ±√{1 − y^{2}} , we only want the left side, so we take the negative result. - Hence − √{1 − y
^{2}} ≤ x ≤ 0 and this area runs from − 1 ≤ y ≤ 1.

_{ − 1}

^{1}∫

_{ − √{1 − y2}}

^{0}f dxdy. Note the order of integration corresponds to the opposite order of the differentials.

_{ − 2}

^{1}5xy dy.

- Taking the integral in respect to y takes any other variable as a constant.

_{ − 2}

^{1}5xy dy = [5/2]xy

^{2}|

_{ − 2}

^{1}= [5/2]x(1)

^{2}− [5/2]x( − 2)

^{2}= [5/2]x − 10x = − [15/2]x.

_{1}

^{3}5xy dx.

- Taking the integral in respect to x takes any other variable as a constant.

_{1}

^{3}5xy dx = [5/2]x

^{2}y |

_{1}

^{3}= [5/2](3)

^{2}y − [5/2](1)

^{2}y = [45/2]y − [5/2]y = 20x

_{1}

^{3}∫

_{ − 2}

^{1}5xy dydx.

- We take the integral in respect to the order of the differentials.

_{1}

^{3}∫

_{ − 2}

^{1}5xy dydx = ∫

_{1}

^{3}− [15/2]x dx = − [15/4]x

^{2}|

_{1}

^{3}= − [15/4]( 9 − 1 ) = − 30

_{ − 2}

^{1}∫

_{1}

^{3}5xy dxdy.

- We take the integral in respect to the order of the differentials.

_{ − 2}

^{1}∫

_{1}

^{3}5xy dxdy = ∫

_{ − 2}

^{1}20x dy = 10x

^{2}|

_{ − 2}

^{1}= 10( 1 − 4 ) = − 30. Note that ∫

_{1}

^{3}∫

_{ − 2}

^{1}5xy dydx = ∫

_{ − 2}

^{1}∫

_{1}

^{3}5xy dxdy.

_{ − 2}

^{0}cos(x) dy.

- Taking the integral in respect to y takes any other variable as a constant.

_{ − 2}

^{0}cos(x) dy = |ycos(x) |

_{ − 2}

^{0}= (0)(cos(x)) − ( − 2)cos(x) = 2cos(x)

_{ − π\protect\mathord/ \protect phantom π2/ 2}

^{0}e

^{y}dx.

- Taking the integral in respect to x takes any other variable as a constant.

_{ − π\mathord/ \protect phantom π2 2}

^{0}e

^{y}dx = xe

^{y}|

_{ − π\mathord/ \protect phantom π2 2}

^{0}= (0)e

^{y}− ( − [(π)/2] )e

^{y}= [(πe

^{y})/2]

_{ − π\mathord/ \protect phantom π2 2}

^{0}∫

_{ − 2}

^{0}e

^{y}cos(x) dydx.

- We take the integral in respect to the order of the differentials.

_{ − π\mathord/ \protect phantom π2 2}

^{0}∫

_{ − 2}

^{0}e

^{y}cos(x) dydx = ∫

_{ − π\mathord/ \protect phantom π2 2}

^{0}( e

^{y}cos(x) |

_{ − 2}

^{0}) dx = ∫

_{ − π\mathord/ \protect phantom π2 2}

^{0}( cos(x) − e

^{ − 2}cos(x) ) dx = sin(x) |

_{ − π\mathord/ \protect phantom π2 2}

^{0}− e

^{ − 2}sin(x) |

_{ − π\mathord/ \protect phantom π2 2}

^{0}= 1 − e

^{ − 2}

_{ − 2}

^{0}∫

_{ − π\mathord/ \protect phantom π2 2}

^{0}e

^{y}cos(x) dxdy.

- We take the integral in respect to the order of the differentials.

_{ − 2}

^{0}∫

_{ − π\mathord/ \protect phantom π2 2}

^{0}e

^{y}cos(x) dxdy = ∫

_{ − 2}

^{0}( e

^{y}sin(x) |

_{ − π\mathord/ \protect phantom π2 2}

^{0}) dy = ∫

_{ − 2}

^{0}e

^{y}dy = e

^{y}|

_{ − 2}

^{0}= 1 − e

^{ − 2}. Note that ∫

_{ − π\mathord/ \protect phantom π2 2}

^{0}∫

_{ − 2}

^{0}e

^{y}cos(x) dydx = ∫

_{ − 2}

^{0}∫

_{ − π\mathord/ \protect phantom π2 2}

^{0}e

^{y}cos(x) dxdy.

^{3}.

- We set up our double integral in respect to the differentials dy and dx. Note that we have dydx.
- So our double integral has the intervals of x followed by those of y. Then

_{0}

^{2}∫

_{ − 1}

^{1}[1/2]xy

^{3}dydx = ∫

_{0}

^{2}( [1/8]xy

^{4}|

_{ − 1}

^{1}) dx = ∫

_{0}

^{2}0dx = 0

- We set up our double integral in respect to the differentials dy and dx. Note that we have dydx.
- So our double integral has the intervals of x followed by those of y. Then

_{ − π}

^{π}∫

_{0}

^{π}cos(x)sin(y) dydx = ∫

_{ − π}

^{π}( − cos(x)cos(y) |

_{0}

^{π}) dx = ∫

_{ − π}

^{π}2cos(x)dx = 2sin(x) |

_{ − π}

^{π}= 0

^{x}.

- We set up our double integral in respect to the differentials dy and dx. Note that we have dydx.
- So our double integral has the intervals of x followed by those of y. Then

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

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

_{0}

^{5}ye

^{x}dydx = ∫

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

^{1 \mathord/ \protect phantom 1 2 2}( [1/2]y

^{2}e

^{x}|

_{0}

^{5}) dx = ∫

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

^{1 \mathord/ \protect phantom 1 2 2}( [25/2]e

^{x})dx = [25/2]e

^{x}|

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

^{1 \mathord/ \protect phantom 1 2 2}= [25/2]( e

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

^{ − 1 \mathord/ \protect phantom 1 2 2})

^{2})] and R be the shaded region described by:

Find dA.

- Note that dA is the area covered by the differentials dx and dy. We shall integrate in respect to that order, that is dxdy.
- Our region is the area between the curve y = x
^{3}and y = x. The intervals of integration for y ∈ [x^{3},x]. - To find the intervals of integration for x, we note that x runs from 0 to 1 along our shaded region and so x ∈ [0,1].

_{0}

^{1}∫

_{x3}

^{x}[y/(x

^{2})]dydx = ∫

_{0}

^{1}( [(y

^{2})/(2x

^{2})] |

_{x3}

^{x}) dx = ∫

_{0}

^{1}( [((x)

^{2})/(2x

^{2})] − [((x

^{3})

^{2})/(2x

^{2})] ) dx = ∫

_{0}

^{1}( [1/2] − [(x

^{4})/2] )dx = ( [1/2]x − [1/10]x

^{5}) |

_{0}

^{1}= [2/5]

Find dA.

- Note that dA is the area covered by the differentials dx and dy. We shall integrate in respect to that order, that is dxdy.
- Our region is the area between the curve y = 0 and y = sin(x). The intervals of integration for y ∈ [0,sin(x)].
- To find the intervals of integration for x, we note that x runs from 0 to p along our shaded region and so x ∈ [0,p].

_{0}

^{p}∫

_{0}

^{sin(x)}kdydx = ∫

_{0}

^{p}( ky |

_{0}

^{sin(x)}) dx = ∫

_{0}

^{p}( ksin(x) − ksin(0) ) dx = ∫

_{0}

^{p}ksin(x)dx = − kcos(x) |

_{0}

^{p}= 2k

^{3}y

^{2}and R be the region bounded by x = 0, y = 1 and y = x.

i) Find dydx. Do not integrate.

- Sketching our region yields
- Note the blue line demonstrating our intervals of integration for y, that is y ∈ [x,1]. We can also see that x ∈ [0,1].

_{0}

^{1}∫

_{x}

^{1}− 2x

^{3}y

^{2}dydx.

^{3}y

^{2}and R be the region bounded by x = 0, y = 1 and y = x.

ii) Find dxdy. Do not integrate.

- Sketching our region yields
- Note the blue line demonstrating our intervals of integration for x, that is x ∈ [0,y]. We can also see that y ∈ [0,1].

_{0}

^{1}∫

_{0}

^{y}− 2x

^{3}y

^{2}dxdy.

^{3}y

^{2}and R be the region bounded by x = 0, y = 1 and y = x.

iii) Verify that ∫

_{0}

^{1}∫

_{x}

^{1}− 2x

^{3}y

^{2}dydx = ∫

_{0}

^{1}∫

_{0}

^{y}− 2x

^{3}y

^{2}dxdy .

- We integrate each side of the equation and see if they are equal.
- First ∫
_{0}^{1}∫_{x}^{1}− 2x^{3}y^{2}dydx = ∫_{0}^{1}( − [2/3]x^{3}y^{3}|_{x}^{1})dx = ∫_{0}^{1}( − [2/3]x^{3}+ [2/3]x^{6}) dx = ( − [1/6]x^{4}+ [2/21]x^{7}) |_{0}^{1}= − [1/14] - Second ∫
_{0}^{1}∫_{0}^{y}− 2x^{3}y^{2}dxdy = ∫_{0}^{1}( − [1/2]x^{4}y^{2}|_{0}^{y})dy = ∫_{0}^{1}( − [1/2]y^{6}) dy = − [1/14]y^{7}|_{0}^{1}= − [1/14]

_{0}

^{1}∫

_{x}

^{1}− 2x

^{3}y

^{2}dydx = ∫

_{0}

^{1}∫

_{0}

^{y}− 2x

^{3}y

^{2}dxdy . Note that the second double integral was simpler to compute.

*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

### Double 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
- Double Integrals 0:52
- Introduction to Double Integrals
- Function with Two Variables
- Example 1: Find the Integral of xy³ over the Region x ϵ[1,2] & y ϵ[4,6]
- Example 2: f(x,y) = x²y & R be the Region Such That x ϵ[2,3] & x² ≤ y ≤ x³
- Example 3: f(x,y) = 4xy over the Region Bounded by y= 0, y= x, and y= -x+3

### Multivariable Calculus

### Transcription: Double Integrals

*Hello and welcome back to educator.com and multivariable calculus.*0000

*Today we are going to start on a new topic. We are going to be talking about double integrals.*0004

*The nice thing about this particular topic is you do not have to learn anything new.*0008

*There is going to be some new notation, but in face the notation itself is not even new. It is exactly the same as what you saw in single variable calculus, just picked up from another dimension -- dimension 1, dimension 2.*0012

*So for all practical purposes, you have done this stuff that we are going to be doing over and over and over again.*0025

*Rather than belabor the point with a lot of theory, I am actually just going to give a couple of quick definitions, and then we are just going to launch right into examples.*0032

*We just want to be able to handle these problems and not worry about what is going on behind the scenes too much.*0041

*You already know a lot of what is going on. Okay. Let us just jump right on in.*0048

*Okay. Now, in single variable calculus what we did was we defined this integral of f(x)/some interval from a to b.*0054

*We had something that looked like this... the integral from a to b of f(x) dx, and this was some number and we learned a bunch of techniques for evaluating this.*0063

*Well, let us go ahead and talk about what some of these things mean.*0077

*This right here is just a value of f at a given x on the particular interval, and this ab is of course the closed interval from a to b, just something like this.*0080

*If there is some function, this is a, this is b, this is the function, so we are just integrating along that length. That is the important part, we are just integrating along a length.*0095

*Now, this dx here it is just a differential length element -- a differential is just a fancy word for small -- so differential length element, and this symbol right here... that is just the symbol for an infinite sum.*0105

*So, what we have done is we have a particular function, we have a bunch of x's in between a and b, we evaluate them at each of those x's, and then we multiply by the actual length, some length element along x and then we take these numbers that we get and we just add them up with this fancy technique called integration.*0137

*Then we get some number. That is the integral, that is all we have done.*0159

*Well, we can do the same thing with a function of two variables, except now, a function of 2 variables is not defined just over the x axis, it is defined over x and y.*0162

*So, instead of a differential length element, what you are going to have is a differential area element.*0173

*Because again, now you are going to have an interval from a to b along x, but you are also going to have an interval c to d along y, so now we are going to basically break up the a to b the way we did before in single variable calculus, but we are also going to break up c to d and what you end up having is a bunch of little rectangles.*0180

*Instead of integrating over a length, we are integrating over a length and another length.*0205

*Well, length × length is area, so we are integrating over the entire area. Everything else is exactly the same, the notation is exactly the same.*0212

*Let us go ahead and write this out. So, for a function of two variables, f(x,y), we can integrate this function over an area, which is just length × length.*0220

*You will see in a minute that is exactly what you are going to be doing. You are just going to be doing one integral at a time, in an iterated fashion, in a row.*0272

*The symbol for it is exactly the same, except you are going to use two integral signs instead of one.*0282

*So, double integral, f(x,y) dy dx.*0288

*Also written as double integral of f, I will leave off the x,y, da.*0299

*dy × dx, well, differential can be x, differential in the y-direction, that gives me a little bit of a square, so this square is a differential area element. That is why we have da for dy/dx, that is it.*0310

*This is the one that we want to concentrate on, but you will see this when you see certain theorems, certain statements along as you understand what is happening.*0327

*This is a symbolic definition, so let us talk about what these mean. This is the value of f at the point (x,y).*0337

*This is a differential area element.*0349

*So, instead of having a very, very, tiny length, like single variable calculus, what we have is a really, really tiny area. One of these small rectangles or squares.*0356

*This, of course, is the symbol for the infinite sum -- the symbol for infinite sum -- and we use two of them to differentiate from the single variable because we are talking about two variables.*0367

*Later, when we do functions of three variables, you are going to see triple integral, three integral signs. It actually exists that you can keep going.*0382

*Okay, that is it. That is literally all that is going on here.*0391

*Let me go ahead and draw out one more time this thing just to make sure we completely understand.*0396

*So, now, because our domain is now in two dimensions, we are going to split up the x length, we are going to split up the y length, and we are just going to add the value of the function over all of these little rectangles, that is what this is.*0405

*Now, I am going to write out how we actually evaluate this. The practical way of finding the integral, and we do it with a single integral at a time.*0423

*So, let us go here. The double integral of f(x,y) dy/dx, is evaluated as -- let me go ahead and write over here -- so from a to b, from c to d, so, our interval along the x-axis is a to b, our interval along the y axis is c to d.*0434

*It is equivalent to taking the integral from a to b, taking the integral from c to d of f(x,y) dy, and then dx.*0478

*What this symbol means is that we are going to be working from the inside out, just like we do in mathematics.*0495

*What we are going to do is we are going back to one variable first, in this particular case, dy, f(x,y) dy, and when we do this we have to remember that y is the variable we are integrating with respect to so we keep x constant.*0500

*When we do this, and we evaluate this integral from c to d, we are going to get a function x.*0515

*Now this function of x is what we integrate from a to b and we get our final answer. That is it, this is called iterated integration and you can actually do the integral in either order.*0519

*You can either do f(x,y) dy dx, or you can do f(x,y) dx dy.*0531

*The problem itself, the particular domain you are going to be working with, will decide for you which you are going to integrate first.*0536

*Again, you are just integrating one variable at a time. let us just go ahead and jump into some examples, and hopefully it will all make sense.*0544

*Again, you have done this over and over and over again. Now instead of just stopping after one integral, you are doing another integral after that. The only thing that you have to watch out for is keeping track of the variables.*0553

*If you are going to be integrating with respect to y, make sure to keep x constant and carry it forward.*0563

*If you are integrating with respect to x, hold y constant, and carry it forward and do the evaluation. That is actually going to be the biggest stumbling block in this. It is not going to be the mathematics itself, it is going to be keeping track of what is going on.*0571

*So, example 1. Find the integral of xy ^{3} over the region x goes from 1 to 2, and y goes from 4 to 6.*0585

*We have this little bit of a rectangle, we have 1, we have 2, let us say 4 is here. Let us say 6 is here, so we have this little rectangle that we are going to be evaluating this function, integrating this function over this domain.*0622

*Well, the integral, I will just call it i. In this particular case I have a choice since I am given the endpoints of the intervals explicitly, I can do dx dy, or I can do dy dx.*0642

*I am going to do dy first and save dx for last. Personal choice.*0657

*So I am going to integrate from 1 to 2. I am going to integrate from 4 to 6 -- we are working inside out -- xy ^{3} dy dx.*0660

*okay. So, I am going to do the first integral. The inside integral first. This is going to equal the integral from 1 to 2, and it is always great to write out everything, do not keep things in your head, do not write things shorthand, write everything out.*0672

*It is not going to hurt if you have a couple of extra lines of mathematics, at least that way if there is a problem, you can follow it -- working your way back.*0686

*Now, when you integrate this, this symbol stays, we are doing this integration. We are integration with respect to y, x stays constant.*0696

*Well, integrate y ^{3}, the integral of y^{3} is y^{4}/4, so it becomes xy^{4}/4 evaluated from 4 to 6 and then dx... so far so good.*0705

*That equals the integral from 1 to 2, well when I put 6 into y and evaluate this, I am going to end up with so 6 ^{4}/4.*0721

*I get the integral of 324x -... and of course I put 4 in for y, take the fourth power, divide by 4, multiply by 6, I am going to get -64x dx.*0736

*That is equal to the integral from 1 to 2 of 260x dx.*0752

*I am going to pull the 260 out because that is just my personal preference. I like to keep the constants outside and multiply them at the end.*0759

*260, integral of x, dx = 260 × x ^{2}/2 evaluated from 1 to 2.*0770

*I just integrated this now with respect to x. I got x ^{2}/2, and when I run through this evaluation putting 2 in, subtracting putting 1 in, multiplying by 260, I end up with 390.*0781

*That is it. Nice and simple. Okay.*0796

*Let us go ahead and give a physical interpretation of what this means and it might help, it might not. In single variable calculus, we had some function and we had the interval from a to b. Well, we interpret the integral physically as the area underneath the graph.*0801

*Okay. We have the same thing in a function of two variables. We said in previous lessons that a function of two variables, since it is defined over a region in the x,y plane, the value of the function can actually be used as a third variable.*0821

*We can graph 3-space. Basically, a function of two variables is a surface in 3-space, so let us go ahead and draw a little -- so this is the x, and this is the y, and this is the z -- so there are some surface above the x,y plane.*0836

*Well, basically, the integral of a function of two variables over a particular region can be interpreted as the volume of everything above that region up to the surface. That is it.*0861

*Again, the area underneath the graph, the volume underneath the graph. We are just moving up one dimension. That is a physical interpretation that is there for you. If you want to think about it that way, that is fine.*0884

*I think it helps to -- you know -- in certain circumstances, but it is important to remember that an integral is a number. It is an algebraic property, not necessarily a physical property. It can be interpreted as such, but that is not what it is.*0894

*Let us go ahead and do another example here. So, example 2.*0908

*This time, f -- let us let f equal to x ^{2}y -- and r be the region such that x runs from 2 to 3 and y is the region that is above, well between, the functions x^{2} and x^{3}.*0918

*Let us go ahead and draw this out. Again, we do not need a precise drawing... that is the x ^{2}. that is the x^{3}, let us say this is 2 and this is 3.*0962

*We have that, we have that, so the region that we are looking at is that region right there. That is the area over which we are going to be integrating this particular function.*0975

*Okay. So, in this particular case, we are constrained by the nature of the problem to do our... to differentiate with respect to y first and then differentiate with respect to x. It is simply easier that way.*0990

*So, let us go ahead and do it.*1005

*So, the integral is equal to the integral from 2 to 3, of the integral, so the lower function is the x ^{2}, the upper in this case is the x^{3}.*1008

*So, we are going to go from x ^{2} to x^{3}, this is perfectly acceptable.*1023

*The function that we are integrating is x ^{2}y dy dx, so we just need to make sure that we keep the order correct.*1027

*Well, this is going to equal the integral 2 to 3. Now when we integrate x ^{2}y with respect to y, we are holding x^{2} constant, so it is going to be y^{2}/2... x^{2} y^{2}/2.*1041

*We are evaluating it from x ^{2} to x^{3}, and then we are going to do the dx.*1055

*This is going to be the integral from 2 to 3 -- oops, let us see if we can eliminate as many of these stray lines as possible -- the integral from 2 to 3, so when we put x ^{3} in for here, we get x into y because we are evaluating with respect to y.*1060

*This is going to be x ^{6} × x^{2}, so we are going to get x^{8}/2 - ... and when we put x^{2} in for here, it is going to be x^{4} × x^{2}, it is going to be x^{6}/2 dx.*1079

*Right? So far so good. Let us go ahead and go to the next page.*1099

*When we do that integration, it is going to equal -- actually, let me write it again so we have it on this page -- 2 to 3 x ^{8}/2 - x^{6}/2 dx = x^{9}/18 - x^{7}/14.*1103

*Right? 6, 7 × 2 is 14, 9 × 2 is 18. We are going to evaluate that from 2 to 3, and then when we do this evaluation we are going to end up somewhere in the neighborhood -- ahh its okay, I am just going to do an approximate... it is going to be somewhere in the neighborhood of about 918 when you actually run that.*1130

*The number itself, I mean it is important, but it is not that important. It is the process that is important. This is where we want to come to.*1153

*Now, let us go ahead and do a third example. Example 3.*1163

*We have f(x,y) is equal to... this time, 4xy is our function, over the region bounded by y = 0, y = x and y = -x + 3.*1172

*Let us go ahead and draw this region out and go ahead and see what it is we are looking at.*1200

*So, y = 0, that is the x axis. y = x, that is this line right here, y = -x + 3, so let us go up 3, and let us go down that way in a 45 degree angle.*1207

*So this is going to hit at 3 and let us go ahead and put a half-way mark here. This is actually going to be 3/2 because these have the same slope.*1224

*In this particular case, our region is right here, this triangle. We are going to be integrating this function over this particular triangle.*1235

*Of course you remember from first variable calculus, sometimes regions have to be split up simply to make the integral a little bit easier to handle.*1242

*In this particular case, I am going to decide to save the integration with respect to x last, the outside integral.*1249

*Since that is the case, I am actually going to split this into two regions. This region r, I am going to split it up into R1 and R2.*1257

*I am going to integrate up to here, and up to there. Then I am going to add those 2 together. The reason that I do that is the upper function from here to here is different. Here it is 0 to x, here it is 0 to -x + 3, which is why I am splitting it up.*1265

*Let me erase this thing. So, in this particular case, our integral over R is going to equal the integral over R1 + the integral over R2 because the integrals are additive, that is exactly right.*1282

*Let us go ahead and deal with the integral of -- I am going to move to the next page -- so, the integral over R1 is going to equal the integral from 0 to 3/2, that is the first half and the y value, let me draw my domain again so I have it here.*1300

*It is going to be that, and that, so the y value is going to go from 0 to x, and my function is 4xy dy dx.*1324

*That is -- excuse me -- equal to the integral from 0 to 3/2 of 2xy ^{2}, I am integrating with respect to y, I am holding that constant, y^{2}/2, the 2 and the 4 cancel leaving me 2xy^{2}, and I am evaluating that from 0 to x dx.*1339

*That is going to equal, well, 2 × the integral from 0 to 3/2... when I put x in for here, it is x ^{2} × x is x^{3}, so it is x^{3} dx.*1359

*This is 0, so it goes away and now that is going to equal 2 × x ^{4}/4 evaluated from 0 to 3/2.*1374

*When I go ahead and do that, I get 81/32, so that is the integral with respect to the first region.*1389

*Now we will do the integral over the second region, this region right here.*1399

*So, the integral over region 2 is equal to -- now we are integrating from 3/2 to 3 -- so, 3/2 to 3, and the y value is going to be 0 to -x + 3.*1404

*Because now the upper function is -x + 3 and again our function is 4xy dy dx.*1423

*Well, that is going to equal 3/2 to 3, it is going to be again 2xy ^{2}, evaluated from 0 to -x + 3.*1432

*When I put this in here, let us see if I have enough room, no I probably do not so let me come down here -- the integral from 0... nope, doing 3/2 to 3 now -- again, keep track of our numbers.*1449

*From 3/2 to 3, it is going to be 2x × -x + 3 ^{2}, because the 0 goes away and I am just putting this into that, dx.*1464

*Ohh... crazy... we do not want these crazy lines all over the place.*1480

*Okay. 0 dx, when I multiply this out, actually I am going ot multiply everything out here so it is fine. We will do 3/2 over 3, it is going to be 2x × x ^{2} - 6x + 9 dx = the integral from 3/2 to 3 of 2x^{3} - 12 x^{2} + 19x dx.*1490

*When I go ahead and evaluate that, I use mathematical software to evaluate that, I ended up with 243 over 32, therefore the integral of f over our original region R is equal to the 81/32, the integral over the first region + 243/32, the integral of the other region = 324/32.*1532

*That is it. Okay. So, in this particular case, I took this region and I divided it into 2 regions and I did 2 separate integrals.*1567

*There is a way to actually do this as one integral by just reversing the order of integration by doing dx first and then doing the dy.*1575

*I will go ahead and show you what that is. I will not do the integration, but I will show you how to approach it.*1584

*So, let me draw the domain again. We had that, and we had this, okay? So this was 3, this was 0, we had 3/2 here, this was 3, this was y = x, or if I wrote it in terms of x, x = y.*1590

*Okay. This graph right here, that is the y = -x + 3. If I write it in terms of x, it is x = 3 - y.*1619

*Now, I am thinking about it this way. I am going to do my final integration with respect to y, which means I am going to go from 0 all the way up to 3.*1630

*In this particular case, there is no interference. The difference -- it says if I turn the graph this way -- the difference, this length right here, okay, that I am going to be integrating this way... this length is the difference between this function and this function.*1644

*In this particular case, I can go ahead and take the difference between this function and this function, and then integrate with respect to y this way, so the integral looks like this.*1662

*The integral from 0 to -- this is not 3, this is actually 3/2 -- 0 to, this value right here is 3/2.*1681

*If you solve simultaneously, this is x is 3/2, this is y is 3/2, so this y value is 3/2.*1692

*So 0 to 3/2, the integral... well, it is going to be... you want to keep this positive, so we are taking this function, this function, this is the lower that is the upper, so let us go 3 - y to y, 4xy, this time we are going to do dx dy.*1699

*That is it. So, you can do this as a single integral as long as you change your perspective and if you are integrating finally along this direction, the first integration that you do, which is going to be this way, well, this difference always stays the same in the sense that it is always going to be a difference between this function and this function.*1726

*That is it. The outer integral covers the entire region, I did not actually have to break this up, but the way that I chose to do it, simply because I like doing x last.*1752

*I had to break this region up into 2, and the reason I broke it up into 2 is because the y value is actually different depending on which function I am working with past this 3/2 mark.*1764

*That is it. That is our introduction to double integrals. Thank you for joining us here at educator.com, we will see you next time. Bye-bye.*1778

1 answer

Last reply by: Professor Hovasapian

Fri Aug 22, 2014 8:27 PM

Post by Denny Yang â™• [Moderator] on August 19, 2014

Also in example 3, you made a minor calculation error. You wrote: The integral of(2x^3 - 12x^2 +19x, x, (3/2), 3 ). It should have been (2x^3 - 12x^2 +18x, x, (3/2), 3 ). The [ x, (3/2), 3 ] is the variable we are integrating and the lower bound and upper bound. Typing it just like you would on a Ti-89 Calculator.

1 answer

Last reply by: Professor Hovasapian

Fri Aug 22, 2014 8:23 PM

Post by Denny Yang â™• [Moderator] on August 19, 2014

Q. ii) Integrate âˆ« âˆ’ Ï€\protect\mathord/ \protect phantom Ï€2 / 20 ey dx.

What is this?

1 answer

Last reply by: Professor Hovasapian

Mon May 13, 2013 12:53 AM

Post by Josh Winfield on May 12, 2013

Hello Raffi,

Can you just clarify why in example 3 you put the limit of integration for dx as (3-y) on the bottom and (y) on the top. I had them the other way around but im wrong but i cant figure out why?

0 answers

Post by Justin Dehorty on February 23, 2013

How do I find the Lower and Upper Sums? Apparently this is another way to calculate double integrals, but I don't understand how to set up the sigmas properly.

1 answer

Last reply by: Professor Hovasapian

Sat Dec 29, 2012 5:01 PM

Post by Alexander Rekitsanov Alexander Rekitsanov on December 15, 2012

I think that the region the prof is integrating is wrong in example 3. in the question it is stating that region that is bounded by y=0 ,y=x and y=-x+3 but the prof ignores the y=0 part. isn't the region suppose to be the left triangle in the drawing ? please if you could clarify it, thank you.