| Start | Are you ready?
Oh, you're ready? Ok, suppose you wanna take the region between, oh, y=1+e^x and y=4. |
| 0:32 | Ok, so let's look at what that region looks like for a second.
That's e^x and that's 4. Ok, and you wanna go around the x-axis. And we're gonna go just from uhh, x=0 to x=1. |
| 1:03 | Cause the intersection's sort of annoying.
We could find the intersection, I take it back. x=0. So where do they intersect? They intersect at natural log of 3. Because e^ln(3) is 3+1 is 4. Ok, that's not the important part. So now, suppose you wanna revolve this around the x-axis. |
| 1:31 | We're gonna have a problem.
Cause this is a upside down region. Ok, what's the problem? Well, let's do it. We get the volume is π integral from 0 to ln(3) of top curve squared minus bottom curve Sorry, I don't know why I said it's a problem. |
| 2:01 | You don't have a problem, you just have a mess.
Ok. So we can integrate that. You wanna integrate this, you have to multiply out 1+e^x. So (1+e^x)^2 This is 16-(1+2e^x) plus e^x squared which is e^2x. |
| 2:30 | I don't really care what that integral is. It's a very simple integral, right?
Can you guys all integrate that? Let's make sure you can. Ok, so this is π integral from 0 to ln(3) 16-1 is 15-(2e^x)-(e^2x) dx. |
| 3:01 | Everybody's so quiet.
So this is 15x. Integral of e^x is e^x. And integral of e^2x is e^2x/2. Ok? From 0 to ln(3) |
| 3:33 | I don't really care what that comes out to. Ok?
How we doing with this so far? Does everybody know how to set this up? Important thing is to be able to set it up. You should be able to plug in your numbers. But the last step, what these numbers actually equal, isn't that important. Ok, you get π times 15ln(3) minus 2(3) is 6 minus 9/2 |
| 4:04 | minus π(0-2-(1/2)).
Because e^0 is 1. e^2ln(3), e^ln(3)^2 is just 3^2. Ok? Anticipating a question. |
| 4:32 | I suspect if we give you these the numbers will be easy to plug in. That's not the hard part. Ok?
The hard part is the set up and doing whatever integral we give you. But, I don't think we're concerned with giving you really hard integrals. So how can we make this more interesting? I don't think we can make it more interesting. It's fascinating the way it is. But, well we could rotate around something other than the x-axis or y-axis. |
| 5:00 | We ok so far? Anybody have any questions?
We sort of did that one quickly because I viewed it as one you should all be able to do. So again, the top curve is 4. The bottom curve is 1+e^x. When you square out 1+e^x you get 1+2e^x + e^2x. Ok, that simplifies to this. Yes. Well, now it's 1+e^x. There's always one in every class. |
| 5:33 | It's ok.
Your idea is correct, it goes to (0,2) not (0,1). That made a thorough mess out of that. Ok, um You should be able to integrate this. You should be able to integrate [inaudible] The integral is 2e^x. |
| 6:00 | 2e^x
The integral of e^2x is e^2x/2.
I'm gonna go get another battery. I'll be right back. Hope you're all here when I get back. |
| 6:38 | There's a whole cool room back there with all sorts of stuff.
Not really. But it does have batteries. Alright. So integral of e^2x is e^2x/2. I'll leave it to you guys to do the plugging in. Now, what we can ask you to do, instead of going around on the x-axis is we could say go around, or about, y=-2. |
| 7:05 | So what changes when you do that?
I'll redraw the picture. Ok. So what changes? Essentially what happens is now when we take each little slice and we rotate it around, we're going around the line y=-2 |
| 7:39 | rather than y=0, which is the x-axis.
So our washer, the hole in our washer just gets extended. Ok? To find the two distances it's now the distance from -2 to 4 is the top curve |
| 8:03 | and the distance from -2 to e^x+1 is the bottom curve.
So what's happened Other than that nothing changes. Alright, we extended this by 2. So this distance, this 4, now the distance is 6. Or (4+2)^2 and the bottom curve is e^x+1 plus 2. |
| 8:36 | Does that make sense?
Cause all we did really was we dropped the x-axis 2. So each of these distances is increased by 2. So the top curve is now 6^2. The distance to the top is 6^2. And the distance to the bottom is (e^x+3)^2 dx. Otherwise, then it's all the same stuff. |
| 9:00 | So if you do around
another axis like that, what you do is you do exactly what you would do before, and you add
the distance to the new axis.
Ok, sometimes it's gonna be subtract though, so we have to figure some of those out. Everybody understand what I'm doing? So far so good? You're waiting for it to get really nasty. Well, suppose instead |
| 9:34 | take the same region
and I went, I went around this line.
Suppose I went about y=6. I have to redraw that actually, because I'm gonna run out of blackboard. Drop the picture a bit. |
| 10:03 | That's 4 and that's 2. So if I went around y=6, now, I'm gonna revolve this way. I'm gonna revolve up.
So now the region would be up here. When I do a slice, my washer there. Ok? So I'm going, my big radius would be from the line y=6 to the curve e^x+1. It would be this distance. |
| 10:32 | And the small would be this distance. But now we're rotating up
around the line y=6.
Rather than where you've normally been doing it, which is down here. Does that make sense? Some of you are gonna be lost on that. Let's do that again. Ok, you're going around this line. So this region's now rotating up. So the big radius is the distance from 6 to 2. |
| 11:00 | And, I'm sorry, that's the small radius.
The big radius is the distance from 6 to e^x+1. And the small distance is from 6 to 4. Still got the π. Still going from 0 to ln3. Ok? Since we're rotating upwards, remember it's always π big radius squared minus small radius squared. |
| 11:32 | The big radius is the distance from this line, y=6, to the bottom curve now, because this is the wider radius.
And from y from 6 to 4 is the smaller radius. Ok? So here, when I went down this way, the bigger radius is from the line -2 to 4. And the smallest radius is from the line -2 to e^x+1. Ok here, the big radius is from 6 to e^x+1 and the small radius is from 6 to 4. |
| 12:05 | Ok, and then it's big radius squared minus small radius squared.
Let's practice a couple of these. Take a simple region, we'll rotate it, revolve it a few different ways. Ok, let's take the region between |
| 13:10 | Ok, so let's take the region between y=x^2+2 and y=6.
Revolve it three different times. First we'll do it about the x-axis. Then we'll do it about y=-4. And then we'll do about y=+4. |
| 13:35 | So that's really nasty to do about a diagonal line.
That'll happen. Or you could do about a curve. Do it about sin, cos revolving, revolving You don't do that. That's not in this course. |
| 14:00 | Unless [inaudible]
Ok.
So let's draw the region. y=x^2+2 is just a parabola. Goes from 2. And y=6 is a horizontal line. And then we care about from x=0 to x=2. So this is my region. Ok, and this is 2. |
| 14:30 | And this is 6.
And I don't mean y=4. That's dumb. Let's do y=10. 4 would give you a mess cause it would be in the middle. Ok so first, for a the bottom would be π integral from 0 to 2. Well what curve is on top? y=6. So 6 is on top. So 6^2. And what's on the bottom? |
| 15:01 | The bottom is x^2+2 squared.
In other words, when we revolve this about this line we have a cut or a slice. The distance to the top curve is just the big radius, that's 6. And the distance to the x^2+2 is the bottom curve, that's the small radius. If instead I wanted to go around y=-4 |
| 15:35 | well now, the distance to the top curve is 6-(-4).
6-(-4) squared, also known as 10^2. So if you have 6 fingers and then you have, you take away -4 fingers you have 10 fingers. |
| 16:00 | Not quite sure how you take away -4 fingers.
I guess you chop them off of somebody else and you add them to you. Then you end up with 14. But if you were missing 4, sure. Alright. And then the distance from -4 to the bottom curve is x^2+2 minus -4. So that becomes a plus. Ok? So that increases the size of the two radii. |
| 16:41 | What about c?
Well c, you got around y=10. That's this line. So now, the distance to the big radius is from 10 down to the curve. |
| 17:02 | 10-(x^2+2)
squared
And the smaller radius is from 10 to 6.
I know some of you will just be looking for a rule. Ok, and the trick is you have to figure out whether your radius, the radii get bigger or they get smaller, depending on what the axis is. |
| 17:38 | Ok, let me give you guys one to do on your own.
Do we understand this so far? And none of these integrals is particularly hard. They're messy. Cause you have to multiply it out and you have the polynomial and all that stuff. It wouldn't be that hard, it's kind of annoying. Some messy arithmetic. I would just ask you to set them up, but I don't know what we're gonna do yet. |
| 18:01 | We should have the exam done sometime today.
Tomorrow the worst. Cause we gotta xerox a ton of copies. Why is the 10 first? Cause this distance is 10 and I'm subtracting this distance which is the distance from the x-axis to the bottom curve. Ok, I'm taking this whole distance and I'm subtracting this distance. And that leaves me with 10-(x^2+2) and that gives me the large radius. |
| 18:36 | To get the small radius I'm taking 10 and I'm subtracting 6.
And that'll leave me with the small radius. Ok? Yeah? You guys all look a little stunned. Alright, let's make up another one. |
| 19:26 | Sure.
|
| 19:50 | Ok. So we'll do 3 of these.
You've got the equations, the region between y=7-x^2 and y=3. |
| 20:01 | From x=0 to x=2.
And we're gonna revolve about the x-axis, the line y=-5, and the line y=10. Ok? See how that works out for everybody. |
| 24:59 | Alright. Let's just set these up.
|
| 25:02 | I'm not that concerned about the integration right now.
I don't know why this isn't [inaudible] my voice. I do know why. That's better. Alright. Now you'll be able to hear me better. Do you hear me? Ok. I figured if she can hear me you can hear me. Ok. Even if she's not listening. Hi. |
| 25:31 | Alright. The region between 7-x^2 and y=3.
7-x^2 is a downwards facing parabola, and I'm gonna move that because I'm not gonna have room otherwise. Rewrite that just a little bit lower. Ok, that's y=7-x^2. That's y=3. And we're gonna go from x=0 to x=2. |
| 26:02 | So if they just wanna go about the x-axis
then this is my slice.
So to find the slice, you take the distance from the x-axis to the curve is farther away, which is 7-x^2. Subtract the distance to the curve that's closer. Which is y=3. And it's gonna be π integral from 0 to 2 of 7-x^2 squared minus 3^2 dx. |
| 26:38 | As I was saying, I don't know if you'll be expected to actually do the integration or if you're just expected to set it up.
This is not incredibly hard to integrate. It's just 49-14x^2+x^4-9, and I'll leave it to you from there. It quickly runs into messy fractions, that's the problem. |
| 27:02 | So without a calculator you guys are gonna get a little lost in the last round of arithmetic.
So generally we're just gonna ask you to set these up. That's my instinct. Alright, B. Now what happens with B? Well This is our region. Now we're gonna go around the line y=-5. So this slice now goes around that line, something like that. |
| 27:35 | So the distance from -5 to the lower curve is the small radius and the distance from -5 to the upper curve is the big radius.
So the volume would be π integral from 0 to 2 of 7-x^2 minus -5. |
| 28:00 | Ok, so you've extended it by 5.
And the distance to the lower curve is 3-(-5) also known as 8. Ok. So far so good? And I recommend that you write things like 3-(-5). Make sure, so if you screw it up, we can see where you messed up. |
| 28:31 | Ok?
Give yourself, don't do it in your head and then later think that we're gonna spot that when we're grading your exams. We're gonna grade hundreds of these. We're not gonna be in a good mood. No. Well when I say we, I wouldn't be the one grading. That's for the TAs. I have to grade my other course. Alright, what about y=10? Well now we're up here. So when we rotate we're rotating up this way. So |
| 29:09 | The distance from 10 to this line, 3 is gonna be 7.
Because this distance from 10 to the x-axis minus the distance from 3 to the x-axis. π integral from 0 to 2 of (10-3)^2. And then the distance for the lower curve is from 10 to the parabola. |
| 29:35 | Because it's the distance from 10 to the x-axis minus the distance from 10 to the parabola.
Which is 10 minus 7-x^2. Squared. Ok? |
| 30:00 | So that would simplify to π from 0 to 2 7^2 -(3+x^2)^2.
As I keep saying, I'm not really interested in whether you can do the actual integration or not. I mean I am, I expect you to be able to. How'd we do on these? Loving this? This isn't so bad. Better than trig substitutions and partial fractions. |
| 30:32 | And improper integrals.
And biology and chemistry. Ok. Well look on the bright side. I already took those classes. It's not my problem. That wasn't a brighter side? Bright for me. No? It wasn't bright for you guys? I actually never took bio or chemistry in college. Smart move. Alright. Now, we're gonna do a whole different type of rotating. Volume |
| 31:02 | Cross sections. There's no rotations on this one.
Once again it gives me a chance to show off my inferior drawing skills. I assume we're gonna do these. Hold on, are we gonna do these? Oh yeah. Ok. |
| 31:31 | Now, we're gonna do one other type.
So you imagine you have a solid, where the base is 4 but yes yeah Why is it 10-3? Well what's the longer distance? The greater distance is from 10 to 3. And the shorter distance is from 10 to 7-x^2. Ok, cause you're going from 10 down. So 3 is farther away, ok. |
| 32:05 | Now imagine you have one of these regions and you slice it vertically.
And every time you slice, the cross section has a shape. The cross section could have a shape like a circle, like a triangle, like a square, etc. And the size of the square or the triangle is determined by where you slice the cross section. |
| 32:31 | For example
What's a fairly straightforward one?
Sure, that works. Ok, suppose we had the curve y=25-x^2. |
| 33:05 | Oh, this is one I didn't give you yesterday or Monday.
Y is the square root of a^2-x^2 is a semi circle centered at the- that's not a semicircle that's a very bad circle. Ok, y=√(a^2-x^2) is a semicircle. |
| 33:33 | And a is the radius. And it's on the origin.
Ok and so let's say our region is formed by √(25-x^2) and the x-axis. And every time I slice, my cross section is a square. So in other words if I slice here, it's 3 dimensional, and I get a square. |
| 34:03 | I can't really draw it very well.
But you imagine that it's got it's base and what's sticking up is a square. So I slice it and I get this square shape, right? Like that. Ok, you know, piece of paper. So I've got my base and every time I slice I get a square and I slice it. But the problem is, the size of the square depends on where I slice it. So when I slice it here, I'll get a very small square, if I slice it here I get a very big square cause the size of the square is where I cut from the curve to the x-axis. |
| 34:36 | Ok, that's the side of the square.
So this slice is the side of the square. Ok, they draw them very nicely in the book. I don't draw them very well. It's hard. Ok? I don't know if you can see that. So, as I said you have this side, I pick a spot and I slice it and then I form a square from that. |
| 35:02 | So this shape would be, would sorta look like a, what would it look like?
Well kind of like a gyoza, or a dumpling. You know, when you go to Chinese/Japanese restaurant, you've got that dome shape. Ok, if you imagine when you slice it that would be slightly parabolic but Ok, the cross section and the size of the shape, the size of the cross section just depends on where you cut it. If you're right by the edge, it's a very little slice. If you're right by the middle it's a bigger slice. |
| 35:34 | Alright, so how will we find the volume of that?
Well, very simple. Volume is integral of the area. It's kind of untrue. Well that's not true but as far as you're concerned it's true. |
| 36:04 | What's the formula for the area of a square?
Side squared. A side is formed by the top curve minus the bottom curve. So the volume would be, so the top curve is √(25-x^2), and the bottom curve is 0. That's a squared. From, ok so, square root of 25 is 5. |
| 36:35 | So you'd go from -5 to 5.
Ok, and remember the area of a square is side squared. And the side is found by the top curve minus the bottom curve. |
| 37:00 | Ok, so the way this is described is you have a region and perpendicular cross sections, or squares.
That's the way we're gonna write, when I write out the full problem and my arm gets tired from writing all of it out. Ok? And then you just integrate this. This is a very straightforward integral. I love when I say that. -5 to 5 √(25-x^2)^2 is just 25-x^2. |
| 37:33 | Ok, I expect you to be able to integrate that.
Cause when you square the square root, they go away. So far so good? Dizzy? Tired? Loving this? Ok. We'll do more. The base of a solid |
| 38:04 | is the region between |
| 38:38 | No, wait.
Make that 9. The numbers work out better. Ok. |
| 39:11 | Ok. You could say every cross section perpendicular to the x-axis is a square. You could say I have perpendicular cross section at r squares.
It's an English language thing. In other words, every time you cut it you get a shape that's a square. It's very simple. You draw your region |
| 39:34 | Try to make a straight line.
We've got y=x^2+1. And y=9-x^2. Where do they intersect? I'd try 2 and -2 if I was you guys. Let's see. 9-x^2=x^2+1 |
| 40:03 | 8=2x^2
x=+/- 2
This is -2, that's +2.
Ok, so we're gonna integrate from -2 to 2. And the cross sections are squares. The volume |
| 40:37 | from -2 to 2
Ok, and the side of a square would then be formed by this slice.
And then the area of a square is just side squared. The side is the top curve minus the bottom curve. So the area, I mean the side is (9-x^2)-(x^2+1). |
| 41:02 | And then the whole thing is squared.
Don't confuse this with what we were doing before. Where we had pi top squared minus the bottom squared. This is top minus bottom, squared. Ok, they all deal with the same thing. They deal with finding what the slice is. Ok, when I say slice, in other words it's, if I look at this I say what's the top curve and what's the bottom curve and I subtract. Ok, and that gives me the slice. |
| 41:31 | So the area of the square is just the top minus the bottom squared.
[inaudible] and then you can figure that out. You have 8-2x quantity to the x^2 quantity squared. FOIL it out. Quantity 2 is -2. You could go from 0 to 2 and double the answer. Could also do that. You guys look a little stunned. Now, how could I make this more annoying? |
| 42:00 | As if it's not annoying enough. Well, ok.
I can give you three formulas which you should know from your childhood. The area of a square. What's the area of an equilateral triangle? Well the area of a triangle in general is 1/2 base times height. But if I have an equilateral triangle it's also the side squared times √3/4. |
| 42:39 | I'm not gonna prove it.
You can derive it on your own time. What else could we give you? We could give you like a semicircle. We could give you circles. We could give you isosceles right triangles. |
| 43:04 | Equilateral triangle, square, square...
Isosceles right triangles.
|
| 43:30 | An isosceles right triangle is hypotenuse squared over 4.
So in each case, the thing we're plugging in is f(x)-g(x). So, if this, instead of being square was isosceles right triangles, you would literally just write √3/4 in front of the problem. Ok? Because, I'm not gonna erase that. Why would I do that? Let's say the cross sections are equilateral triangles. |
| 44:10 | Ok?
Now, the volume is the same stuff as before, ok, the volume is found by √3/4 times the side squared. So this is the side. The side is found by taking the top curve minus the bottom curve. |
| 44:31 | So from -5 to 5.
That'd be the only difference. If I said it was equilateral triangles. Then I have the √3/4. Because the area of an equilateral triangle is √3/4 times side squared. Ok? You all have this look of despair. If I said they were isosceles right triangles you're looking at the hypotenuse. |
| 45:00 | Ok?
So, sorta like that. Where that's a right triangle. That's your slice. Ok? That's a right triangle. If you can't tell, that's a right angle symbol. Ok, then instead of √3/4, I would just have 1/4. |
| 45:31 | Ok?
Got the idea? Good, let's do one. Yes. Well, the... -2 to 2, sorry. You knew that. Ok? Alright, should we do one as a team? Sure. |
| 46:56 | Ok.
The base of a solid is the region between y=24-x^2 and y=8. |
| 47:06 | And it has perpendicular cross sections that are a)squares, or b) equilateral triangles. Find the volumes.
There's only a tiny difference between a and b. Some of you are already finished. You're not done, you just finish. |
| 50:55 | How are we doing on this?
Good? |
| 51:00 | Not so good.
You good? Alright, let's do this together. I recognize that some of you will just sort of memorize this and do it as a formula. That's fine. I mean, I feel sad because you don't explore the deeper beauty of mathematics, but in the end the goal is to get the question right. |
| 51:34 | Ok, you can explore the deeper beauty after you get the question right.
So we have a solid. 24-x^2, parabola. That's 24. And y=8. Oh, I guess about there. So if the cross sections are squares, every time we slice we get a square shape |
| 52:05 | and the side of the square is formed by our slice. Ok?
So the volume, well we just find where they intersect. Where are these two intersections? Well, 24-x^2=8 x= +/- 4 I hope you all got that. Good. You get something for that. |
| 52:30 | Pat on the head.
There you go. You both got pats, that's pretty good. Alright. Then we just need to figure out what the slice is, cause it's gonna be that slice squared, cause the volume is a side squared. So top curve is 24-x^2. Bottom curve is 8. Squared. That's it. All you had to do. |
| 53:00 | Ok? That's a.
B, volume is -4 to 4 (24-x^2-8)^2, that's the side squared. √3/4 Ok? It is that easy. Good. It is that easy. |
| 53:32 | The whole test will be this question.
Ok, this exact question. Some of you take me literally and you're gonna wonder why the whole test is not exactly this question. That's because I am being, what's the word, sarcastic. Ask your parents. I don't know if we're doing cylindrical shells so I'm nervous about teaching another topic. So I don't wanna end early with the test next week. |
| 54:04 | Are we voting for ending early?
Alright, I'll end 10 minutes early. |