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Current time:0:00Total duration:18:48

Video transcript

if we're just dealing with two dimensions and we want to find the area under curve we have good tools in our toolkit already to do it and I'll just remind us of our tools so let's say that's x-axis that's the y-axis let me draw some arbitrary function right here that's my function f of X and let's say we want to find the area between X is equal to a so that's X equal to a and X is equal to B we saw this many many many videos ago that the way you can think about is you take super small widths of X or super small changes in X we could call them Delta X's but because they're so small we're going to call them a DX we're going to call them a DX super infinitesimally small changes in X and then you multiply them times the value of f of X at that point so you multiply it times the height at that point which is the value of f of X so you get f of X times each of these infinite tails infinitesimally small basis that will give you the area of this infinitesimally narrow rectangle right there and since each of these guys are infinitely small you're going to have an infinite number of these rectangles in order to fill the space you're going to have an infinite number of these right and so the tool we used was the definite integral the definite integral is a sum it's an infinite sum of these infinitely small areas or these infinitely small rectangles and the notations that we use which they would go from A to B and we've done many videos on how do you evaluate these things but I just want to remind you conceptually what this is saying this is conceptually saying let's take a small change in X multiply it times the height at that point so small change in X multiply it times the height at that point and you're going to have an infinite number of these because these X's are super small they're infinitely small so you're going to have infinite number of those so take an infinite sum of all of those from X is equal to a the X is equal to B and that's just our standard definite integral now what I want to do in this video is extend this broaden this a little bit to solve a I guess that maybe you could say a harder or a broader class of problems let's say let's say that we are let's go to three dimensions now and draw the xy-plane first maybe I'll keep this just to kind of make the analogy clear so let me draw it I'm going to kind of flatten this so we have some perspective so let's say that this right here is the y-axis kind of going behind the screen and you can imagine if I just pushed on this and knocked it down so that's the y-axis and that is my x-axis right there that is my x-axis and let's say I have some path in the XY plane so in order to really define a path in the XY plane I'll have two parametric matter eyes I have trouble saying that both the X and y variables so let's say that X let's say that X is equal to let me switch colors I'm using that Orange too much let's say that X is equal to some function of some parameter T and let's say Y is equal to some other function of that same parameter T and let's say we're going to start we're going to have t go from T is going to be greater than or equal to a and then less than or equal to B now this will define a path in the XY plane and if you if this seems confusing you might want to review the videos on parametric equations but essentially you know when T is equal to a when T is equal to a you know you're going to have X is equal to so T is equal to a you're going to have X is equal to G of a and you're going to have Y is equal to H of a so you're going to have this point right here so maybe it might be I don't know I'll just draw an ordinary random point here maybe that's when T is equal to a you're going to have you're going to plot the coordinate point G of a that's going to be our x coordinate this is G of a right here and then our y coordinate is going to be H of a alright you just put T is equal to a in each of these equations and then you get a value for x and y so this coordinate right here would be H of a and then you would keep incrementing T larger and larger until you get to B but you're going to get a series of points that are going to look something like let's say it looks something like that that right there is a curve or it's a path in the XY plane and you know what you're saying how does that relate to that right now you know what what are we doing well let me just write a here for saying that's our curve or that's our path now let's say I have another function that associates every point in the XY plane with some value so let's say I have some function f of XY f of XY what it does is associate every point on the XY plane with some value so let me plot X f of X Y so let me make a vertical axis here we could let me do a different color call it the f of XY axis maybe we can even call it the z axis if you want to but it's some vertical axis right there and for every point so if you give me an X and a Y and you put into my f of XY function it's going to give you some point so let me so I can just draw some type of a surface that f of X Y represents and this will all become a lot more concrete when I do some concrete examples so say let's say that f of XY looks something like this I'm going to try my best to draw it let's do it a different color let's say f of XY is some surface I'll draw a part of it it's some surface that looks let's say it's it looks something like that that is f of XY f of XY and remember all this is is you give me an X and you give me a Y you pop it into f of X Y it's going to give me some third value that we're going to plot in this vertical axis right here I mean examples f of X Y it could be I'm not saying this is a particular case it could be X plus y it could be f of X Y these are just examples it could be x times y if X is 1 Y is 2 f of X Y will be 1 times 2 but let's say when you plot for every point on the on the XY plane when you plot f of X Y you get this surface up here and we want to do something interesting we want to figure out not the area under this curve this was very simple when we did it the first time I want to find the area the area if you imagine a curtain or fence that goes along this this curve that we'd this pet you can imagine this being a very straight linear path going just along the x axis from A to B now we have this kind of crazy curvy path that's going along the XY plane and you could imagine if you drew a wall or curtain or a fence that went straight up from this to my F of XY z-- let me do my effort to draw that so let me draw it so it's going to go up to there and maybe this point corresponds to there and when you draw that curtain up it's going to intersect it something like that let's say it looks something like that so if this point right here corresponds to that point right there so if you imagine you have a curtain you could f of XY is the roof and this is a what I've drawn here this curve this kind of shows you the bottom of a of a of a wall this is some kind of crazy wall and let me say this point it corresponds to well actually let me let me draw a little bit different I want to draw this point will correspond to some point up here so when you when you trace where it intersects it'll look something like maybe like that I don't know something like that and I'm trying my best to help you visualize this so this side you know maybe I'll shade this in to make it a little solid let's say F of XY is a little transparent you can see but that you have this curvy looking wall here and the whole point of this video is how can we figure out the area of this curvy looking wall right this curvy looking wall that's essentially the wall or the fence that happens if you go from this curve and jump up and hit the ceiling at this f of X Y so let's think a little bit about how we can do it well if we just use the analogy of what we did previously we could say well look you know this let's call this you know let's call it let's make a little change in distance of our of our curve let's call that D s D s that's a little change in distance of my curve right there and if I multiply that change in distance of the curve times f of X Y at that point times f of X Y at that point I'm going to get the area I am going to get the area of that little rectangle right there right so if I take D s my change in my you can imagine the arc length of this curve at that point so let me write you know D s is equal to super small super small change change in arc length and arc length of our path or of our curve that's our D s so you can imagine the area of that little rectangle right there along my curvy wall is going to be DS D I'm akin to capital s D s times the height at that point well that's f of XY f of X Y and then if I take the sum because these are infinitely narrow these DS's are infinitely they have infinitely small width if I were to take the infinite sum of all of those guys from T is equal to a to T is equal to B right from t is equal to a I keep taking the sum of those rectangles - T is equal to B right there that will give me my area you know I'm just using the exact same logic as I did up there and I'm not being very mathematically rigorous but I want to give you the intuition of what we're doing we're really kind of just bending the the base of this thing to get a curvy wall instead of a straight direct wall like we had up here but you're saying Sal you know this is all abstract and how can I even calculate something like this this makes no sense to me I have an S here I have an x and a y I have a T what can I do with this and that's let's see if we can make some headway and I promise you when we do it with a tangible problem the end product of this video is going to be a little bit hairy to look at but when we do it with an actual problem it'll actually I think make me very concrete and you'll see it's not too hard to deal with but let's see if we can get all of this in terms of T so first of all let's let's focus just on this D s so let me reap it up the X Y axis so if I were to reflect the X Y let me switch colors it's just getting a little monotonous so if I were to reflux Y axis like that I'm actually doing that same green so you know we're dealing with the same X Y axis so that's my y-axis that is my x-axis and so this path right here if I were to just draw it you know straight up like this it would look something like this it would look something like this right that's my path my my arc you know this is when T is equal to a so this is T is equal to a this is T is equal to B same thing I just kind of picked it back up so we can visualize it and we say that we have some change in arc length let's say this let me switch colors let's say that this one right here let's say that's some small change in arc length and we're calling that D s now is there some way to relate D s to infinitely small changes in X or Y well if we think about it if we really if we and this is all a little bit hand wavy I'm not being mathematically rigorous but I think it'll give you the correct intuition if you imagine this is you can figure out the length of D s if you know the lengths of these super small changes in X and super small changes in Y so if this if this distance right here is DX infinitesimally small change in X this distance right here is dy infinitesimally small change in Y right then we could figure out D s from the Pythagorean theorem you can say that D s D s is going to be it's the hypotenuse of this triangle it's equal to the square root of DX squared plus plus dy plus dy squared so that seems to make to make things a little bit you know we can get rid of the DS all of a sudden so let's rewrite this little expression here using this sense of what dsds is really the square root of DX squared plus dy squared and I'm not being very pretty rigorous and actually it's very hard to be rigorous with differentials but intuitively I think it makes a lot of sense so we can say that this integral the area of this curvy curtain is going to be the integral from T is equal to a to T is equal to B of f of XY instead of writing DX we can write this times the square root the square root of DX squared plus dy squared now we at least got rid of this big capital s but we still haven't solved the problem of how do you solve something you know an integral a definite integral that looks like this we have in terms of T here but we only have it in terms of X's and Y's here so we need to get everything in terms of T well we know x and y are both functions of T so we can actually rewrite it like this we can rewrite it as we can rewrite it as from T is equal to a 2 T is equal to B and f of X Y we can write it f is a function of X which is a function of T and Y is a function f is also a function of Y which is also a function of T so you give me a T I'll be able to give you an X or Y and once you give me an extra Y I can figure out what F is so we have that and then we have this part right here the scope do it in orange square root of DX squared plus dy squared but we still don't have things in terms of T we need a DT someplace here in order to be able to evaluate this integral and we'll see that in the next video when I do a concrete problem but I really want to give you a sense for the end product where the formula we're going to get this at the end product this video where it comes from so what thing we can do is if we if we allow ourselves to algebraically manipulate differentials what we can do is let us multiply and divide by DT so one way to think about it you could rewrite so let me just take a do this orange part right here let's do a little side right here so if you take this orange part and write it in pink and you have DX squared and then you have plus dy squared and let's say we just multiply it times DT over DT all right that's you know a small change in T divided by a small change in T that's one so of course you can multiply it by that if we were to bring this part inside of the square root sign right so let me let me rewrite it this this is the same thing as 1 over DT times the square root of DX squared plus dy squared and then times that DT right I just wanted to write it this way to show you I'm just multiplying by 1 and here I'm just taking this DT writing it there and leaving this over here and now if I wanted to bring this into the square root sign this is the same thing this is equal to and I'll do it very slowly just to make sure I'm not allow you to believe that I'm not doing anything shady with the algebra this is the same thing as the square root of 1 over DT squared let me make the radical a little bit bigger times DX squared plus dy squared and then all of that times DT right I didn't do anything you could just take the square root of this and you get 1 over DT and then if I just distribute this this is equal to this is equal to the square root the square root and we have our DT out at the end of DX squared or we could even write DX over DT squared plus dy over DT squared right DX squared over DT squared is just DX over DT squared same thing with the Y's and now all of a sudden this starts to look pretty interesting let's substitute this expression with this one we said that these are equivalent and I'll switch colors just just for the sake of it so we have the integral from t is equal to a and let me get our drawing back if i from t is equal to a to T is equal to B of f of X of T times or f of X of T and F of or and Y of T they're both functions of T and now instead of this expression we can write the square root the square root of well what's DX what's what's the change in X with respect to time or whatever this parameter is what is DX DT DX DT is the same thing DX DT let me write this down DX DT DX DT is the same thing actually I should write it here is the same thing as G prime of T right X is a function of T is of the function I wrote it is G prime of T and then this dy DT dy DT is the same thing as H prime of T we could say that you know this function of T so I just want to make that clear we know these two functions so we can just take their derivatives with respect to T but I'm just going to rewrite leave it in that form so the square root and we take the derivative of X with respect to T squared plus the derivative of Y with respect to T squared and then all of that times DT and this might look like some strange and convoluted formula but this is actually something that we know how to deal with we've now simplified this strange that you know this arc length problem or this line integral right that's essentially what we're doing we're taking an integral over a curve or over a line as opposed to just an interval on the x-axis we've taken this strange a line integral that's in terms of the arc length of the line and X's and Y's and we've put everything in terms of T and I'm going to show you that in the next video right everything is going to be expressed in terms of T so this just turns into a simple definite integral so hopefully that didn't confuse you too much I think you're going to see in the next video that this right here is actually a very straight forward thing to implement and just to remind you where it all came from I think there's I got the parentheses right this right here was just a change in our arc length that whole thing right there was just a change in arc length and this is just the height this is just the height of our function at that point and we're just summing it doing an infinite sum of infinitely small lengths so this was the change in our arc length times the height this is going to have an infinitely narrow width and you're going to take an infinite number of these rectangles to get the area of the scent of this entire fence or this entire curtain and that's what this definite integral will give us and we'll actually apply it in the next video