15.2: Double Integrals over General Regions

okay so 15.2 covers double integrals over general regions so 15.1 we looked at how to calculate a double integral and what it meant but we looked at rectangular regions so I'm going to give us a picture of what we did in 15:1 and a picture of what we are doing today so here is the idea in 15:1 you had some surface let's call it Z is some function of x and y and if then if you remember we looked at some rectangular region R in the XY plane and what we did is we mapped that region R okay it's kind of off the figure but you get the point we mapped it up to Z and what did the double integral will calculate for us do you remember that it calculated the volume between our surface and the xy-plane okay here's what we are doing today it's still the volume problem we still have some surface Z which is a function of x and y here's the difference which makes things much tougher R is no longer a rectangle our is gonna be some different figure and it's the same general idea you're gonna map our up onto your surface and then you're gonna calculate the volume between your surface and our but as soon as our is not a rectangle things get much trickier okay so first thing we need to know is that there are two types of regions are not including the rectangle the rectangle is one type and then within these general regions there's two types aptly-named type 1 and type 2 okay and again these are regions in the XY plane here's the idea with type 1 or here's what type 1 looks like type 1 has a vertical - vertical boundaries 1 is x equals a and x equals B and then the top two boundaries are functions of X type 2 region is the other way you have two horizontal boundaries at y is equal to two constants y equals C in y equals D your vertical boundaries are then curves that are functions of life the type of our that you have will determine how you set up the volume problem so when we did 15.1 where we had a rectangular R we talked about it doesn't matter if you integrate with respect to X or Y first and then we talked about sometimes it's easier one way or another that's not going to be the case now now it's gonna be specifically you have it's much easier to integrate one versus the other if you have a type 1 region if you're looking at double integral over the region R of Y or f of X Y here's how that's gonna look still the double integral of f of X Y whatever your constants are that's what you integrate last so in this case the upper and lower boundaries for Y those are functions so we want to integrate with respect to the functions first so your lower and upper boundaries for Y are actually going to be functions not constants for a type 2 region it's going to be a similar idea the constants are always what you integrate last so we have y is a constant so why will we will integrate with respect to Y last we will do X first X is those two functions so our inner limits will be functions of Y not constants okay this is gonna involve a little bit of trial and error so for me personally I don't look at each and think okay do I have a type one or a type two I try to set up the limits so that's what we're gonna do we're gonna do a few examples where we set them up we're gonna do a few examples where we actually solve and as you guys do more problems with this you'll get better at it so let's try one we are going to evaluate double integral over the region d that's just are just using a different letter our function is gonna be X plus 2y bless you da and then our region D is going to be the region that is bounded by y equals 2x squared and y equals 1 plus x squared first step is always to draw a picture of your region so I don't mean draw a picture of this function I mean just the two-dimensional region so before I called it R now it's called D that's the first thing I want us to do is draw that region so y equals 2x squared look something like this 1 plus x squared [Music] don't look like this [Music] okay so the region that we are referring to is this region right here okay here's my suggestion to you my suggestion is forget about the type 1 and type 2 I don't think they're very helpful what I do is I look at my region and I look at if I integrate one way what happens if I integrate another way what happens so we're gonna set up our limits we have to figure out do we integrate with respect to Y first or X first so here's where you're going to look at if I integrate with respect to Y first I'm going up and down no matter where I am in the region is the lower bound always the same and is the upper bound always the same function that is yes if I go left to right no matter where I am in the region is the left function always the same and as the right function always the same do you notice if you're down here you go from this function to this function but if you're up here you go from this one to this one that changes we don't want to go left to right first then we want to go up and down first because top always the same function down always the same function ok this bottom function is that y equals 1 plus x squared so that is our bottom limit our upper function is y equals 2x squared those are in terms of why so we are integrating with respect to Y first right because I have them backwards thank you sorry guys thank you for saying that okay that looks better thank you now we need to integrate with respect to X first your outer limits are always constant so leftmost point has an x value of negative one right most point has an x value of positive one this then is our what we're gonna be integrating so we're gonna integrate exactly the same as we did in fifteen point one we're gonna do the inner integral first so you're gonna integrate with respect to Y first so if we integrate with respect to Y we get XY plus y squared we are integrating from y equals to x squared - y equals one plus x squared remember I suggested to you that you write y equals so you know what you're substituting it in for so this then becomes x times the quantity 1 plus x squared plus 1 plus x squared squared minus x times 2x squared minus 2x squared squared so you guys see how I got all of those okay I would like to tell you that there's like some nice way to do this but you have to multiply it everything out and combine all the like terms I'm fairly confident you guys could do that so can I just tell you what it ends up being okay ends up being negative three X to the fourth subtract X cubed add 2x squared add X add one DX now just looks like it normally would okay so now we're gonna integrate again this time with respect to X so we get negative 3/5 X to the fifth minus 1/4 X to the fourth add 2/3 X cubed add 1/2 x squared plus X evaluating from x equals negative 1/2 x equals positive one if you trust me you get 32 over 15 okay at this point I'm your website I don't care if you use a calculator but I will tell you right now you will not going to calculator on your test so I'm sorry I'm not the one that makes that decision for your final Eastern doesn't let you use a calculator so that's what we're gonna do for our tests do we get the general gist of how to evaluate these okay in my opinion the setup is significantly more difficult than solving so I would like to focus more on off setting these up the saw mean you'll figure out next example we're gonna do is just we're just setting up so we are going to evaluate double integral over the region R of XY da R is the region bounded by y equals negative x plus 1 y equals x plus 1 and y equals 3 okay so with the last example we're gonna approach this the same way start by drawing out that region our okay this is the region that I got you guys agree okay it can be very misleading sometimes to look here so if we look here a lot of people are gonna think well I'm going to integrate with respect to Y first because all the functions are in terms of Y sometimes that that's the case sometimes that's not the case gun look here if we go up down first is the top function always the same is the bottom function always the same no if we go left right there's the left function always the same it's the right function always the same okay so in this case we want to integrate with respect to X first we could integrate with respect to Y first that would be fine if we integrated with respect to Y though we would need two integrals we would need one for over here and one for over here does that make sense and sometimes that's going to be the case sometimes you can't avoid that bounds for y are gonna be the easiest so let's do those first because there are constants the bottom value of y is 1 top Y value that we hit is 3 ok the left boundary is that y equals negative x plus 1 has to be in terms of X though so it has to be some function of Y it has to be x equals so if you solve for x you get 1 minus y that is your lower bound then you're right function is that y equals x plus 1 has to be x equals though so again solve for x so then this would be your limits of integration okay how do we feel about this okay am i right in saying that the type 1 and type 2 is not very helpful okay I think you guys need some practice setting these up so the next one I'm gonna give you I'm gonna give you two or three minutes just to set up that work for us okay you are gonna find the volume of the solid that lies under Z equals XY above the region D where D is the region bounded by y equals x minus 1 and y squared equals 2x plus 6 okay I'm gonna start you out but then you guys are gonna find the limits if we want volume that's gonna be the double region double integral rather over the region D of our function which is X Y and then da so you're gonna have to figure out the limits and you're gonna have to figure out if you're integrating with respect to X or Y first we're gonna draw the picture together and then you guys will go from there okay y equals X minus one you guys know what that looks like y squared equals 2x plus 6 do we know what that looks like what shape is it let's start with that parabola opening sideways if you plug in an x value of negative 3 you get Y to be 0 so that's your vertex of your parabola and then it'll open sideways okay so here's what I want to want you to do now I'm gonna give you 2 or 3 minutes talk to someone next to you first thing I want you to decide is are we integrating with respect to X or Y first and then find your limits I want you to do this with someone though don't do it individually so turn around make friends with someone next to you okay do we have a setup with limits almost like another minute and we will okay in a minute we're going to check it go over this up and then you guys are actually okay let's look up here did we decide we wanted to integrate with respect to X first or Y X okay are we all confident on Y its X okay so next thing that you should have done is you should have solved both of these to get them in terms of X did we do that so we switch them both to be x equals the leftmost bound is this parabola if you solve for x that ends up being x equals 1/2 Y squared minus 3 right most bound is this line if you solve for x it ends up being x equals y plus 1 okay then we need to know the lower bound for Y and the upper bound for why did you all it sounded like you guys figured out we needed to find these two intersection points did we do that okay this top one is the point 5 4 and the bottom one is the point negative 1 negative 2 did we all get that do we want to talk about how to find those or work both of these for X so added one over this one subtracted 6 divided by 2 so then we should have gotten from negative 2 to 4 people got that great any questions okay you guys do at least have to go through one of actually calculating these so I'm gonna give you like three four minutes now so now go through I'm actually calculating this volume good luck there should be some squaring of binomials we have any answers yet when you are ready to look up here or if you want to check where you are my work is at the front [Music] hey when you get to here like use your calculator that's fine I will not give you something like this on your test so I will give you something that's much more reasonable to work with but my concern was did we get the gist of this right dropping a negative or a constant that's not a big deal to me but we understand the process of how to go about this okay any questions on this before we go on to the next things we need to talk about okay this is kind of the idea for the rest of the chapter so we're also going to get into triple integrals with the same idea it's really really important that we know how to set up the limits because you've subs limits wrong you're totally okay so moving on next problem this time I'm gonna be kind and I'm gonna give you the limit okay so um this time I'm Thomas Stanton great with respect to Y first and then X what's the issue do we know how to integrate sine of Y squared no okay so another idea from this section is we need to be able to reverse the limits or to switch the order of integration so what we're gonna do is we're gonna integrate with respect to X first and then Y that does not just mean you get to switch those okay so here are the skills that we need to have in order to approach a problem like this first thing that we need to do is we need to be able to draw the region based off of those limits next thing you have to be able to do is reverse the order okay so here's the idea if we look at the inner limits first these are the limits from for Y so we are going from y equals x to y equals 1 okay so that's what we're gonna draw this is our y equals 1 this is our y equals x then we are going from x equals 0 to x equals 1 x equals 0 to x equals 1 so this is the region that we're looking at okay so in this case top and bottom limits are constant left and right limits are constant so it really isn't easier one way or another except that we can't integrate Y squared or sine of Y squared so we are going to integrate with respect to X first and then Y if we look at the limits for Y what is the bottom limit for Y so this is the lowest point for Y what is Y here zero and then it goes to 1 for X so our inner limits we're going left to right the left function is x equals 0 the right function is y okay so then we're gonna do this one integrate with respect to X first so we end up with x times the sine of Y squared we're evaluating from x equals zero to x equals y zero to one if we plug in Y we get Y times the sine of Y squared and then we get zero now this so she should be much easier to integrate so you can you do u substitution if that's what you need to do or you can do it in your head should end up being negative one half cosine of Y squared evaluating from y equals zero to y equals one are you guys okay that I just did it without the so this ends up being negative one half cosine of one minus negative one half cosine of zero negative one half cosine of one plus one okay so on your website I don't know if it specifically tells you you need to reverse the order but if you don't know how to integrate what you're already given if you can't that's one option is to reverse the order okay we got have one last thing to talk about question than this okay great last thing we need to talk about so far we've been using double integrals to calculate volume you can also use double integrals to calculate area so that's the last thing I want to talk about okay well sketch ourselves a little picture let's say that we're at Z equals one and we have some region R that we're mapping up on to that Z equals one we can all agree that the general idea with volume is that you are calculating the area of the base and then you're multiplying it by the height of the figure and that gives you volume in this case the one that I drew what is the height of our figure one because we're at Z equals one so this is the special case where volume is actually equal to area so this is volume using a double and a girl that's an R we want that to be equal to the area of our and that'll be the case when Z is one so if you take the double integral over the region R of 1 times da that gives you the area of our okay surround we've got one last example we are going to use a double integral to find the area of the region enclosed between y equals x cubed and y equals 2x specifically in the first quadrant okay first thing we need to do obviously draw a picture so that's our y equals 2x as our y equals x cubed y equals x cubed though it doesn't go through that point that's just a poor figure this is the area that we are calculating we're gonna do a double integral over the region R of 1 times da we need to figure out if we're integrating with respect to X or Y first okay is the top function always constant bottom function always constant okay left function right function yeah so in this case doesn't really matter I would suggest that you integrate with respect to Y first why would I suggest that yeah both functions are already y equals so I would suggest that you do Y and then X that's just my suggestion our bottom limit for Y is X cubed our top limit for Y is 2x okay and we are going to have to find this point of intersection that one is just x equals zero so if we set that X cubed equal to 2x factor out an X so we get our x equals zero which we already knew and then plus or minus root 2 so at this point X is root 2 so that would be your upper limits questions

Channel: Alexandra Niedden

Views: 28,187

Rating: 4.8888888 out of 5

Keywords:

Id: ss_2Rn6AOU4

Length: 41min 28sec (2488 seconds)

Published: Thu Oct 17 2019