For more information, please see full course syllabus of AP Calculus AB

For more information, please see full course syllabus of AP Calculus AB

### 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
- Volumes II: Volumes by Washers
- 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
- Example II: Find the Volume of the Solid Obtained by Rotating the Region Bounded by the Following Functions around the Given Axis
- Example III: Find the Volume of the Solid Obtained by Rotating the Region Bounded by the Following Functions around the Given Axis
- Example IV: Find the Volume of the Solid Obtained by Rotating the Region Bounded by the Following Functions around the Given Axis

- 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

### AP Calculus AB Online Prep Course

### 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 just *0499

*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 area*0635

*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

*Outside radius, inside radius, outside radius, inside radius.*1204

*Let me go back to blue here.*1211

*Let me go to black.*1213

*Outside radius is that radius.*1215

*Inside radius is that radius.*1220

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

*The original graph is here.*1231

*Let me erase that.*1233

*That is our outside radius, that is our inside radius.*1245

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

*The function is here.*1258

*Our outside radius, distance is what we are worried about.*1260

*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 region*1510

*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 go ahead and identify our outside radius and our inside radius.*1710

*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

*Let us go ahead and find our outside and inside radius.*1813

*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

*Our outside radius is 2.*1835

*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 radius *1885

*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

*That is your washer.*2245

*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 x *2474

*+ 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 integration*2570

*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

*Just go ahead and use your calculator, in this case.*2591

*But make sure that your calculator is in radian mode.*2594

*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

*Let us go ahead and talk about our inside radius and our outside radius.*2880

*Let us talk about our outside radius first.*2884

*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 radius *2953

*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

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