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INSTRUCTORS Raffi Hovasapian John Zhu
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Lecture Comments (7)

2 answers

Last reply by: Professor Hovasapian
Wed May 11, 2016 3:05 AM

Post by Sazzadur Khan on April 28 at 11:36:11 PM

on exammple 1, why are the bounds for the integral 0 to 6?

3 answers

Last reply by: Professor Hovasapian
Wed May 11, 2016 2:59 AM

Post by Acme Wang on April 27 at 04:26:30 AM

Hi Professor,

When you introduce the idea of slicing the solid like a loaf of bread, I am a bit confused about the meaning of dx and dV, can you explain that? Thank you in advance.

Acme

Volumes I: Slices

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Volumes I: Slices 0:18
    • Rotate the Graph of y=√x about the x-axis
    • How can I use Integration to Find the Volume?
    • Slice the Solid Like a Loaf of Bread
    • Volumes Definition
  • Example I: Find the Volume of the Solid Obtained by Rotating the Region Bounded by the Given Functions about the Given Line of Rotation 12:18
  • Example II: Find the Volume of the Solid Obtained by Rotating the Region Bounded by the Given Functions about the Given Line of Rotation 19:05
  • Example III: Find the Volume of the Solid Obtained by Rotating the Region Bounded by the Given Functions about the Given Line of Rotation 25:28

Transcription: Volumes I: Slices

Hello, welcome back to www.educator.com, and welcome back to AP Calculus.0000

Today, we are going to be talking about volumes.0004

We are going to be taking volumes and we are going to be generating these solids from graphs.0007

And then, we are going to find ways of finding volumes using integration -- let us jump right on in.0013

Here is the problem that we are concerned with.0021

Let me work in blue.0023

Our problem is this, rotate the graph of the function of y = √x from 0 to 9.0031

We want you to rotate this graph around the x axis.0054

This rotation, it sweeps out a surface.0066

Our issue is what is the volume of the region that is contained by that surface?0083

Here is what we have going on.0110

We start with our graph of y = √x from 0 to 9.0112

I’m going to get something that looks like that.0116

Let us just say that 9 is here.0119

What we are going to do it now, we are going to take this and we are actually going to spin around the x axis.0127

We are going to spin this, we are going to rotate it.0133

What we are going to end up doing is generating this surface.0137

We are spinning it this way.0145

We have this surface that we are looking at sideways, we are spinning it this way. 0147

Basically, what you have is this cup, imagine a cup that has been turned on its side.0153

This is a side view of it.0158

If I were to draw a little bit in perspective, what you will get is something like this.0161

The axis passes right through the center, you have this cup.0173

It generates this surface.0177

Our goal is how we can use integration to actually find the volume of what is inside that cup, the solid.0183

Imagine, this is just one solid concrete cone, what is the volume of that cone?0190

How can I use integration to find the volume?0198

Volume by definition of any object, something like this, is equal to cross sectional area.0227

It is equal to its cross sectional area × its height or length.0243

In this particular case, this would be the cross sectional area.0250

If I multiply that by the height, I would actually end up getting the volume.0256

You know for example that if I have a cylinder, if I take this area or this area of the circle,0261

and I multiply by the height, I get the volume of the cylinder.0272

This is not a cylinder but I can approximate it by a bunch of little cylinders, by slicing it up like a piece of bread.0275

Finding the volume of that one little cylinder, and then integrating, adding up all the little cylinders.0284

That one, that one, that one, that one, that is what we are going to do.0291

Volumes by slicing, let us go ahead and start that process.0299

We are going to slice the solid like a loaf of bread.0304

Let me go ahead and redraw this thing.0319

I’m going to draw it this way.0321

I have got that, I’m going to go ahead and draw.0323

This is that, I’m just going to take one little slice.0331

One little slice of that is going to look like this.0335

If I take a slice, what I have got is this little slice.0349

This little slice, the height is the radius.0368

The area of this little slice is equal to π × the radius².0376

The thickness of that slice is my differential dx.0383

Therefore, my differential volume element is just equal to π r² × my differential height.0386

This little cylinder of this thickness dx is equal to the cross sectional area × that little bit of thickness.0398

Now I'm going to add up all of these going down.0408

I'm going to integrate that.0413

In this particular case, the area of the circle is equal to π r².0419

We said that r is just the height.0428

The height was √x, r = √x.0431

The area as a function of x is equal to π × √x² which is equal to π x.0437

The volume is equal to the area × the height which is equal to π x dx.0450

I'm going to add up all those little things.0462

Our volume is going to be the integral of π x dx from 0 to 9, because we said that we are interested from 0 to 9.0466

That is going to equal π × the integral from 0 to 9 of x dx.0480

It is going to be π × x²/ 2, 0 to 9 that is going to equal 81 π/ 2.0486

If I were to take the function y = √x from 0 to 9, rotate that around the x axis,0500

the surface that is generated is going to contain a volume, that volume is 81π/ 2.0506

I have done it by, because I'm actually rotating, I'm creating a circle.0513

The area of a circle is π r².0518

π r² × a little differential length element gives me the differential volume element.0520

Now I just add up all the volume elements from whatever to whatever.0525

That is it, let us go ahead and write down the definition here.0530

We will say let f(x) be a continuous function on the interval ab.0545

Let s be the solid of revolution, that is what we call this.0570

What you generate when you take a curve and rotate about a given access is something called a solid of revolution.0580

Be a solid of revolution and generate it by rotating f(x) around a given line.0588

It does not necessarily need to be x axis or the y axis, it could be any line.0610

We just have to see what the function is, where the line is, 0615

to see what solid is being generated around a given line.0618

Let a(x) be the cross sectional area of the solid perpendicular to l, perpendicular to the line l.0624

Then, the volume of our solid = the integral from a to b of a(x) d(x).0657

Since we are rotating about a given line, here is a line, rotation automatically generates a circle.0670

Rotating about a given line, a(x) will be π r², where r the radius is an appropriate function of x.0691

All of this will make sense, when we actually start doing some problems, function of x.0726

Examples will make it clear, let us do some examples.0734

Find the volume of the solid obtained by rotating the region bounded by the given functions about the given line of rotation.0740

I have a function y = 4 - x², I have y = 0, and I have x = 0.0747

It is going to give me a bounded region.0754

I’m going to take that region and I'm going to rotate it about the x axis.0756

What solid do I generate and what is the volume of my solid?0760

Let go us go ahead and do this.0766

Let us go ahead and draw this out.0767

4 - x² - x² + 4.0775

This graph is parabola that passes through 2.0780

I’m not going to go through the details of where it hits the stuff, that is all pre-calculus stuff.0797

Hopefully, you are comfortable with that.0801

y = 0, that is this line, x = 0 is this line, this is our region.0803

Let me go ahead and go to red.0814

Our region is this right here.0815

I'm going to take this region, I’m going to rotate it about the x axis.0819

I’m going to rotate it this way.0822

I’m going to rotate it that way.0825

The solid that I'm going to generate is going to be the solid right here.0829

This is the side view, this is some solid that I have generated.0839

If I were to turn this around and look at it, it would look like a bowl.0845

What is the volume of this region?0854

If I took a cross sectional area which means if I would to actually take, 0856

what you are looking at is something like that, from your perspective.0862

If I were to turn this around and look at it that way, what you would see is a circle.0865

If I turn it back that way, what you would see is this thing that looks like a bowl.0872

That is what we are talking about.0879

Let me make this a little bit smaller because I'm actually going to be using this space.0882

If I were to turn this around, what I would get is something which is circular.0890

If I were to turn it back like this, what I’m going to get is exactly what I'm looking at, 0895

this is the region that we are talking about.0900

We want to find the volume of this solid.0905

We know that the volume =, let us do it over here.0910

Volume = the integral from a to b, we said of the a(x) dx.0916

We are going to be integrating along the x axis.0926

We are going to be integrating from 0 to where it hits there which is 2.0928

It is going to be the integral from 0 to 2.0933

The area that I’m taking is going to be the cross sectional area of a rectangle, like this.0936

In other words, I’m going to be taking a little rectangle like that and I’m going to be adding them all up.0943

When I turn this around, that is this.0949

The circle or radius of that circle, it is going to be π, this area, π r².0956

The radius, if I go at distance x, my radius is 4 - x², it is the height.0965

As I move along x, my y value changes.0977

If I go to some random x value, my radius, that circle right there, if I take this is this, my radius is my function.0983

It is going to be 4 - x²², that is π r² dx.1000

I hope that make sense.1014

π × 4 - x² is my radius of the circle that I generate, when I take this thing and I rotate it this way.1016

My area × my differential element, this thing right here is a little disk.1029

That is r and that is dx.1039

The area is π r² dx.1043

This becomes the integral from 0 to 2, 4 × 4 is 16 – 8x² + x⁴ dx.1047

I integrate, it is 16x - 8/3 x³ + x/5⁵.1070

I evaluate this from 0 to 6 whatever that number happens to be.1083

At this point, I’m going to let you handle the integrations, it is not a problem.1087

What is important is being able to set up this.1091

You need to take a look at your solid, take a look at what a characteristic circle might be from the side.1096

Turn that circle around, realizing that from the axis of rotation up to the function that is your radius, always.1104

From the axis of rotation up to where it meets the function, that length is going to be the radius, 1112

that length, whatever that length is, now matter what the line is.1119

You are always going to be measuring the radius from the line of rotation to where the function is.1122

In this particular case, we just straight put in the function.1129

That would not always be the case, you have to look at just what that length is.1132

In this case, the length is the function, that is r.1136

It is π r² dx, the rest is just basic integration.1139

I hope that make sense.1145

Let us try another one.1146

This time we have y = e ⁺x, we have x = 0, we have y = 4.1148

This time, we want you to rotate this region around y = 4.1153

Let us see what we have got going here.1159

Do I have an extra? Yes, I do.1164

Let me go here and here.1167

I know y = e ⁺x, is it something that looks like this.1173

This is 1, 2, 3, 4.1178

y = e ⁺x, x = 0, that is this line.1188

y = 4, 1, 2, 3, 4, that is that line.1195

This is the line around which we are going to rotate.1202

The region that we are looking for is this region right here.1206

We want to take this region and we want to rotate it around that line.1210

The region that I’m going to generate is that, like the head of a bullet, like a bullet, if you will.1217

I want to know the volume of that region.1227

I want to draw it in slight perspective.1230

I’m looking at a region that looks like this, all solid.1234

What is the volume of that solid?1244

We said that we always measure the radius from the line of rotation.1252

A slice of this thing is here.1275

The radius is measured from a line of rotation.1280

The radius is this length right here.1282

What is that length?1287

This length is this – this.1293

This height is equal to 4, this height is equal to my e ⁺x.1302

This height, I’m going to do this in blue.1313

This height which is 4 - this height which is e ⁺x gives me this distance which is my r.1319

My r is 4 – e ⁺x, I hope that make sense.1325

Now I have got my area which is a function of x is just equal to π × 4 – e ⁺x².1336

I have that, that is not a problem.1347

Now my question is, I’m going to integrate this from 0 to that point.1351

What is that point, I need to know what the x value of that point is so that I can actually integrate.1360

I need my upper limit of integration.1365

What is the x value of where y = 4 and y = e ⁺x meet?1376

We set them equal to each other.1396

4 = e ⁺x which implies that x = natlog of 4.1397

Our volume is going to equal the integral we said from a to b of a ⁺x dx, 1407

which is the integral from 0 to ln of 4 of π × 4 - e ⁺x² dx, 1416

which = π × the integral from 0 to ln 4 of 4 × 4 is 16 – 8e ⁺x + e ⁺2x dx.1427

There you go, which is equal to π × 16x - 8e ⁺x + ½ e ⁺2x evaluated from 0 to ln 4.1446

When you evaluate this which I will leave to you, simple arithmetic,1468

which is always simple but seems always be the issue, at least with me.1473

Notorious for arithmetic mistakes.1479

I have got 16 ln 4 - 24 + 15/2.1481

Again, the important part is being able to set up that integral.1487

That is what is important, that is what we want to know.1491

Volume, in this case was equal to the integral from 0 to ln of 4 of π × 4 – e ⁺x² dx.1494

From here, it is just a question of integration or pure calculate on your software, 1506

whatever it is, being able to set this up from the description given.1511

Remember, whenever you are measuring a radius, you are always measuring the length of that radius from the axis of rotation.1516

That is it, I hope that make sense.1525

Let us do this one, y = sin(x) + 2, x is from 0 to 3π/ 2, and y = 0.1530

I want you to rotate this above the x axis.1539

Let us take a look at our region, y = sin x + 2.1544

It is the regular sin x curve moved vertically up 2.1552

It normally starts at 0, I will start at 2.1557

I will go ahead and put π here, I will put 2π here which means this is π/2 and this is 3π/ 2.1563

We have got, this is π/2, this π, and this is 3π/ 2.1570

It touches 0 here and 0 here, it goes up by 2.1580

At π/2, it goes up to 1.1585

Now I will go up to 1 and I will drop down here 1.1587

My function is going to look like this.1592

This region, I'm going to rotate around the x axis.1597

Therefore, I'm going to get a region that looks like this.1601

That is my solid.1610

My solid, I'm going to take a slice of that solid, I’m going to turn it around.1616

This is going to be my circle, this point is the x axis.1624

This height is my function.1632

If I go out x, my height is going to be sin x + 2.1637

In this case, my radius = sin x + 2.1643

0, 3π lower limit, upper limit.1652

I did not have to worry about where, setting functions equal to each other to find the x value.1655

I was given the x values explicitly.1660

In this case, the volume is equal to the integral from 0 to 3π/ 2.1662

Let us make this a little more clear.1670

To 3π/ 2 of the area of x dx which is equal to the integral from 0 to 3π/ 2.1673

π × sin x + 2² because r is this, it is π r² × dx.1686

I get π × the integral from 0 to 3π/ 2 of sin² x + 4 sin x + 4 dx,1695

which I can separate out into π × the integral from 0 to 3π/ 2 of sin² x dx + 4 ×1717

the integral from 0 to 3π/ 2 of sin x + 4 × the integral from 0 to 3π/ 2 of dx, three integrals.1729

At this point, again, it is best to just go ahead and use your calculator, if you can.1744

But if you cannot, for any reason on the AP exam this is the part where you are not allowed to use your calculator 1748

or your teacher is not letting you use the calculator, you are just going to go ahead and solve these integrals.1754

Let us go ahead and deal with these integrals one at a time.1759

The first integral, our first integral was π × the integral from 0 to 3π/ 2 of sin² x dx.1762

This particular type of integral, you have not met yet.1779

It is part of a future unit that we are going to be discussing called techniques of integration.1782

When we deal with certain types of trigonometric integrals, however, I'm going to go ahead and do it for you now.1788

But realize that if you come up with something like this in the problems, just go ahead and use your calculator1794

because the particular techniques for solving this type of integral you have learned yet, but you will very soon.1799

I’m going to rewrite this as π × the integral of 0 to 3π/ 2 sin² x.1805

There is an identity where sin² x is actually equal to ½ 1 - cos 2x.1815

Wherever I see this, I’m just going to put that in.1824

It is going to be π × ½ 1 - cos of 2x dx which is equal to π/2.1827

I’m going to pull the ½ out, 0 to 3π/ 2 dx.1840

I’m going to separate these out, -π/2 × the integral from 0 to 3π/ 2 of cos 2x dx.1847

What I end up with here is π/2 × x from 0 to 3π/ 2 - π/2 × the integral, 1859

this is going to be ½, the integral is going to be ½ sin of 2x evaluated from 0 to 3π/ 2.1876

I will go ahead and let you take care of that, that is going to give you some number, that is the first integral.1894

The second integral, the second and third, you can deal with, that is not a problem.1899

The second integral, let us just go ahead and go through it.1904

We said that that was 4π × the integral from 0 to 3π/ 2 of sin x dx.1907

That is just going to equal 4π × - cos x from 0 to 3π/ 2.1916

That is going to be some other number, I will call this a, I will call this b.1925

That is the second integral, whatever it happens to be.1931

I would not worry about evaluating it.1936

The third integral is also something that you can do.1939

That was 4 × the integral from 0 to 3π/ 2 dx, that is equal to 4x evaluated from 0 to 3π/ 2.1945

This one is easy, this is just 6π.1959

Therefore, our final volume, were just going to add up the ab and the 6π.1964

It is going to be a + b, whatever you got, + 6π.1968

Again, the whole idea is based on finding the solid, taking a slice, turning that slice around.1977

Once you have turned that slice around, find what the radius is, set up.1988

The area is nothing more than π × the radius².1996

It is going to be some function of x.2000

Therefore, the volume is just going to be the integral from a to b of π a(x) dx.2002

The rest is just solving the integral.2015

I'm going to go ahead and actually stop just with those 3 examples.2018

I hope that made sense.2022

The pattern is the same as you see.2024

We break things up into little bits, we analyze that little bit, and then we integrate all of the little bits.2027

That is the whole idea behind the calculus.2033

Calculus in Latin means little stone.2036

If you want to find out what the weight of the big stone is, you break up the big stone into a bunch of little stones, 2039

you add the weight of the little stones, you add them all up, that gives you the weight of the big stones.2046

That is why it is called calculus.2050

Thank you so much for joining us here at www.educator.com.2052

We will see you next time, bye.2054