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### Volumes II: Volumes by Washers

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 II: Volumes by Washers 0:11
• Rotating Region Bounded by y=x³ & y=x around the x-axis
• Equation for Volumes by Washer
• Process for Solving Volumes by Washer
• Example I: Find the Volume of the Solid Obtained by Rotating the Region Bounded by the Following Functions around the Given Axis 15:58
• Example II: Find the Volume of the Solid Obtained by Rotating the Region Bounded by the Following Functions around the Given Axis 25:07
• Example III: Find the Volume of the Solid Obtained by Rotating the Region Bounded by the Following Functions around the Given Axis 34:20
• Example IV: Find the Volume of the Solid Obtained by Rotating the Region Bounded by the Following Functions around the Given Axis 44:05

### Transcription: Volumes II: Volumes by Washers

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

Today, we are going to talk about taking volumes of solids by the method of washers.0004

Let us jump right on in.0011

In the last lesson, we introduced this notion of taking a function, rotating that function around a certain axis.0013

Generating a solid, a solid of revolution, and then taking slices.0020

And then, taking the cross sectional area, finding that little differential volume.0024

And then, integrating, adding up all of those volumes.0030

Now let us continue that, let me go ahead and work in blue here.0033

Sometimes when we rotate a region around a given axis, we do not get a complete solid of revolution.0039

In other words, we do not get a full… Like if I take a cup and fill it full of water, that whole cup full of water is a solid.0085

Sometimes there are some hollows, there are things that are not quite complete.0094

Let us take a look at an example, I think it will make this clear.0098

We have the region bounded by the functions y = x³ and y = x.0103

What we are going to do is we are going to rotate this region that is bounded by those two curves around the x axis.0116

Let us see what kind of solid we get.0132

Our question is what is the volume of the resulting solid?0136

Let us go ahead and draw.0153

We have the y = x³ is going to look something like that.0158

The y = x line is going to look something like this.0162

This is our y = x, this is our y = x³, and the region that we are concerned about is this region right here, the one that bounds those.0166

This, we are going to rotate around the x axis.0175

We are going to spin it this way.0179

The thing that we actually end up generating is the following.0181

I’m going to make the axis just a little bit higher.0191

The solid that we ended up generating is going to be the following.0194

It is going to have this one, it is going to be this way.0198

This is a rough drawing.0205

There is this solid, notice this is not a complete solid, this is a hollow inside.0207

What we have done is take this region, spit it out, generate this.0217

If I were to take this thing and turn it a little bit this way, you would see something like that.0223

This is all hollow, this sort of triangular part, it is the other volume.0232

How are we going to do this, how are we going to find this volume?0237

This is how we do it, let us go ahead and draw another picture of it to make it a little bit smaller.0242

Once again, we have our region that looks something like this.0250

I will go ahead and draw this, something like that.0255

Again, we do the same thing, we are still going to slice.0261

Around the axis that we rotate, we are going to take a slice perpendicular to that axis.0266

We are going to take a little bit of a slice, it is going to look like this.0272

The slice is not going to be a single piece.0277

It is going to be a single piece, in a sense that it is one slice.0280

Again, there is this little gap right in between here.0285

I take this slice and I’m going to rotate it out, and see what it is that I look.0288

It is going look like a washer, instead of a solid disk, it is just going to be a disk with a hole in it, what we call a washer.0293

Let us go ahead and go out here.0301

This is going to be an outer disk, it is going to be that one.0309

This right here, it is going to occupy an inner circle.0313

That is what you get when you splice it.0323

When you turn it, you are going to actually get a washer.0325

This is a washer of thickness dx.0335

This is the thickness from here to here.0344

That is the thickness of that washer, that is the dx.0347

What we are going to be doing is we are going to be integrating along the x axis.0349

In other words, we are going to be integrating along the axis of rotation, along the x axis.0354

Exactly the same as before, the only thing that is different now is we want the area of this disk.0363

The area of the disk is the area the outer circle - the area of the inner circle.0368

Because the area that we are looking for is that one.0372

We are going to multiply by its thickness which is dx.0379

That is going to give us a differential volume element.0387

And then, we are going to add all of the volumes, all of the washers along the x axis.0389

Our differential volume element is nothing more than the area of this × dx.0396

The area of this is π × the outside radius² - π × the inside radius².0402

This is the outside radius, this is the inside radius × dx which is the thickness of the washer.0413

That is the volume of that washer, we want to add them all up.0422

Let me we write this, dv = π × just pull out the π, outside radius² - inside radius² dx.0426

When we integrate both sides, the integral of dv is going to be just v.0438

The volume = the integral from a to b of π × the outside radius² - the inside radius² dx.0442

That is the general formula for finding the volume of a solid,0455

that has gotten a solid of revolution generated by rotating a particular region around a given axis.0462

When that particular slice that you took is a washer, it is not quite complete.0468

It is this solid disk but it is a disk with a hole in it.0473

In this particular case, we have this function which is y = x.0477

This function y = x³, and of course, this is the x axis, that is the axis of rotation.0483

What are a and b?0493

The lower and upper limits of integration, they are just going to be just0499

from here we are integrating along the x axis to here.0502

These two graphs, we know that they meet at 1.0508

a is equal to 0 and b is equal to 1.0512

Let us go ahead and fill this in.0516

We have the volume = the integral from 0 to 1 of π × the outside radius.0519

The outside radius is this one, it is from 0 to x, it is the y value.0528

However far we go along the x axis, it is the y value.0537

It is just going to be x, the outside radius is x and we square it.0541

The inside radius, it is just this point right here.0547

We go along x and it goes up to the x³ part.0551

This is going to be x³² dx.0557

That is going to equal π × the integral from 0 to 1 of x² - x⁶ dx.0565

This is really what we want.0575

The most important thing in all of these problems is actually finding the integral, establishing the integral.0577

Solving it is important but it is the secondary concern.0583

We need to take a situation, convert it into an integral.0587

My students, when I teach this course, I generally have them stop here.0591

I do not want them to solve the integral.0596

This is what I'm interested in, I'm interested that they actually can form the integral.0597

The rest is just techniques of integration, or software, if you need.0602

Let us go ahead and solve this though.0608

The volume is equal to π × this is going to be x³/ 3 – x⁷/ 7 from 0 to 1.0610

The final answer is 1/3 - 1/7, that is it.0625

That is all we are doing here.0634

Same exact thing that we did before, except now we are just looking for the cross sectional area0635

which happens to be outer circle - inner circle, because we are dealing with a washer.0640

All of these problems are dealt with the same way, in some form or another.0646

That is all we are doing.0650

We are finding some area, we are multiplying it by a differential length dx.0651

That is our differential volume.0656

We are going to add up all those volumes, that is all it is.0657

We just need to make sure that we see what is happening pictorially.0660

That is going to be the hardest part in all of these problems.0664

What is it look like, how does one visualize this?0666

Let us go ahead and, a little bit further before we start our examples.0672

When you are faced with a region that generate washers upon slicing,0679

the volume is equal to the integral from a to b of π × the outside radius² - the inside radius² dx.0706

If you are going to be integrating along the dy, then this is just outside radius - inside radius dy.0725

The problem itself will tell you whether you are generating along x or generating along y.0733

I will do this in red, this really is the most important part about all of these things.0741

Not just this particular section but all of these, generating volumes and things like that.0746

The most important thing you always measure the radii or any length for that matter from the axis of rotation.0753

Whatever the axis of rotation is, x = 5, y = -1, it is from that point that you are measuring.0781

From there, the distance from that to the particular functions that we are dealing with.0788

This is the generic, once you have a picture, you are going to decide what our outside is, what our inside is.0795

Again, you always measure radii from the axis of rotation.0802

Let me go back to blue.0807

Examples will make this clear.0809

Our general process, number 1, always draw a picture.0822

Two, identify your solid, identify the solid.0833

In other words, once you actually perform the rotation, make sure you have an idea of what the solid looks like.0838

It does not have to be a 100% full graph but you should at least be able to see this side view, what it actually looks like.0845

Three, identify the washer.0854

In other words, take a slice, draw it out, identify the washer.0861

This identification of the washer, whether it is this way or this way,0869

this will decide whether you are integrating with respect to x or y.0875

Instead of saying with respect to, integrating along x or y.0901

The final part, number 4, measure radii from the axis of rotation to the original graph.0910

Once you generate your solid, you are going to have that other half.0940

From the axis of rotation, measure to the original graph, that is going to give you your length.0945

Again, it will make more sense when we start our example problems.0949

With that, let us go ahead and start our example problems here.0953

Find the volume of the solid obtained by rotating the region bounded by the following functions around the given axis.0959

We have the function y = x³, y = x, x greater than or equal to 0.0967

We want you to rotate it around y = 3.0972

Same functions as before, different axis of rotation.0976

Let us draw this out, see what it looks like.0982

Let us go down here.0987

We are rotating this around y = 3.1000

I’m going to go this way, there we go.1003

y = x is that line, y = x³ is going to be that thing.1008

What we are going to do is going to rotate it around this line.1024

This is our y = 3 line.1034

When we rotate that way, now what we are going to do is we are going to get something that looks, it is going to be over here.1036

Let me extend this out, it is not a problem.1046

We are going to get something like that.1050

We are going to rotate that way.1052

What we are interested in is this region right here.1054

That solid, whatever it is, is going to have a volume.1059

We want to find that volume.1063

We draw our region, let me see, 1, process, we drew our region.1069

Two, we identify the solid.1079

Three, we identify the washer.1085

We are going to take a slice of this, we are not going to go this way, nothing there.1093

Let me go to red actually.1099

We are going to slice it this way.1102

We are just going to take a slice that way.1107

When we take this and turn it around, what we are going to end up getting is right over here.1109

Our view is slightly to the side, not fully turned so you see it this way.1121

But just a little bit so you can see what is actually going on.1125

Here is this washer of a given thickness.1130

The region that we are interested in is that region right there.1141

There we go, we know the washer is a vertical washer.1146

We are going to be adding all of the washers this way.1150

We are going to be integrating along the x axis, integrate in the x direction.1152

We will just say integrate along x.1167

This is going to be from 0 to 1 again, because from 0 and 1.1170

From here, we are going to integrate that way.1176

We are going to integrate that way, along the axis.1178

Part 4, measure the radii from the axis of rotation to the original graph.1185

Let me go back to blue here.1211

Let me go to black.1213

We said to measure the radii to the original graph.1227

The original graph is here.1231

Let me erase that.1233

Measure it to the original graph because that is actually where the function is.1253

The function is here.1258

This distance is 3, this distance here is x³, because if this is x, this height here is x³.1267

Our outside radius is 3 - x³.1280

The whole distance - this little distance, that gives me this distance.1285

The inside radius, it is the whole distance 3 - this distance, from here to here that is x.1292

Now that we have our outside and inside radius, the rest is really simple.1304

We just have volume = the integral, let me write the formula one more time.1309

a to b π × the outside radius² - the inside radius² dx.1318

It is always a nice idea to write the equation over and over again.1326

That is a way of memorizing it, before you start the problem.1330

In this case, it is going to be the integral from 0 to 1 of π × outside radius 3 - x³² - inside radius 3 - x² dx.1333

That is the integral that we want to solve.1354

Volume = π × integral from 0 to 1.1358

In other words, this is technically where you can stop.1364

You have converted the problem to the integral.1370

Now it is just an integration problem.1372

0 to 1, we got 9 – 6x³ + x⁶ - 9 - 6x + x², all of that dx.1375

9 – 9, the - - this function².1411

The minus is going to distribute over everything.1418

We are going to get = π × integral from 0 to 1.1424

We are left with -6 x³ + x⁶ + 6x - x² dx,1445

is equal to the π × -6x⁴/ 4 + x⁷/ 7 + 6x²/ 2 - x³/ 3 evaluated from 0 to 1.1461

When I add it all together, I get 1.31 π.1481

If I have done my arithmetic correctly, which is always a tossup.1485

In any case, that is it, getting the integral is the most important part.1491

Changing the actual problem itself, the physical thing that we are looking at, into some sort of integral.1496

The rest is integration.1505

Let us take a look at example 2, find the volume of a solid obtained by rotating the region1510

bounded by the following functions around the given axis.1513

Again, we have y = x³, this time we have y = √x, x is greater than or equal to 0.1517

We want to rotate this around x = 2.1524

Let us go ahead and draw our region.1528

I probably do not want to make it quite so big.1537

Let me spare myself a little bit of room here.1539

I have got y = x³, it is going to be like that.1546

y = √x, it is going to look like that.1550

They happen to meet at 1.1554

Basically, y = x³ is going to go off that way.1556

y = √x is going to go off that way.1560

They are going to meet here.1562

The region that I'm interested in is this region right here.1564

Of course, x is greater than or equal to 0, I’m interested in this region.1567

I’m rotating it around x = 2.1572

If this is 1, this is 2, I’m rotating it that way which means I'm going to have that and I’m going to have that.1575

This is the region than I'm interested in, this thing.1594

I’m looking at it straight on, that is my solid.1604

Draw the region, we have drawn the region.1611

Identify the solid, we have identified the solid.1616

Region is this way, rotating around this axis.1628

The slice that we take is going to be perpendicular to the axis.1633

If you are ever not sure how you are going to slice, you are going to slice perpendicular to the axis.1637

This direction is not perpendicular to the axis.1643

This direction is perpendicular the axis.1647

Our washer is going to be, that is a characteristic washer.1650

Since that is a characteristic washer, we are going to be adding washers in the vertical direction from 0 to 1.1661

This is 0, this is 1.1669

Our lower and upper limit of integration are 0 and 1.1672

We are integrating along the y axis.1675

When we are integrating along the y axis, we need functions of y.1679

Identify the washer which means we integrate along y.1683

You know what, I think I should probably write up my words a little bit better than this.1696

Let us integrate along y and we are going to be going from 0 to 1.1719

We know that the volume is equal to the integral from a to b of π × the outside radius² - the inside radius² dy.1727

That is going to be our general equation.1740

dy, we need functions of y not functions of x.1742

We express these functions, that function, and that function.1751

We express these functions as functions of y.1764

y = x³ that is going to become x = y¹/3.1774

y = √x that is going to turn into x = y².1785

These are the functions that I’m going to be dealing with.1794

I’m integrating along y which means my formula is going to have a dy.1800

The differential is in the y direction.1806

Therefore, my functions have to be in terms of y, that is what is going on.1808

Outside radius is, we measure from the axis of rotation to the original function which is this one right here.1816

Our outside radius is that one.1825

Our inside radius is this one.1831

This distance 2 - the distance from here to here.1840

The distance from here to here is this, it is y².1844

This x distance, whatever y is, our distance is y², that is the whole idea.1849

We have 2 - y² is our outside radius.1855

Our inside radius is the distance, it is from here to here.1860

This distance is from here to here which is 2 - this distance which is the y¹/3, 2 – y¹/3.1864

There we go, now we have got it.1878

Our volume is equal to the integral from 0 to 1 of π × the outside radius1885

which is 2 - y²² - the inside radius which is 2 – y¹/3² dy.1893

That is the most important part, coming up with this integral.1908

The rest is just integration.1911

We have got π × the integral from 0 to 1.1914

Let us go ahead and multiply all of these out.1919

We have got 4 - 4y² + y⁴ - 2 × 2 is 4 - 4 × y¹/3 + y²/3 dy.1921

Our 4’s go away, the negative sign distributes.1946

I have not done the integral yet, be very careful.1954

The integral from 0 to 1 of -4y² + y⁴ + 4y¹/3 – y²/3, all of that dy.1958

In your AP exam as you know, there are going to be some sections where you are going to have to do the integration by hand.1981

There are going to be some sections where you are going to do the integration,1986

where you are going to be allowed to use your calculator.1989

In this particular case, it depends.1992

Sometimes, it is nice to use a calculator.1995

Sometimes, it is nice to do it by hand.1996

You will have both options.1998

The idea is to be able to integrate this by hand, if you need to.2000

But again, on the calculator section you are just going to be plugging this in and getting an answer.2004

You are not going to be doing the integration like we are doing here.2009

We have got π × this is going to be -4y³/ 3.2014

This is going to be + y⁵/ 5.2023

This is going to be + 4y⁴/3 / 4/3 - y⁵/3 / 5/3.2026

Again, it is just my particular habit of not simplifying these and making this 3y⁵/3/ 5.2032

It is up to you, the rest is just arithmetic.2048

I ended up with 1.27 π, when I actually put this into my calculator.2051

Let us try our third example.2061

Find the volume of a solid obtained by rotating the region bounded by the following functions around the given axis.2064

We have y = 2 sin x, y = 2 cos x.2069

The x from π/4 to 5π/ 4.2073

We want to rotate this around y = -3.2076

Let us see what the heck is going to be going on here.2082

Let me work in blue for my graph.2086

I will go this way.2100

This is my x and y axis.2101

y = 2 sin x, it is a normal sin function, except the height is 2.2104

Down here this is -2.2116

Cos function starts here, it passes here, here, here, and here.2121

The cos function looks like this.2132

Not the best drawing in the world but I think we understand what is happening.2138

π/4 to 5π/ 4, that is this point and this point.2143

Our integration, if we decide to go along the x axis, it is probably going to be from π/4 to 5π/ 4.2148

Let us generate the solid and see what we have.2157

We have taken care of this.2160

Let us rotate around the line y = -3.2161

y = -3, let us go ahead and put it like right there.2164

This is -3, we are going to rotate this region right here around this axis right here.2168

Let us see what we get.2182

We end up with something that looks like this.2183

Again, I do not promise the best graph but here is something.2189

Something like that.2194

We have this nice region, this thing that we have rotated around there, rotated around there.2196

We have this solid of revolution.2206

Now we are going to take a slice.2210

We take a slice perpendicular to the axis of rotation.2212

The axis of rotation of this way, we are going to slice it this way.2216

Our slice is going to be vertical.2220

Our washer is that.2223

When I take this, rotated it out a little bit like that, what you are going to get is a side view of your washer.2232

It is going to have a certain thickness, that is going to be our dx.2247

In this particular case, we are going to be integrating along the x axis.2252

We are going to integrate along the x axis and we are going to go from π/4 to 5π/ 4.2258

Lower and upper limit of integration, respectively.2268

A little bit of an issue here with sins and cos and other functions that tend to occupy both above the axis and below the axis.2275

We said we are going to measure our radius.2284

We are going to have an outside radius.2287

We are going to have an inside radius.2289

Let me go to black, our outside radius, we are going to measure from the axis of rotation to the original function.2294

Our outside radius is that one.2301

Our inside radius is this one.2305

Distance, what is this distance?2311

It is the distance from here to here which is 3, because from -3 to 0 is 3, + this distance here.2315

That is the function, that is the 2 sin x function.2326

Our outside radius is going to be 3 + 2 sin x.2332

Now inside radius, this is where it gets a little interesting.2341

Again, it is going to be this distance 3 - this distance.2344

3, it is actually going to be +, here is why.2355

Wherever you are from π/4 to 5π/ 4, the cos is going to take on positive values and it is going to take on negative values.2360

The function itself accounts for the negative sin here.2371

Because sometimes it is going to be this + that, sometimes it is going to be this – that.2376

The function itself accounts for it.2386

It is just 3 + 2 cos x.2388

In this particular case, this particular slice, the cos x function is below the axis.2393

It is going to be 3 - 2 cos x.2398

I will leave it as + because the 2 cos x is going to be negative itself.2404

It is going to be 3 – a number.2408

I hope that make sense.2411

There we go, we have our outside radius, we have our inside radius.2414

Now our volume is really simple.2417

Our volume is going to be the integral from π/4 to 5π/ 4.2420

We are going to integrate from π/4 to 5π/ 4.2427

It is going to be π × outside radius 3 + 2 sin x² - the inside radius which is 3 + 2 cos x².2434

We are integrating along x.2459

This is the integral that we want.2463

Sometimes that will be enough.2468

Let us actually integrate it.2471

We have volume = π × the integral from π/4 to 5 π/4 of 9 + 12 sin x2474

+ 4 sin² x - 9 + 12 cos x + 4 cos² x dx.2495

This is going to equal, the 9’s cancel, I always have a thing with 0, I do not know where it is.2512

π/4 to 5π/ 4 of 12 sin x + 4 sin² x - 12 cos x - 4 cos² x.2523

There we go, this is dx.2547

This one, I just decided to go ahead and use a calculator to solve this, as opposed to doing the actual integration.2554

Part of the reason is, we can handle this one and this one, we have already learned techniques for that.2564

But in the lessons that follow, we are going to be discussing techniques of integration2570

for other types of functions that we have to integrate.2575

One of them is going to be the sin² function and the cos² function.2580

Technically, as far as the progress of this course is concerned, we have not learned how to deal with these integrals formally.2583

Otherwise, you will get the wrong answer.2602

Make sure your calculator is in radian mode, when you do any of the integrals.2604

Definitely check that before you go in the AP exam or any other exam,2613

when you are doing these and you are going to be using your calculator, radian mode.2616

Everything is in radian mode in calculus.2620

When I actually plug this in and do the integral, I get a volume equal to 33.94.2627

Again, finding the integral is the important part.2637

Let us do one more example.2640

Find the volume of a solid obtained by rotating the region around the given axis.2647

This time we have x² + y² = 9.2651

We are given an equation in implicit form not explicitly.2654

x is greater than 0, y is greater than 0, rotate around x = 3.2658

We have to decide what we are going to do.2661

Let us go ahead and turn this into an equation that we actually recognize.2666

This is going to be x²/ 3² + y²/ 1² = 1.2669

This is the equation of an ellipse.2681

We are going to take the x is greater than or equal to 0 and the y greater than or equal to 0.2687

What we are looking at is here.2691

What we are looking at is an ellipse.2697

The x direction, we are going to go 1, 2, 3.2702

y direction we are going to go 1.2705

But we want only the region that is above the x, axis above the y axis.2708

It is just this, region right here.2716

It continues on this way but the region that we are interested in is this region right here.2719

They say rotate around x = 3.2727

That is this axis right here.2730

When we rotate around that axis, we are going to end up getting that region.2734

We are going to take a slice of that region.2748

We are going to take a slice perpendicular to the axis of rotation.2749

The axis of rotation is vertical, the slice we are going to take is horizontal.2752

Our representative washer, when we slice this solid, our solid is this.2757

When we take a slice of it, we are going to end up getting a washer.2767

When I take that washer and turn it this way, it is going to look exactly like you think it is going to look.2771

It is going to look like this.2777

It looks like that.2782

It is going to have a thickness which is going to be dy.2783

When we add up all these washers, we are adding them vertically.2789

We are going to be integrating along the y axis.2791

We are going to be integrating from 0 to 1, nice perfect.2801

So far so good, now we are integrating along y.2809

We need to express these functions as functions of y.2813

We are going to turn this into a function of y.2815

Let me just write that out.2831

Integrating along y means expressing functions as functions of y.2833

Now I have got x² + 9y² = 9.2853

I have got x² = 9 – 9y².2860

I have x = √9 - 9y².2866

Do not worry about simplification, just leave it as is.2873

Pulling out 9 and 3, it is not even worth it.2876

I will go to black.2886

We are going to measure from the axis of rotation to the original function.2887

The outside radius is that one, the inside radius is this one.2894

The outside radius is just 3.2904

Nice because it stays always 3.2908

The inside radius that is the radius that is going to be short or long.2914

That is the one that changes.2919

Inside radius, it is going to be 3.2921

It is going to be this distance - this distance because we want this distance right here.2925

This whole thing - this thing.2934

This thing is that.2937

It is 3 - √9 - 9y².2940

There you go, now that we have that, very simple.2949

We have our volume is equal to the integral from 0 to 1 of π × the outside radius2953

which is 3² - the inside radius 3 - √9 - 9y²² dy.2963

This is the integral that we were looking for.2982

Let us go ahead and see if we can solve this.2988

We have volume is equal to π × the integral from 0 to 1 of 9 - 9 -2992

6 × 9 - √9y² + 9 - 9y² dy, which is equal to π × the integral from 0 to 1.3006

This 9 and this 9 cancel, negative sign becomes 6 × √9 - 9y², this becomes -9 and this becomes +9y² dy.3026

I just go ahead and plug this into my calculator.3046

I get myself 8.14, that is my volume.3049

Once again, you should note that some of these integrals, you have not learned how to deal with yet.3058

Do not worry about it, that is absolutely fine.3075

Some of these integrals, you have not dealt with formally.3078

We have not dealt with formally.3085

We will, very soon.3087

That is it, that is volumes by the method of washers.3097

Thank you so much for joining us here at www.educator.com.3100

We will see you next time, bye.3102