AP Calculus AB 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 + 4x²0048

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 horizontally0321

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 dx1514

= 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 region1563

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 happen2312

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 dx2581

= 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