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 IV: Volumes By Cylindrical Shells

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 by Cylindrical Shells
- Find the Volume of the Following Region
- Volumes by Cylindrical Shells: Integrating Along x
- Volumes by Cylindrical Shells: Integrating Along y
- Volumes by Cylindrical Shells Formulas
- Example I: Using the Method of Cylindrical Shells, Find the Volume of the Solid
- Example II: Using the Method of Cylindrical Shells, Find the Volume of the Solid
- Example III: Using the Method of Cylindrical Shells, Find the Volume of the Solid
- Example IV: Using the Method of Cylindrical Shells, Find the Volume of the Solid
- Example V: Using the Method of Cylindrical Shells, Find the Volume of the Solid

- Intro 0:00
- Volumes by Cylindrical Shells 0:11
- Find the Volume of the Following Region
- Volumes by Cylindrical Shells: Integrating Along x
- Volumes by Cylindrical Shells: Integrating Along y
- Volumes by Cylindrical Shells Formulas
- Example I: Using the Method of Cylindrical Shells, Find the Volume of the Solid 18:33
- Example II: Using the Method of Cylindrical Shells, Find the Volume of the Solid 25:57
- Example III: Using the Method of Cylindrical Shells, Find the Volume of the Solid 31:38
- Example IV: Using the Method of Cylindrical Shells, Find the Volume of the Solid 38:44
- Example V: Using the Method of Cylindrical Shells, Find the Volume of the Solid 44:03

### AP Calculus AB Online Prep Course

### Transcription: Volumes IV: Volumes By Cylindrical Shells

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

*Today, we are going be talking about finding volumes by the method of cylindrical shells.*0004

*Let us jump right on in.*0010

*How can we find the volume of the following region?*0015

*Let me work in blue, I think.*0020

*How to find the volume of the following region?*0027

*We can go ahead and say, we have the function y = -3x³ + 4x²*0048

*and we want x to be greater than or equal to 0.*0059

*We want y to be greater than or equal to 0.*0063

*We want to rotate this around the y axis.*0066

*Let us see what region we are actually looking at.*0076

*When we draw out this region, this function right here, when we draw it out,*0079

*it is going to look something like this.*0089

*It is going to end up hitting it at 1.33.*0090

*x greater than or equal to 0, everything over here, everything in here are.*0095

*We are looking at this region here.*0100

*We are going to take this region and we are going to rotate it around the y axis.*0103

*We are going to rotate this way.*0108

*The solid of revolution that we are going to generate is going to be something symmetrical.*0110

*This is -1.33.*0116

*We are looking for the volume of this region, how can we do that?*0119

*We have already dealt with washers, it is possible to do something like this.*0124

*It is possible to basically take a washer, we will take a slice perpendicular to the axis of rotation.*0131

*And then, we will add all the washers up along the y axis.*0141

*We are going to integrate along y.*0145

*Let us write this down.*0149

*We could use washers, we try washers, I should say.*0150

*We could try washers which mean we are going to integrate, in this case, along the y axis,*0160

*or I will just say along the y direction.*0167

*Since we are integrating along y, we need x equal to some function of y.*0180

*In other words, we need to express this function as a function of y.*0192

*For this particular function -3x³ + 4x², that is not going to be a very easy thing to do.*0198

*Solving this function for x, in terms of y explicitly, it is not going to be very easy to do at all.*0204

*If in fact if it is possible, I do not even know if it is, in this particular case.*0213

*Here is where we run into a little bit of problem.*0218

*It is doable, theoretically, but expressing this in terms of y is going to be hard.*0222

*Because it is going to be hard, we ask ourselves is there another way of doing this?*0228

*There is, let me actually write out.*0234

*We are going to integrate along y so we need x = f(y).*0239

*For this function, expressing x as a function of y is difficult.*0245

*Is there another way, the answer is yes, there is.*0268

*The answer is using things called cylindrical shells.*0282

*Let us go ahead and redraw our little thing here.*0291

*We have that, we have this, that is one region, that is another region.*0296

*Let me go ahead and make this.*0309

*Here we have 1.33, we have -1.33.*0313

*Here is what I'm going to do.*0320

*Instead of taking a washer which means slicing this solid horizontally*0321

*perpendicular to the axis of rotation which we said is the y axis,*0326

*that is the axis of rotation, I'm actually going to take a piece of it and go perpendicular to the axis of rotation.*0330

*I’m going to pick this little slice right here.*0341

*Not really a slice though, you will see why in a minute.*0343

*Now I'm going to come over here.*0348

*What I’m going to do, I have this solid.*0352

*Essentially what I'm going to do, I’m going to take this little region.*0355

*Now from your perspective, that region, I'm going to end up turning it towards you.*0359

*I’m going to take this solid and you are going to end up taking this little, and actually turn it this way.*0366

*When we rotate around the y axis, we are going to get a circular object.*0373

*I’m going to basically bore into the object from the top, I’m going to turnaround this way.*0377

*I’m going to get a cylindrical shell.*0383

*When I take this down, what I end up getting is something like this, this region right here.*0385

*What I actually done is I have taken this solid.*0402

*When I view the solid from the top, I have actually bore into it and pulled out a cylindrical shell.*0406

*All of this is the solid part, that is what this is, this is from the side.*0414

*It is going to be of some radius here.*0425

*Of course this is going to be some differential length dx.*0428

*What this actually looks like is the following.*0434

*This and this, when I turn it this way, it looks like this.*0435

*The perspective drawing is actually this.*0440

*This is h, this is h, it is the height of the cylindrical shell.*0460

*This is r, that is this right here, that is r.*0472

*The thickness is dx.*0479

*Instead of taking washers and adding them up along the y axis,*0483

*I'm going to take concentric circular cylindrical shells working my way out, from this point to this point.*0487

*Shell, shell, shell, these are all a bunch of concentric cylindrical shells.*0502

*Now I can integrate along the x axis.*0507

*My function was a function of x.*0511

*All I have done is instead of dealing with washers or disks, now I’m dealing with cylindrical shells.*0514

*We need the volume of this one cylindrical shell.*0521

*When I have the volume of this one cylindrical shell, I integrate all of the volumes.*0525

*We need an expression for the volume of the cylindrical shell.*0531

*What is that?*0535

*What is the volume of this shell?*0540

*Cut it and unroll it, in other words, slice it here and unroll it.*0549

*What you end up having is a rectangle, what you end up having is a slab.*0564

*That is what you are going to get.*0570

*When you unroll this shell, you cut it, you unroll it, you are going to get a slab.*0580

*This is c, that is going to be the circumference.*0592

*It is going to be the circumference of the shell.*0597

*This is h, that is the height.*0601

*This right here, this, that is the depth.*0606

*We know that the volume of this is equal to the circumference.*0613

*This is length × width × height, which is circumference × height × depth.*0618

*The circumference × the height × the depth.*0623

*The circumference is equal to, this is r, the circumference is 2π r, 2π × the radius.*0630

*The height is just the height of the shell and the depth is your dx.*0639

*Our volume is going to equal 2π r h dx, circumference × the height × the dx.*0655

*That is the volume of our one little shell.*0669

*Let me draw real quickly again, let draw out in red here.*0673

*We have this and we have this.*0680

*We have this little shell right there.*0685

*This is our shell right, that is our height, this is our radius.*0689

*The radius is just a function of x.*0699

*r is just the x value, the height is f(x).*0706

*Our volume element of our little shell is equal to volume element as a function of x is equal to 2π × x × f(x) × dx.*0717

*I just add up all the shells, as we are working out.*0734

*As we are looking for the top, add up all the shells, working out in concentric circles.*0736

*Now just add up all of the volumes of these individual shells, these differential shells.*0744

*Our integration is going to go from 0, we are adding from here to here.*0777

*We are adding up all of these shells, we are looking it sideways.*0785

*We are going out from the center.*0789

*Integrate from 0 to 1.33.*0795

*Looking at this from the top, that is the solid looked at from the top.*0802

*You are taking concentric cylindrical shells.*0816

*You are integrating, integrating, you are adding them all up.*0820

*Cylindrical shell from 0 to 1.33, that is all we are doing here.*0827

*In general, let us go back to blue.*0838

*Volumes by cylindrical shells, the volume = the integral from a to b of 2π x f(x) dx,*0846

*if we are integrating along the x axis.*0870

*This is for integrating along x and we have v = the integral from a to b of 2π f(y) dy.*0877

*This is if we are integrating along y.*0900

*Again, very important measure, the radius of the shell from the axis of rotation.*0910

*Now because we are measuring from the axis of rotation, it is not always going to be just x and f(x), y f(y).*0934

*This is not just, they give you an equation, plug it into the formula.*0944

*These come from the actual physical situation that we described.*0949

*Finding the shell, finding what the radius is from the axis of rotation.*0955

*There are better formulas than these, more general, that you should actually concentrate on.*0960

*These are just thrown out there because it is what you are going to see in your book.*0965

*You have to let the situation decide what the radius is going to be and what the function is going to be.*0969

*It is better if we write the following.*0979

*Let me do this in blue.*0984

*Better formulas are, volume = the integral from a to b of 2π × the radius of the shell × the height of the shell × dx.*0987

*Or the integral from a to b of 2π radius of the shell, the height of the shell × dy,*1016

*depending on whether we are integrating with respect to x or respect to y.*1030

*You have to find r as a function of x and h as a function of x.*1035

*r as a function of y, h as a function of y.*1039

*I will write where r and h are functions of x.*1046

*Here where r and h are functions of y.*1059

*Probably, we want to avoid those.*1071

*They are accurate for a given situation, when you are revolving around the x axis or y axis.*1073

*But we are always going to be revolving around those axis.*1079

*We might be revolving around any other line x = 5, y = -6.*1082

*These are the general equations.*1087

*Once you have decided based on the situation that you are going to be integrating along x,*1089

*your r and your h, that your going to get from the picture are going to be functions of x.*1095

*If you decided that you are going to integrate along y, then your r and your h,*1100

*the radius of the shell and the height of the shell have to be functions of y.*1105

*These are the equations that you want to use.*1110

*Let us do examples because I think that makes everything clear.*1117

*Use the method of cylindrical shells, specifically cylindrical shells.*1119

*Find the volumes of the solid generated by rotating the region bounded by the following expressions about the given line.*1126

*x² – 5x + 8, 0 to 4, we want to rotate this around the y axis.*1132

*We need to know what this thing looks like, that is the whole idea.*1141

*We need to see what that looks like.*1145

*I’m going to go ahead and complete the square, turn it into a form where I know where the vertex is,*1150

*and then find f(0), f(4), and then, we will rotate that region.*1154

*Let us start off by going, I have got y is equal to x² - 5x + 8.*1160

*I think I’m going to go over here.*1173

*I’m going to take y - 8 = x² - 5x.*1175

*I’m going to complete squaring this, I’m going to take half of the 5/2 and I’m going to square it and add it.*1180

*I'm going to add 25/4 over here which means I'm going to add 25/4 over on the left hand side to retain the equality.*1187

*I get y – 7/4 is equal to x – 5/2² which implies that my vertex is at 2.5 and 1.75.*1197

*When I do y(0), I'm going to get y(0) is equal to 8.*1216

*I'm going to get y(4) is equal to 4.*1227

*Now I have a picture that I can work with.*1231

*Now I got this, now my vertex is at 2.5, 1.75.*1234

*Let us go 1, 2, let us go 1, 2.*1245

*I will keep this as 1, 2, but I will make my 1, 2, 3, 4, 5, 6, 7, 8.*1268

*I have different scales on my x and y axis.*1276

*My vertex is 2.5 and 1.75.*1281

*0 and 8, f(3) to 4.*1290

*I’m at 1, 2, 3, 4.*1296

*That is my region, this is the region that I’m talking about.*1303

*I need to be able to draw it.*1309

*Now I’m rotating this around the y axis.*1310

*It is going to look something like that.*1315

*This is my region.*1322

*Cylindrical shells, the axis of the shell, the axis of rotation is the center of the shell.*1324

*The little sliver that you draw is going to be parallel to the axis of rotation.*1335

*In this particular case, my shell is going to be like this.*1340

*I rotate that, this is going to be the other side of the shell.*1346

*When I take this and I turn it towards me, now I'm going to see my circle like that.*1350

*This thing rotated in perspective drawing, it is going to look like,*1361

*This is the height, that is this right there.*1378

*Let me work in black.*1386

*This is my radius, that is x, the x value.*1392

*This is my height, that is my f(x), that is what is going on here.*1400

*The radius is equal to x and the height is equal to y which is equal to x² - 5x + 8.*1409

*I have gotten the picture telling me what is going on, I’m not just putting it in.*1426

*I know my general formula is volume = the integral from a to b of 2π r h dx.*1429

*In this particular case, r is x, I put that in there.*1440

*h is y which is the x² – 5x + 8.*1443

*I put it in there and then I integrate but it is based on this.*1447

*Once again, the axis of rotation which in this case was the y axis, is the center of the shell.*1450

*It is the axis of the cylindrical shell.*1458

*Side view of the shell looks like that.*1460

*This is the top view of the shell, this is the perspective view of the shell.*1463

*I’m integrating from 0 to 4.*1467

*I’m taking shells concentrically out.*1475

*Let us write it all out, let us go back to blue.*1481

*I have got the volume = the integral from a to b 2π r h dx.*1482

*The integral from 0 to 4, 2π x x² - 5x + 8 dx, this is what is important.*1496

*The rest is just integration.*1512

*This is equal to 2π × the integral from 0 to 4, x³ - 5x² + 8x dx*1514

*= 2π × x⁴/ 4 - 5x³/ 3 + 8x²/ 2 from 0 to 4.*1530

*When I do this, I get 21.33, that is all.*1544

*Again, this is what is important, being able to form the integral.*1551

*The rest is just integration problem.*1555

*Let us do another example.*1557

*Using the method of cylindrical shells, find the volume of the solid generated by rotating the region*1563

*bounded by the following expressions about the given line.*1567

*Let us see what we have got.*1571

*We got y = 1/8 x³.*1572

*Let us see what we have got.*1579

*I have got this and we are going to rotate about the x axis this time.*1581

*Let me make this a little bit smaller, actually.*1591

*I will use a little bit more room.*1594

*Let me go here.*1597

*This is my y axis, this is my x axis, y = 1/8 x³.*1606

*Something like that.*1613

*y = 10, it is going to be up here.*1615

*Let us just say that is the line y = 10 and x = 0.*1623

*This is the region that I'm interested in.*1630

*I’m going to rotate this around the x axis this way.*1632

*This is going to be like this, then, I’m going to get a region like that.*1638

*This is my region.*1642

*They are saying specifically, use cylindrical shells.*1645

*The shells are going to be the length.*1648

*The sides of the shell are going to be parallel to the axis of rotation.*1654

*In other words, the axis of rotation is going to be the center of the shell.*1657

*The axis of rotation is the x axis.*1660

*Our shell is going to be parallel to that, that is this way.*1663

*If I were to take this and turn it that way, looking at it straight on, I will be looking at something like this.*1672

*I hope that make sense.*1685

*The radius of this shell is that value, it is y, that is the radius.*1688

*The height of the shell is that, that is x.*1697

*Let us see what we have got.*1707

*We are going to be integrating along the y axis.*1709

*We integrate along y.*1718

*Because we are integrating along y, we need our functions to be functions of y.*1723

*This is a function of x.*1729

*Let us go ahead and see what we can do, radius = y.*1731

*In this particular case, the radius we said is equal to our y value.*1737

*We have radius is equal to y.*1741

*We need our height, the height is x.*1746

*We need x, in terms of y.*1750

*We have got y = 1/8 x³.*1753

*We have 8y is equal to x³.*1769

*We have x is equal to 2y¹/3.*1775

*x is equal to h, it is the height.*1781

*Therefore, the height is equal to 2y¹/3.*1788

*Now that we have our radius and we have our height, we can go ahead and do this problem.*1794

*Volume = the integral from a to b, 2 × π × the radius × the height dy.*1802

*We are integrating from 0 to 10 shells.*1814

*Looking at it from the top, I see these concentric shells going outward from a radius of 0 all the way up to a radius of 10.*1824

*Shell, shell, shell, until I have covered all of them, that is what is happening.*1837

*We have got 2 × π × y × 2y¹/3 dy.*1843

*This is going to equal 4π × the integral from 0 to 10 of y.*1856

*y × y¹/3 is going to be y⁴/3 dy.*1866

*It is going to be y⁷/3 / 7/3 from 0 to 10.*1876

*Our final answer is going to be 3/7, 10⁷/3, that is all.*1886

*Example number 3, let us go back to blue here.*1900

*Cylindrical shells, we have y = √2x, we have y = 0, and we have a line x = 2.*1903

*Let us go ahead and draw this out.*1912

*Let us do x = 2, we want to rotate along x = -2.*1916

*Let us go ahead and draw this out this way.*1927

*We have got this.*1930

*This is our y axis.*1935

*I do not need to make it this big and this one I might need to.*1938

*This is our 0,0, this is our x, the y = √2x.*1947

*Some functions is going to look like that.*1952

*y = 0 that is this line, x = 2.*1956

*Let us go over here.*1963

*This is the region that we are interested in, that is our region.*1969

*We are going to rotate about the line x = -2.*1973

*Here is -1 and -2.*1977

*This is our axis of rotation.*1979

*When we rotate about that lines, that means we are going to go two more over here.*1982

*We are going to have that.*1993

*We are going to have this region right here.*1996

*This, this, this, this is our region, this is the y axis, this is the x axis.*2004

*We are going to take this region, we are rotating it around the line x = -2.*2011

*This is our solid of revolution, we are going to use cylindrical shells.*2018

*The axis of rotation is the center of the shell.*2026

*The sides, the walls of the shell are parallel to the axis of rotation.*2031

*It is going to be here and here.*2038

*Looking at it from the side, cylindrical shell is opening out.*2045

*Now we are integrating along the x and we are going to integrate from 0 to 2.*2049

*We have taken care of the lower and upper limit of integration.*2070

*Now we need to find the radius, we need to find the height.*2072

*We measure the radius from the axis of rotation.*2077

*The axis of rotation is over here.*2081

*It is at the point -2, distance is what we are worried about.*2083

*Let me work in black.*2094

*The radius from the center, from the axis of rotation which is the center of the shell to the wall of the shell,*2096

*that is going to be this distance which is 2 + this distance which is x.*2106

*Our radius is equal to 2 + x.*2113

*Notice, it is not just x, our axis of rotation has changed.*2116

*It is the radius, the length, that matters.*2121

*Our height that is going to equal this which is f(x).*2125

*Our height is equal to √2x, now we have everything that we need.*2136

*Let me go ahead and erase this.*2144

*This is our y = -2.*2146

*Notice, distance is what we want.*2148

*Even though this is -2, this is not negative, this is 2.*2152

*2 + the x, in order to take me from the center of rotation of the shell to the actual wall of the shell, that is the radius.*2156

*I think I will do it in red, I love changing the colors.*2168

*Volume = the integral from a to b of 2π r h dx, that is our general formula,*2172

*= the integral from 0 to 2 of 2 × π × 2 + x × h which is √2x × dx.*2183

*This is what we want, that is the integral.*2198

*The rest is just integration.*2203

*Let us go ahead and go over there.*2209

*We have got 4 √2π from 0 to 2 of x ^ ½.*2218

*I pulled out the √2 and I just left the √x in there.*2227

*dx + 2 √2π, the integral 0 to 2, x³/2.*2231

*I distributed and just separated the integral.*2250

*I get 4 √2 × π × x³/2 / 3/2, from 0 to 2 + 2 √2 × π x⁵/2 / 5/2 from 0 to 2.*2261

*When I work all of this out, I end up getting 64π/ 3 + 128π/ 5, that is my total volume.*2288

*Again, the rest is just arithmetic which I will leave to you.*2305

*Finding the integral was the important part.*2308

*Integration is important, it is actually where a lot of the problems happen*2312

*because you are dealing with arithmetic issues, +, -, √, this and that.*2316

*In any case, that is the nature of the game.*2321

*Let us take a look at example 4, let us see what we have got here.*2325

*Cylindrical shells, y = x⁴, x = y⁴, we want to rotate about x = 1.*2331

*Let us go back to blue here.*2337

*Let us draw this out.*2341

*y = x⁴ is going to look something like that.*2352

*x = y⁴ is going to look something like that.*2354

*They are going to meet at 1, rotate about the line x = 1 which is this line.*2358

*Now we have got that and we have got that.*2366

*This is our solid that we want to find the volume of, rotated about x = 1, that is the center of the shell.*2371

*The walls of our shell, our representative shell is going to be something like this.*2383

*We are going to be integrating along x.*2392

*We are going to be integrating from 0 to 1, lower and upper limits of integration.*2400

*Let us see, what do we want to do next?*2412

*We need to find r and h.*2423

*Our radius from the axis of rotation is going to be, my radius is going to be that.*2427

*This is my radius and my height is going to be this.*2455

*My radius is going to be this distance - that distance.*2464

*1 - the x value, our radius is 1 – x.*2477

*My height that is going to equal the top function of x - the bottom function of x, that is my height.*2486

*It is going to equal the top function of x which is x¹/4.*2505

*I need functions of x here because I’m integrating along x.*2520

*This one is fine, I need to convert that to a function of x.*2523

*x = y⁴, this is the same as y = x¹/4, that is my top function.*2528

*It is x¹/4 - my bottom which is x⁴.*2537

*Now that I have my r and my h, volume is simple.*2546

*Volume = the integral from a to b of 2π r h dx which = the integral from 0 to 1 2π 1 - x × x¹/4 – x⁴ dx.*2552

*That is what I want.*2578

*When we solve this, we end up with 2π, the integral from 0 to 1 of x¹/4 – x⁵/4 – x⁴ + x⁵ dx*2581

*= 2π x⁵/4 / 5/4 – x⁹/4 / 9/4 – x⁵/ 5 + x⁶/ 6, from 0 to 1.*2608

*When I work that out, I get 0.322 or whatever numbers you happen to put in there, when you put the 1 and 0 in.*2631

*Let us go ahead and try one more example here.*2644

*Cylindrical shells, we have the function y = x + 3/x, y = 10.*2649

*We want to rotate about the line x = 12.*2657

*Let us see what we have got here.*2662

*As far as the graphing is concerned, you can use your graphing calculator,*2681

*you can use any graphical tool that can find on the internet.*2684

*You can go ahead and express this as a rational function and use the techniques of differential calculus to graph it.*2688

*Suffice it to say that y = 10, we are looking at that.*2693

*The graph itself actually looks like this.*2700

*It goes down and it comes up like that, whatever technique that you need to use in order to graph it,*2708

*that is what the graph of this looks like.*2715

*y = 10, that is this line right here.*2720

*That is y = 10, this is the region that we are concerned about.*2723

*It is this region that we are going to be rotating.*2727

*They say rotate around the line x = 12.*2729

*It turns out that x = 12 is actually right about there.*2733

*This is 12, therefore, our region is going to be something like that.*2738

*I probably draw a little bit better than that.*2744

*It is going to come down to about right there.*2750

*Something like that.*2756

*This is a solid that we are dealing with.*2758

*Our axis of rotation is 12 which mean that the sides of the shell are going to be parallel.*2761

*When we measure r and h, r again, we are measuring from the axis of rotation to the original function, that is r, that is h.*2773

*r is equal to this length 12 - this length which is x.*2790

*It is 12 – x.*2803

*The height is 10 – f(x), it is this height - that height.*2805

*It is 10 – f(x) which = 10 - x - 3/x.*2820

*Now we need the a and b.*2834

*We have our r, we have our h which is this thing.*2838

*The question is what are a and b?*2845

*I'm going to integrate from where the two graphs meet, the x value.*2851

*This one and this one.*2857

*Where do the two graphs meet?*2863

*Just set them equal to each other, x + 3/x = 10, and solve.*2872

*Let us do x + 3/x + 10, x + 3/x = 10 which gives us x + 3/x -,*2887

*Let us get this right, 3/ x - 10 is equal to 0.*2907

*We are going to get x² + 3 - 10x = 0.*2912

*We have x² - 10x + 3 = 0, this is a quadratic.*2920

*When I solve this, I get x = 0.31 and I get x = 9.69.*2925

*I'm going to be integrating from 0.31 to 9.69.*2933

*My volume is equal to the integral from a to b 2π r h dx.*2940

*The integral from 0.31 to 9.69 2π, my radius we said was 12 – x.*2948

*Our function, our h, our height was 10 – x - 3/x.*2961

*We are integrating along dx.*2971

*Plug this into your calculator, this is going to be one of those situations where you definitely going to need a calculator.*2973

*My final answer for volume is going to be 2π × 301.29, that is all.*2979

*The important thing is coming up with this.*2991

*Thank you so much for joining us here at www.educator.com.*2999

*We will see you next time, bye.*3001

1 answer

Last reply by: Professor Hovasapian

Fri Feb 26, 2016 4:18 AM

Post by Jessica Lee on February 23, 2016

Hi! I just wanted to make sure. For example one, isn't 21.33 before multiplying by 2pi?