Type I and II regions
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Double integrals of type I and type II regions calculus video example.
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25:20:24 of on-demand video • Updated November 2020
- Partial Derivatives, including higher order partial derivatives, multivariable chain rule and implicit differentiation
- Multiple Integrals, including approximating double and triple integrals, finding volume, and changing the order of integration
- Vectors, including derivatives and integrals of vector functions, arc length and curvature, and line and surface integrals
English [Auto] Today we're talking about Type 1 and Type 2 regions and how to use double integrals to find the area of regions defined as type 1 or Type 2. In this particular problem we've been given the double integral of x squared and this figure here which is an area defined by four separate areas made up of each of these four triangles here and I've labeled the different regions. Someone is up to d of three and ds of four so we can distinguish them from each other. The way that we're going to find the area of this figure is by finding the area of each triangle individually and then adding the area of the four of them together. But before we do that let's talk about type 1 regions and type 2 regions so Type 1 regions you can think of as regions that are easy to define by an upper and lower bound. So for example if we had two parabolas like there's one parabola here and then another parabola that was more shallow maybe like this we could say that this is a type 1 region because it's really easy to see that this shallow parabola is an upper bound and the narrower a parabola is a lower bound. And if we wanted to take slices of this it would be really easy to take vertical slices like this. It wouldn't be as easy to take horizontal slices as you can imagine if we tried to take a slice right here. We'd run into this and then you know we continue to slice over here it doesn't really make as much sense it's not as easy so we'd be wanting to take vertical slices we'd have an upper bound a lower bound. This is a type 1 region. On the other hand if we have a figure like this maybe we have a parabola here that's opening out to the right and then we have a line like so. And these two graphs intersect one another. This region is easier to defined by a right. Most found here the line is the right bound and a leftmost bound where the parabola is a left about and it's easier to take slices of this region that are horizontal like this. If we take slices like this it's real easy to do that because of that this would be a type 2 region and sometimes depending on the kind of region it is it's possible to treat the region as either a type 1 region or a type 2 region. This particular problem that we're dealing with here is interesting because the areas that we have the regions Daesung 1 2 3 and 4 can be treated either as type 1 regions or type 2 regions. I want to show you guys how to treat them as type 1 and Type 2 and how that changes our double integral. So if we wanted to find for example the area of the region we saw one. So this triangle here in the upper right if we treat it as a type 1 region we're going to be integrating our double integral first with respect to y then with respect to x. So that's the other thing about type 1 and 2 regions type 1 regions are associated with an order of integration. First with respect to y then with respect to x type 2 regions are associated with the opposite order of integration we would integrate first with respect to x and then with respect to y. So if I treat this region here dece have one as a type 1 region. Let's go ahead and draw a picture of it that is larger. So this is our region here. Sub 1. And this point here is at 0 1 this point here is that one zero coordinate point 1 0. And this here is the sub one. If I integrate First with respect to why I remember my order of integration is going to be with respect to y first then with respect to x. If I integrate First with respect to y that I'm looking at vertical slices like this like we do here in our type 1 region diagram. Well those vertical slices each one the upper limit of that vertical slice is defined by the top of this triangle here which is at the line y equals 1. So the upper limit of integration here would always be y equals 1. And no matter where I take a vertical slice whether we take it here on the left or all the way over here on the right the upper coordinate point here is always going to be 1. So my upper limit of integration is one my lower limit of integration though is going to change right this point here has a different y value than this point here which has a different y value than this point here. So I can't give a concept for the lower limit of integration with respect to y I have to instead give the equation of this line so that I can say no matter where I take my slice right here. The lower the minimum integration is defined by the value of the function that is this line here. Well the line I've already written the equation on a line it's y equals negative x plus 1. If I'm treating this as a type 1 region then I want this equation defined for y in terms of x my lower limit of integration is then negative x plus one. Then of course my limits of integration with respect to x here are going to be 0 and 1 because the left most value attained in this region for x is at zero. Right right here the leftmost value is X equals zero. The rightmost value is this line here which is at X equals 1 this point here so we have 0 and 1 0 being the leftmost value X can attain and 1 being the rightmost value that x can attain. So those would be right. Limits of integration. If I treat this instead as a type 2 region and I'm taking horizontal slices like this and integrating First with respect to x here and notice that my upper limit of integration the rightmost value here that X obtains is always going to be X equals 1. Just like we said up here that Y was always going to be y equals 1 that was the highest value it attained over here on the right hand side the highest value had ever tains and it's constant it's always going to be one the lower limit of integration again is defined by this slanted line here because I'm defining it in terms of a type to region and then integrating first with respect to x. I would need to solve this equation here y equals negative x plus 1 in terms of x and the way I do that is by subtracting 1 from both sides I'd get negative x equals Y minus 1. I'd multiply both sides by negative 1 and I get X equals 1 minus Y and I want to say that my lower limit of integration here is 1 minus Y which is just the equation this line. But it needs to be in terms of x. So that would be here for dx d y here. The lowest value that Y will ever attain is zero. Right here. So my lower limit of integration with respect to y is zero. The highest value it'll ever attain is this line here which we know is why. Equals 1. So the highest value there is 1. So I can write the area of this region as a type 1 integral or a type 2 integral. Let's write it first as a type 1 integral the integral for type 1 of this particular region D of one will be the integral from 0 to 1 of the integral from negative x plus 1 to 1 of our equation here. X squared that we were given and of course we have d y DX the type 2 integral and we'll come back to these in a second. But I want to show you the contrast of the two. The integral if we treat this as a type 2 region will be the integral from 0 to 1 of the integral from 1 minus y to positive one again of our equation x squared and then D. X d y. So what we notice here is that if we have a type 1 integral then we integrate First with respect to y then with respect to x so the order is d y then dx that means because d y here is on the inside. That means that our inside integral has to have limits of integration with respect to y. And we found that those were negative x plus 1 to 1 so we put that there then because D X is on the outside the integral with respect to x is on the outside and we know that our limit of integration for X and the type 1 region were 0 to 1. And then for Type 2 the opposite case we have the same equation but we reverse the order of integration we have Dayaks then D-y we integrate First with respect X. Then with respect to why. Because DX here is on the inside. We have our limits of integration with respect to x which we found here as one minus y to 1. So we have that here. And then our limits of integration here with respect to y on the outside because D-y is on the outside and we know that those are from 0 to 1. We found those over here. Now it's not always easy or possible to express every region as a type 1 and Type 2 region but in this case because we have a triangle in this way we can express the integral as type 1 or Type 2. So this is remember just the integral for the area d sub one we could use either integral type 1 integral or the type to double integral to evaluate this and we would get the same answer. Let's go ahead and evaluate one of them we'll go ahead and evaluate the type 1 and ignore the type 2 integral. So if we evaluate this type 1 here double integral or a member will evaluate first with respect to y. If we evaluate with respect to y take the integral with respect to y then x squared is just a constant which means we need to add a wise will get here is 0 to 1 of x squared Y. And we'll be evaluating that on the interval. Negative X plus 1 to 1. And remember that this is going to be something we plug in for y. So we'll get y equals negative x plus 1 and y equals 1. I like write the y equals in there so I remember that I plugged these limits of integration in for Y instead of in for x. So now if I plug in my limits of integration I'll get zero to one of x squared times or put in my upper limit of integration first which is 1 minus X squared times my lower limit of integration. Negative X plus 1. And then I've got dx if I simplify here 0 to 1 of x squared negative x squared times and negative x is plus x cubed negative x squared times a positive one is minus x squared. Notice that I'll get my x squared minus X squared those will cancel and I'll just be left with x cubed. So we'll have if I integrate one fourth X to the fourth evaluated on 0 to 1. Because here we're integrating with respect to x we have the DX here. Then when I plug in my limits of integration I'll get one fourth times one to the four minus one fourth times zero. So the for this is obviously going to go away one to the fourth is just one and I'm left with one fourth as a value for my area of the region. Ds. Now if you wanted to you could independently find DS up to three and four and you could treat all of them as type 1 integrals or type 2 integrals. It would work either way. But of course as you can see the sub one is the same area as DSM 2 3 and 4. So for this particular problem all I really need to do is multiply this by 4 to get the area of the entire region and I can say d sub 1 plus D so plus the sub three plus the subform is really just equal to the sub 1 times 4 because they are all equal regions. And I know that that's going to be equal to 1 4 times 4 which is just one that's the area of the entire region.