| Start | We going. Press play.
Where's Bonnie? Bonnie's with Michelle. Put that pressure on me, man c'mon. Somewhere between New Jersey and here. Some people seem to view the first day back from vacation as optional, Michelle. I don't really think that that's right but apparently they do. Just saying. Alright, so remember, you take a little slice of this and you add up all of those slices and that gives you the area. To get the area if you were to add the slices you're gonna start at some spot a, you're gonna finish at some spot b. |
| 0:38 | And the length of each slice is f(x)-g(x)
dx. Ok, and that's how you find the area, you just add all of them up.
So we did this on Wednesday and of course you can watch this in the miracle of video. Not a YouTube channel. Actually, I think there is a YouTube channel. |
| 1:01 | I think we put all these videos somehow in YouTube you can find them.
But you'd have to have like, privileged access. I'm not sure. Alright, um now suppose I want to find what happens if I spin this volume around, this area around the x-axis. I get a volume. In other words If I have some shape like this, this is the slice of the shape. |
| 1:31 | And I spin it around the x-axis.
I am gonna get something like that. In other words, I'm gonna get, you know, a washer. Ok, or a disc. The difference between a disc and a washer is a disc and a washer is a disc doesn't have a hole on the inside and a washer does. So I'm gonna get something that looks like that. If you use your 3 dimensional imagination. I'm a terrible artist. |
| 2:00 | Ok, but basically you take that slice and you spin it around. And why would you do that?
Ok, that will give you a volume. You would have something in the shape of a disc. And then you add up a whole bunch of discs and you'll get the volume. So if you have some region Let's just take a simple region like something like that, and I wanna find out what happens when I spin that around the x-axis. |
| 2:30 | I'm gonna get something that sort of looks like a lemon or a football.
And I wanna find the volume of it. Ok, what I can do is I can slice it. And every cross section will be a circle. Ok, and then I add up all the circles and that'll give me the volume. The circles are gonna have different radii. Ok, it depends on where you slice. If you slice right here you get a big radius and if you slice right here you get a tiny little radius. And then when you add all of those up you get the volume. Remember we add up by using the integral. |
| 3:01 | Ok?
So what does each slice look like? Well each slice looks something like that. Ok? If you take a very thin slice it'll be a circle with a hole on the inside of the circle. So how do you find the area of something like this? Cause that's what the cross section is gonna look like. If you could turn it sideways you would get this circular region. Well |
| 3:38 | Ok, if you call the little radius little r and the big radius big R then the area of this is πR^2-πr^2.
Or if you factor out the π(R^2-r^2). |
| 4:01 | Ok?
So look, that looks suspiciously like this because big R is gonna be the distance from the axis to the top curve and little r is gonna be the distance from the big axis to the bottom curve. So basically, this again except now we're spinning it, right, so we're getting a circle instead of just a line. Ok, we add up all the circles to get the volume. So let's do an example. It's very hard to picture this. |
| 4:30 | And I put pages from my book, you guys saw the announcement yesterday?
So they're up with lots of explanation and pictures and all that stuff. So if I wanna find the volume it's gonna be π big radius squared minus little radius squared from a to b. So for example |
| 5:52 | Ok, r is formed by y=9x-x^2 and the x-axis.
|
| 6:00 | Find the volume that results when r is revolved about the x-axis.
So, it's actually gonna look pretty much like this picture, but I'll redraw it. So what does 9x-x^2 look like? Well y=9x-x^2 is x times 9-x. |
| 6:31 | So that's equal to 0 at 0.
That's equal to 0 at 9. And it's an upside down parabola. Ok? And it's kinda your job to know how to graph something like 9x-x^2. So I'm gonna take that region and I'm gonna spin it around the x-axis. Why would you want that by the way? Because if you wanna make shapes other than the simple sort of spheres and things, basically what happens is you take the cross sections and add them up. |
| 7:07 | If you think about a 3D printer, that's what it does. It lays out all the cross sections and just adds them up.
Ok, so when you wanna make these complex volumes, like you imagine like a Coca Cola bottle, you just have to describe the boundary of the region and then you spin it around and now you have the volume of the region. Ok, if you've ever used [inaudible] in shop you can understand |
| 7:30 | how you have something spinning then you play with it and you'll get a, you can get something really cool.
Ok? So baseball bat, Coca Cola bottle. Pretty much any of these shapes, that's how you do the math behind them. So mechanical engineers, brace yourself. You're gonna have more of this, ok? So what does this look like? Well if you spin it around the axis it's gonna look like that. Exactly the same upside down. |
| 8:00 | And every cross section is a circle.
And you just add up all the cross sections. Ok, now, sometimes this is referred to as discs, sometimes it's referred to as washers. People seem to be confused. They're really the same thing. A disc is just a washer with no hole in it. Ok, so if you wanted to find the volume it's gonna be π and you're gonna do the integral from 0 to 9. |
| 8:33 | Ok, the top curve squared, so that's (9x-x^2)^2
minus the bottom curve. Well the bottom curve is just 0.
Minus 0^2 dx Ok, because the top curve is y=9x-x^2, the bottom curve is the x-axis. |
| 9:03 | Ok, this is the bottom.
This line. Not this dotted line, that's just the image when it goes around. Ok, so the bottom curve, the inside curve is just the x-axis which is x=0. Ok? Kapeesh? Ok, so we can integrate that very easily. |
| 9:32 | FOIL that out, you get 81x^2-18x^3+x^4 dx. Cause you guys can all FOIL.
Maybe some faster than others. Ok. That is 81x^3/3 minus 18x^4/4 |
| 10:00 | plus x^5/5 from 0 to 9.
I would never ask you to plug 9 in. 9^5 is kind of a big number to work out in your head. Ok? I'm sorry? Oh, and then, right well ok. And I have to multiply this by 5. Sorry. Good thing I brought you. Ok? We might just ask you to set these up we might not ask you to actually do the integrals. |
| 10:33 | We might ask you to do the integrals. It depends on our mood.
Ok, these integrals can be algebraically difficult. Of course when you go from 0 to 1 it's not so bad. But 0 to 9, as I said 9^5 is a big number. It's the same size as 3^10. Which is a lot. Ok? So far so good? Does everybody understand what I did? Let's do some more of these. Cause I believe that you get good at them by doing several of them. |
| 11:09 | [inaudible]
Ok.
|
| 11:39 | Ok.
Let's say we were gonna do the region between y=x^2 and y=x. |
| 12:06 | y=x is just a straight line.
We're pretty much just gonna stick with the first quadrant unless we tell you otherwise. Ok, we'll make sure you can understand what the region is. y=x^2 looks like that, so we're looking for that region. |
| 12:32 | Ok, this is the line y=x.
And underneath is the parabola y=x^2. We're gonna spin that around. So, when it goes around you get something that looks like that. So every slice looks like that. |
| 13:00 | And then you're gonna add them up. Kinda looks like the bell of a trumpet, maybe?
It's hard to describe how it looks like. A speaker. Ok? Not a headphone. A speaker. Ok where do they intersect? They obviously intersect at x=0. Anywhere else? Well you just set x^2=x. Right, that's where those points equal. So x^2=x. x^2-x=0. |
| 13:31 | That's x(x-1) equals 0.
They intersect at (0,0) and (1,1). So again, if you wanna find the intersections, which we may give you, we may not give you. If you wanna find where they intersect you have to set the curves equal to each other. And then the volume |
| 14:10 | is π integral from 0 to 1 the top curve squared minus the bottom curve squared.
Yes. |
| 14:32 | How do you determine which one's the top and which one's the bottom? You go to the x-axis.
The first curve you get is gonna be the inner radius. And the second curve you get is gonna be the outer radius. Ok? So top minus bottom. Some of these pictures, as I said, the top and the bottom will switch each other. Ok? So that's π the integral from 0 to 1 x^2-x^4 dx. |
| 15:01 | I would have no trouble expecting you to do this all the way out.
So that's π times x^3/3 -x^5/5 from 0 to 1 Which is π((1/3)-(1/5)). Or 2π/15 cause when you plug in 0 this is just 0. |
| 15:39 | How we doing?
Totally lost. Where are we lost? Ok so, how do you do these? First, you have to draw a picture. Makes your life easier. You wanna find where they intersect and then you're gonna integrate from the left side of the intersection to the right side of the intersection. Ok, so in this case from 0 to 1. |
| 16:02 | You would take the top curve, x, and square it
minus the bottom curve, x^2, and square it.
Times π. So the integral of x^2 is just x^3/3 and the integral of x^4 is x^5/5. You plug in 1, you plug in 0, and you subtract. Let's give everybody one to work on. You're gonna learn how much fun these are. |
| 17:44 | Ok. Why don't you try that?
Nice, straightforward graph. Let's see if you guys can find the volume. |
| 18:01 | You're gonna take the region between y=x^2+4 and y=x+5.
Revolve that around the x-axis. Find the volume. |
| 22:42 | See this is the miracle of video. So we can just fix this, just make it x+6 instead of x+5.
And then it's easy to factor. Sorry. There you go. I mean you could do it with 5, you get 5. |
| 23:05 | 1+√5/2, 1 minus the square root.
|
| 25:22 | We'll get going on this.
First |
| 25:33 | Drew that a little off center.
Ok, that's y=x^2+4 y=x+6 You wanna find where they intersect. You get x^2-x-2=0. |
| 26:03 | And that's (x-2)(x+1).
So this is -1 and this is 2. Ok? Gazuntite. So if you were gonna revolve that then this will go around the axis and you would get a shape roughly like that. |
| 26:33 | As I said, I'm a terrible artist.
Ok? It doesn't really matter what it looks like. You don't have to be able to draw what the 3 dimensional shape looks like. The key is... Yes. Ah, how do you know which one is on top? Well when you graph them you should be able to tell. If you're not sure you pick a number between the intersection points. So 0. And you say which one gives you a bigger value? You plug 0 into x+6, you get 6. |
| 27:01 | You plug into x^2+4, you'll only get 4.
That tells you that the line is above the circle. The parabola. Yes. Well, you have to know what y=x^2+4 looks like. You have to know that's a parabola. Ok. Parabola centered on the y, on the y-axis. Ok? Shifted up 4 from the origin. You should practice drawing these. Ok? So then the top curve is gonna be x+6, the bottom squared curve is gonna be x^2+4. |
| 27:41 | I heard somebody saying something.
Questions? So the top curve squared minus the bottom curve squared |
| 28:03 | dx.
So that's messy but not ridiculously. That's equal to π integral from, not from 0 to 1. Integral from -1 to 2. How do I know I'm going from -1 to 2? I'm going from x=-1 to x=2. That's where I'm integrating. Right? I'm slicing here |
| 28:32 | and I gotta add up those slices.
So (x+6)^2 is x^2+12x+36 minus (x^2+4)^2 is x^4+8x^2+16 dx. Remember when you were in algebra class and you said when am I ever gonna need this stuff? Now. |
| 29:02 | This is when you need this stuff.
This is why copying off your best friend wasn't the best solution. Although it worked just fine in the 8th grade. So if we simplify that you get -x^4 -7x^2+20. Is that what this equals out? Guys, is that right? I don't know. |
| 29:33 | I have +12x, I apologize. It's not 12x^2. See?
That's uh -8x^2+12x+20. Right? No no no no. No no. x^2-7x^2. Thank you. Got it. |
| 30:02 | Oh, you know, you get old.
You're lucky you're wearing your pants. Ok. dx Alright, can we integrate that? Oh, sure. I'm gonna come over here. That's π times (-x^5/5)-(7x^3/3) |
| 30:36 | +6x^2+20x from -1 to 2.
So far so good? Then you gotta plug in 2 and -1 and figure out what that is. It's really annoying so I'm not going to. If I were testing this I would say when you get to this step you can stop. |
| 31:02 | Ok?
When I teach this I don't. When I'm in charge. I'm not in charge but when I'm in charge I think that's good enough. Ok, cause after that the arithmetic gets a little messy. Ok? Although I have faith that you could do it. Cause you guys are in 126. Alright, should we practice another one? Then we're gonna move on to other types of volumes. |
| 31:59 | Ok, this time our region is y=x^3, between y=x^3 and y=√x.
|
| 32:07 | And I'm in the first quadrant.
Just in case there's any confusion. I'm gonna spin that around the x-axis. |
| 32:36 | Ok?
|
| 34:27 | For those of you who are graph impaired |
| 34:33 | y=x^3 looks about like that
and y=√x
looks a lot like that.
|
| 38:19 | Alright, you ready? There's lots of buzz so I assume that means we're ready.
|
| 38:32 | Ok, so that's our region.
So you guys should get comfortable with what these graphs look like. I think I'll take a minute and go over graphing before we go on. So y=x^3 is this graph. And y=√x is that graph. So that's our region. We're gonna spin that around the axis. And each of the cross sections will look like that. |
| 39:02 | So the volume is gonna get me π top curve squared minus the bottom curve
squared dx. Now you just have to find where they intersect.
Well, if you're comfortable with the graphing they obviously intersect at 0 and 1. Or you can set x^3=√x. |
| 39:30 | Square both sides.
And solve it from there. Ok? Once you square both sides it should be fairly easy to solve. You get x^6-x=0. Pull out an x. You get x=0 or 1. So that's (1,1) back to the origin. So you're gonna integrate from 0 to 1. |
| 40:09 | Ok? So that's π integral from 0 to 1 of x-x^6 dx.
So, that is π((x^2/2)-(x^7/7)) from 0 to 1. |
| 40:39 | Which is π((1/2)-(1/7)) minus 0 minus 0.
Which is 5π/14. You guys, I showed you how to do the fractions. Right? I showed you that earlier in the course. |
| 41:00 | Right? If you want to do (1/2)-(1/7) you do 7-2 over 7 times 2.
Ok? So make sure you can handle these fractions. But if you can't, set it up all the way to where you plug in the arithmetic. Ok that way you won't have trouble in the last step. So far so good? Ok so I'm gonna take a detour for a minute. A digression. |
| 41:31 | Just make sure everybody can graph.
So There's families of graphs you wanna make sure you're comfortable with as we do these. This is something I usually do for the first day of 125 but I might as well go over this. So I'm gonna erase this, ok? Everybody set? To erase? Say goodbye. |
| 42:04 | Ok.
y=x Straight line through the origin. 45 degrees. There's no name of this topic. This is just review of basic graphing. But I can call it [inaudible] for a name Alright. y=x^2 |
| 42:31 | is a parabola
through the origin.
Ok? So if I give you y=x^2+1 you just shift it up 1. If I give you y=10-x^2 you turn it upside down and then shift it up 10. Alright so 10-x^2 goes somewhere over there. Ok? y=x^3 |
| 43:00 | looks something like that.
So far so good? y=√x looks like that. |
| 43:37 | Alright, let's see.
We'll do y=1/x. We'll usually only ask about the first quadrant part of the graph but you never know. y=1/x gets interesting cause there's that vertical asymptote in the middle. Right? So if it's y=1/x-2 you'd move it 2 to the right. |
| 44:03 | With y=1/x+2 you'd move it 2 to the left.
Ok? If you had y=1+(1/x) you'd move it up 1. Down 1, left, right, you should be able to do all of those. Ok? So far so good? Some other basic ones. y=sin(x) That's a π, 2π, 1, -1. |
| 44:43 | What else? Cos(x)
That's π so that's π/2.
|
| 45:01 | That's 1.
That's -1, that's 2π. And so on. 2 more. So far so good? So you should review these. Make sure you can do the simple graphs, we're not gonna give you anything too crazy. Can't integrate. That's the other thing. When it's done you have to be able to do the integral, or you just have to set it up. |
| 45:33 | y=e^x
Looks like that. Goes through point (0,1).
|
| 46:09 | y=ln(x)
goes through (1,0) cause it's the inverse of e^x.
Am I forgetting anything? I don't think so. We're not gonna have you graph inverse sine. That kind of stuff. Of course you could if you wanted to but we're not gonna have you do that. |
| 46:34 | Yes.
You wanna know what tangent looks like? Tangent looks a lot like the cubic graph. But sure. If we were feeling really nasty we could give you a tangent graph. |
| 47:03 | That's terrible.
So they kinda look like cubic graphs, but they have vertical asymptotes. Ok? And that's what the tangent of x looks like. Probably enough. Close to the origin. So you have to- if we're gonna do this we have to give you something to do with it. Remember you're squaring and stuff. |
| 47:31 | So we can't give you anything too messy.
Ok, so if we give you y=log(x) and you square it you have to be able to integrate (log(x))^2. Ok, which is not an obvious integral. Ok? So you should make sure you get comfortable with some of these basic graphs. And of course, how can we make this harder? Well, instead of doing y in terms of x, we could have x in terms of y. |
| 48:01 | That's always entertaining.
Suppose I wanted to take the region between x=y^2 and x=2y and revolve that around the y-axis. |
| 48:42 | First, you have to figure out what that looks like.
Now, we know what y=x^2 looks like. So x=y^2 is gonna do the same thing except you're gonna turn it sideways. So it's gonna be a parabola like that. |
| 49:04 | And x=2y if you divide by 2 is the same as the line y=x/2.
So it's just gonna kinda go like that. It's actually a little flatter. So this is our region. But now we're going around the y-axis and not around the x-axis. So |
| 49:31 | that's what our rotated region looks like.
So if we wanna do our washers ok They're gonna be with horizontal slices. So as we saw on Wednesday, it's just the same thing as when you're doing the vertical slices, except everything's in terms of y. And if you wanna find the radius you can take the outside curve minus the inside curve. |
| 50:04 | For the outside is this line x=2y.
And the inside is x=y^2. How do I know that? Well you pick a y and see which one gives you a bigger x. Ok? You imagine you're here, you go out to the line it's farther than going to the curve. So the volume is gonna be π, outside curve squared minus inside curve squared dy. |
| 50:38 | Now we just have to find where they intersect.
And they intersect where y^2=2y. So you subtract 2y from both sides and factor out the y. |
| 51:08 | Ok? And you get y=0
or 2.
And you check, plug 2 in both equations you get 4. So they intersect at (4,2). But when you slice them horizontally, so you wanna go from 0 to 2. You wanna do the y values, not the x values. |
| 51:51 | So that's gonna be the integral from 0 to 2 π(4y^2-y^4)dy.
|
| 52:07 | Which is π((4y^3/3)-(y^5/5))
from 0 to 2. Notice when you plug in 0 you're gonna get 0.
So we just care what happens when you plug in 2. And that's gonna be π((32/3)-(32/5)). |
| 52:34 | Which is 64π/15.
Ok? There's a trick to that. Did we hate that one? We didn't hate that one. Really? I would've hated that if I were you guys. You wanna try one? |
| 53:00 | Let's give you one to try.
|
| 53:37 | Ok, let's take the region between
between x=y^3 and x=y.
|
| 54:00 | In the first quadrant.
And we're gonna go about the y-axis. Ok? |
| 58:20 | How we doing?
Good? We loving this? |
| 58:48 | Alright.
|
| 59:08 | x=y is the same as y=x
If y=x^3 goes this way, then x=y^3 is gonna go that way.
|
| 59:30 | So as long as we're in the first quadrant, we only have to do this region.
Now if we were not just doing the first quadrant we would have to find both intersections. You'd have to do 2 integrals. Ok, you'd have to do the integral of what's here in quadrant 3 and you'd have to do the integral of what's in quadrant 1. But we only care about the part in quadrant 1. That's why I wrote that. Did you all find the intersect at 0 and 1? Ok. |
| 60:16 | So we do π, you're gonna go from 0 to 1
the outside curve squared minus the inside curve squared.
And if you're not sure which one is the outside curve, take a number between 0 and 1, like 1/2, and see which one gives you a bigger value. |
| 60:34 | So when y is 1/2 x is 1/2, x is 1/8.
So since x is 1/2 at x=y that's gonna be the outside curve. So that's gonna go in front. So it's gonna be y^2-y^3 squared. Now what happens if you get it backwards? You'll just get the negative answer. Ok, so if you get negative something, then you just probably did it backwards, and change it from negative to positive. |
| 61:05 | So this is gonna be π integral from 0 to 1
y^2-y^6 dy which is π((y^3/3)-(y^7/7))
from 0 to 1.
Which is 4π/21. Right? |
| 61:31 | Positive.
If you get -4π/21, you had them backwards. Ok? How do we feel about these? Starting to get the hang of it? Now of course, how do we make it harder? Well we don't go by the x-axis and y-axis. You go about a line other than the x-axis or y-axis. I'm not sure if we're covering those. The other is you could slice the other way. Which is what we're gonna do on Wednesday. I have to check with the professor about what else we're gonna do. Yes. |
| 62:07 | You have to come ask me, I can't hear you. Alright.
Explain why the y is on top. When you plug in a value for x, for y between 0 and 1 this number is bigger than this number. So that's why it's on the right, minus the left. Ok. |