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

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

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### 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
- Volumes I: Slices
- 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
- Example II: Find the Volume of the Solid Obtained by Rotating the Region Bounded by the Given Functions about the Given Line of Rotation
- Example III: Find the Volume of the Solid Obtained by Rotating the Region Bounded by the Given Functions about the Given Line of Rotation

- 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

### AP Calculus AB Online Prep Course

### 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 calculator*1794

*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

1 answer

Last reply by: Professor Hovasapian

Wed Jan 18, 2017 7:46 PM

Post by Richard Kennesson on January 2, 2017

Hello Professor,

For example III would the pi carry through to each integral when you split them up.

So the order would be pi, 4pi, 4pi at 29:11 (you left of the pis on the other integrals)

When doing to 2nd integral you made it 4pi and when doing the third you left it at 4.

Would the 3rd integral be 6pi^2 / 2 after evaluating at 3pi/2?

I could be wrong though so I thought I would ask.

2 answers

Last reply by: Professor Hovasapian

Wed May 11, 2016 3:05 AM

Post by Sazzadur Khan on April 28, 2016

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, 2016

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