AP Calculus AB Volumes III: Solids That Are Not Solids-of-Revolution

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

Today, we are going to be talking about solids that are not solids of revolution.0004

Let us jump right on in.0010

So far, we have been talking about regions by taking a given function.0013

Let me write all of this down.0017

We have taken regions in the xy plane bounded by given functions and rotated them around a given axis.0026

That is what we have been doing, rotated them around a given axis.0048

Doing, we generated these things called solids of revolution.0059

We can take something like this.0083

We will take the xy axis and we will take a function, let us just say y = √x, something like that.0086

And then, we rotate it, in this case, let us say around the x axis.0093

We rotate this axis and we end up generating this solid like that.0097

What we do is we end up actually… I have drawn it slightly off so that you can actually see the rotation part.0108

Let me actually erase that.0115

Let me just draw it fully sideways.0117

Just a straight, it goes like that, something like that.0118

What we did is we took a slice of it.0124

That slice, when we turn this slice this way, what you end up getting is a circle.0129

You get a circle, it is a solid disk.0138

This is what we have been doing so far.0143

We have exploited the circular quality that comes from revolution around an axis.0146

Any rotation is going to generate some thing where the cross section is going to be a circle.0175

The area of the circle is very easy, it is just π r².0180

We exploited that axis to find the cross sectional area.0183

In other words, the area of the cross section.0199

Take a cross section which is always going to be perpendicular to the axis of rotation.0202

We turn this way and the figure that we get, that is the cross section.0206

Let us say that again, a cross section is obtained by slicing a solid perpendicular to a given axis.0213

In this case, we sliced perpendicular to the axis of rotation.0241

In the case above, area is equal to π r².0267

Our other types of objects, where they are not quite solid,0278

let us say you have this function and this function, and we rotate it around the x axis.0282

You are going to generate a different type of solid.0289

Not completely solid, it is kind of empty.0292

In this case, when we rotate it, what you end up generating is,0298

Let me do this in red.0303

If we take a slice and a slice, what you end up with is not a full circle but you end up with something which is a washer.0305

It is the same thing, it is still an area.0316

In this case, we are still exploiting the circular quality.0319

Now the area is just π × the outer radius - the inner radius², that is it.0325

The area of the outer circle which is π r, outer radius².0335

Let me actually write it out.0341

It is going to be π × the outer radius² - π × the inner radius².0346

The area = π × the outer radius² - the inner radius².0353

The squared is on the outside, I apologize for that.0361

In each case, we added up all of the slices.0364

We added up which is integration.0374

We integrated the volumes for each slice.0379

That is it, that is all we did.0389

We found the volume of one slice and then we integrated it.0390

We added up all of the volumes.0393

In each case, we added up the volumes for each slice whether it was a disk or a washer.0397

The volume is just the integral from some beginning point to some endpoint.0406

In other words, some beginning point to some endpoint of the area which is a function of x dx.0411

Or if we were integrating along the y axis, the area which is a function of y dy.0419

That is it, that is all we have been doing.0425

Find the area and you integrate.0426

I will write it out.0436

In each case, the cross section was circular.0440

The question is, what happens if the cross section is not circular?0453

What if we have a cross section which is not circular?0462

The general formula is still the same.0480

The general formula still holds.0487

In other words, the volume, our total volume is going to equal the integral from a to b of some area function,0499

whatever the area of that cross sectional shape is now.0510

In the case of rotation around an axis, we get a circle.0514

It does not have to be rotation, there can be another cross section.0517

You can have a square, triangle, trapezoid, whatever shape, but it is still just the area × the little width of the slice.0520

That does not change.0530

Our job is to find this, the area as a function of x or y depending on which way you are integrating.0532

That is really all we are doing, ay dy.0539

And then, you add up all of the individual volumes.0543

ax or ay, depending on our choice, ay are still cross sectional areas but of various shapes.0549

We have area formulas for all the different shapes, the basic shapes in geometry that we deal with, it is not a problem.0578

We must use our knowledge which we got from pre-calculus, geometry, algebra,0586

whatever it is, to derive a formula for a(x) or a(y).0597

That is really all we are doing, that is the hardest part of these problems.0615

It is kind of like the max/min problem or the related rates problem.0616

The hard part is not the calculus, the hard part is coming up with a formula0622

from the information that you are given in the physical situation.0625

You must use our knowledge to derive a formula for a, x, y, then just integrate.0631

And that part at this point should not cause any problem.0639

The integration, that should be the least of our worries.0641

It is the least of our worries, we want to find an integral, that is we want to do.0644

Let us go ahead and do an example, and see what we can do with that.0650

Find the volume of a pyramid whose base is a square of side length s and whose height is h.0656

Let us draw this out in perspective.0662

Let us go ahead and draw a square base, put this up there.0665

We have this pyramid, like that.0673

Side length s, we will call that s.0676

Let me draw a fully straight on view of this.0680

This is going to be not a perspective drawing.0686

Side length is s and this is h.0693

What we are going to do is were going to take a slice of this.0701

We are going to take a slice which is a square.0705

We are going to take the area of the square.0716

We are going to multiply by dx.0718

We are going to find that, that is going to be the volume of that one slice.0719

We are going to find the volume of the slice which is like that, if we are looking straight on.0722

And then, we are just going to add up all of the slices.0728

That is it, that is all we are doing.0730

Except in this case, the cross sectional area is a square, it is not a circle.0731

You just need to arrange the problem, in such a way on your coordinate axis.0741

That will give you the answer that you want.0746

We are going to take slices, find the area of the squares.0750

Find the volumes of the slices, and we are going to integrate.0776

In other words, add up.0785

That is it, that is what we are going to do.0787

Let us set this problem up.0789

I’m going to set it up this way.0791

There is more than one way to do this.0792

Please do not think that this is the way to do it.0794

You are going to arrange it in a way that makes the most sense to you,0797

based on the coordinate system or what it is that your particular mind sees.0801

Clearly, you have discovered by now, even back in pre-calculus,0806

that as these problems become more sophisticated and more complicated, there is more than one approach to the problem.0808

What your approach works, by all means, use it.0816

I'm going to set up my coordinate system in such a way.0820

I’m going to take this pyramid and I’m going to turn it on its side.0823

I'm going to run the vertical axis along the x axis because I want to integrate along x.0826

I’m going take this.0832

When we are looking at side view of this thing, it is going to look something like that.0834

This right here, that is our s, that is our side length.0840

This right here, that is our h.0845

The slice that we are going to take is this slice right here.0849

That is our slice.0856

This slice, I’m going to rotate it this way.0858

This is a slice that we are taking, perpendicular to an axis that runs through the center of the pyramid.0861

This, when I turn this around, the square that I'm looking at, the slice that I'm looking at is going to be that.0870

The axis, this axis, passes right through that.0885

It is right through the middle.0888

This thing right here, I’m just going to go ahead and call this y.0891

This is s and I’m going to call it m for the time being.0897

M, because it is not really s, it is a different smaller size than s.0902

I’m just going to call it m for the time being.0909

The area of this slice is equal to m².0913

We want to find m², in terms of the things they gave us, s, h, and x,0921

because that is the variable that I’m integrating along.0944

I’m going to be adding up all the slices this way.0946

That is what I'm doing.0949

We have a relationship here.0952

Let me work in red here.0955

If this value is x, this value is going to be y.0958

This is my y and this is my x.0965

There is a relationship here.0968

x is related to y, as h is related to this length right here which is s/2.0970

I set up a proportion.0989

This implies that x s/2 is equal to hy.0992

I’m going to solve for y here because I'm looking for y.1002

I’m looking for y so that I can multiply it by 2 and have them be my m.1004

Then, I’m going to square that m and that is going to be my area.1010

y is equal to x × s/ 2h.1013

m is equal to 2y which is equal to 2x s/ 2h.1022

The 2's cancel and I'm left with xs/h, that is my m.1033

I found my m, that is just x × s/h.1040

The area we said is equal to m², that is equal to xs/h², that is equal to s²/ h² x².1049

That is the area of our square.1067

Now I'm going to multiply it by the differential.1071

This width is dx, it is my old differential width.1075

The volume of each slice = s²/ h² x² dx.1083

Now I add them all up.1097

The total volume = the integral from, we had 0.1098

This length was h, 0 to h s²/ h² x² dx.1111

You are just adding up all the different slices.1124

Volume = the integral from 0 to h.1132

S²/ h² is a constant so it becomes s²/ h² × the integral from 0 to h of x² dx.1139

The rest is easy.1151

s²/ h², x³/ 3 from 0 to h.1155

We put in h, we should get s²/ h² × h³/ 3 – 0.1166

We are left with the final answer of s² × h/3.1184

Side length s, height h, the volume is s² h/3.1192

1/3 size² × the height, which you already knew back from geometry days.1198

Calculus is how we actually derive that formula that we use in geometry.1207

That is it, nice and straightforward.1211

You just need to set it up properly, in a way that makes sense to you and just sort of figure out the rest.1214

Set up the coordinate system, take a slice, find the area of a cross section of that slice, and then integrate.1222

I think that was the last of that one.1234

Example 2, a certain solid has a circular base of radius 2, cross sectional areas perpendicular to the base are equilateral triangles.1243

What is the volume of the solid?1252

The hardest part of this problem, all of these problems, is visualizing what is going on.1256

Just follow what is says, draw it out, and everything should fall out.1260

Let me go back the black here.1267

I have got this, the certain solid has a circular base of radius 2.1270

I have got a circular base and my radius is 2.1278

This is 2, this is -2, this is 2, and this is -2.1283

Cross sectional area is perpendicular to the base.1290

If this is my base, perpendicular means hit it that way.1295

In other words, I’m going right down into the base.1300

If I just drop something straight down into this, cross sectional area perpendicular to the base.1304

When I take a slice like that, are equilateral triangles.1311

Cross sectional areas, I hit the base perpendicularly and I turned that around.1318

Basically what is happening here now, they are equilateral triangles.1325

This actually looks like this.1332

When I take this and I turn around, I get something that looks like this.1338

If I call this point A and this point B, this is A and this is B.1348

We are going to rotate it so I look at it a little bit better.1358

Do I really need to rotate it so I look at it a little bit better?1362

This axis point is right there.1366

This axis is right there, all I have done is hit it, turn it around.1368

Now I'm looking at this figure.1374

When I rotate, I’m going to bring this just because I'm used to looking at triangles this way.1376

This is A, now this is B, my axis is here.1388

That is it, we need to find the area of the triangle, the volume, in terms of,1394

in this particular case, this is my y axis, this is my x axis.1401

My slice is this way, I'm going to be integrating along the x axis.1405

I need to find some function of x.1409

We will be integrating along the x axis.1420

We need an area function for the triangle as a function of x.1434

In other words, we need some a(x).1460

Let us see what we can do.1466

Let me draw it again.1472

I have my circle, I have my slice.1475

I turned it, rotated it.1486

Now that I’m looking at a triangle that looks like this, equilateral triangle.1488

Here is my axis.1495

If this is x value, this is my y value.1496

That y value is this, that is my y value.1502

That is my height, it is equilateral.1510

This angle is 60°.1513

I know the fundamental formula for the area of the triangle.1517

It is equal to base × height/ 2.1521

Let us see what the relationship is here, between the x and y.1525

This is a circle of radius 2, I have got x² + y² = 2.1529

I have got y² = 2 - x².1537

Therefore, y is equal to √2 - x², that is y.1542

The base is twice y.1550

The base is equal to twice y, the base is equal to 2 × √2 - x².1557

I found my base, what about my height?1570

My height, this is a 60° triangle.1574

This is 30, 60, 90.1577

If this is y and height is just y√3.1581

It is just y√3, the height is equal to √3 × √2 - x².1587

I have got my height and I have got my base.1599

Plug them into this equation.1603

The area is equal to base × height divided by 2.1604

It is equal to 2 × √2 - x² × √3 × √2 - x², that is h/2.1611

My 2’s cancel, this × that.1629

I’m left with an area function.1632

My area at some function of x is going to equal √3 × 2 - x².1636

That is my area, that is my area of the triangle.1648

When I look at it from that end, that is my area.1652

I’m going to find the volume and I’m just going to add up everything,1656

then I’m going to integrate from -2 to 2 because I’m integrating along the x axis.1659

We have our area function which is equal to √3 × 2 - x².1669

My differential volume element is just √3 × 2 - x² × dx.1677

My total volume is equal to the integral from, once again, we have our circle.1685

This is -2, this is 2.1693

Our slice is here, we are adding up all the slices this way, along the x axis.1695

It is going to be from -2 to 2 of the area function or of this thing right here, √3 × 2 - x² dx.1701

That is it, that is my integral, this is what I was seeking.1714

The volume is equal to √3 comes out, from -2 to 2 of 2 - x² dx.1719

In the problems that I have written now, I actually went ahead and I solved this integral.1731

But I think at this point, I hope you will forgive me.1734

I’m just going to go ahead and leave these integrals to you because they are easy enough to solve.1736

This is just going to be √3 × 2x – x³/ 3 taken from -2 to 2.1740

I will leave that to you.1751

The difficult part was this.1752

This is what we wanted, this is what is important.1754

Finding the integral, the rest is just techniques of integration, whatever function you happen to be dealing with.1758

That is example 2, let us see what we have got.1765

Find the volume of a pyramid whose base is an equilateral triangle of side length a and whose height is h.1769

Let us go ahead and draw a perspective of this.1776

Let us go ahead and do this in black first.1778

Our perspective diagram is going to be something like this.1781

I will put that there, I will put a little point here.1787

This is side a, this is side a.1796

This is side a, we have this pyramid whose base is an equilateral triangle.1802

The base is an equilateral triangle.1807

I will go ahead and draw it straight on.1810

We know what we are looking at.1814

We have this little triangle, I’m going to draw it along,1817

We are looking at it this way, this edge.1822

From your perspective, the triangle is coming out like that.1826

This length is a and the height is h.1834

Not altogether different than the first example that we did.1841

This is our perspective view and this is straight on.1845

Let us go ahead and draw this out.1855

Again, I’m going to use the same coordinate system.1857

I’m going to set it up like this.1860

I'm going to have it so that the axis, my x axis again runs right through the center of the triangle.1863

I’m basically going to take this thing, I’m just going to turn it this way onto the axis.1873

There is that, there is this.1879

There is that, I’m going to take a slice.1892

This is going to be my x, this is going to be my y.1901

I will call this my a and this my b.1910

When I turn it so that I’m looking at my triangle, I have got my triangle this way.1916

I have got that, the axis is right through the middle.1934

The base of the triangle, the base are equilateral triangles.1941

This angle is 60°, I did not rotate it this time.1946

That is the height, this is the base of the triangle.1954

We want to find the area of the base.1962

We know that area = base × height/ 2.1966

If this is x and this is y, that means this right here, half of it, from here to here is y.1976

Therefore, the base is actually equal to 2y.1983

Let us go ahead and find y, same way as before.1988

We have x/y, x is to y, as h is to side length is a.1992

a/2 which implies that y is equal to x × a/ 2h.2006

b is equal to 2y, b is equal to 2 of these.2019

Therefore, we have just x × a/h, that is our base.2026

Let us redraw our triangle.2037

This was our h, this was our y, this was our b.2043

Our h, this is a 60° angle.2049

Therefore, if this is y, this is y√3.2052

It = √3 × ax/ 2h.2060

We said that the area is equal to the base × height/ 2 = the base which is ax/h2080

× the height which is √3 × ax/ 2h.2094

I will go ahead and put this ½ over here.2112

Therefore, our area is √3 a² x²/ 4h².2115

There you go, we have our area function which is a function of x.2130

We are going to b; remember this is r, we put it this way.2136

We are going to be integrating from 0 to h, from here to here, adding up all of our slices like that.2140

Therefore, our volume is equal to the integral from 0 to h of our area function × dx,2149

which is the integral from 0 to h of this thing √3 a²/ 4h² x² dx.2160

This is what we want and the integral is easy after that.2178

This is all a constant, it comes out of the integral.2184

This, when you integrate it, just becomes x³/ 3.2187

I’m going to leave the integration to you.2190

I hope you forgive me for that, very simple integration.2192

That is the formula that we are looking for.2197

Let us see what else what we have got here.2201

Let us see what we can do with this one.2209

Let me go back to black.2216

Find the volume of the solid whose base is given by the equation, 16x² + 4y² = 64.2217

We are looking at a base that is an ellipse.2223

And whose cross section is perpendicular to the y axis are isosceles right triangles,2226

with a base of the triangle being the hypotenuse.2236

A lot going on here, let us see what we have,2241

whose base is given by the equation 16 x² + 4y² = 64.2246

Let us go ahead and take care of this first.2250

We have got 16x² + 4y² = 64.2253

We have got x²/ 2² + y²/ 4² = 1.2259

Let me draw it a little bit over here, in fact.2271

I will draw it over here.2276

Our base is this, we go 1, 2, 3, 4, 1, 2, 3, 4.2282

We go 1, 2, we go 1, 2.2290

We are looking at an ellipse, something like that.2293

They say this cross section is perpendicular to the y axis.2302

We are going to hit the y axis, in other words, we are going to hit the y axis.2305

The cross sectional area is our isosceles right triangles.2314

I take a cross section along the y axis.2320

When I pull this cross section out and I flip it this way up, from your perspective, I have a cross section.2326

I’m going to flip it this way, these are isosceles right triangles.2334

What it is going to look like is the following.2340

Let me actually marks some points here.2347

If this is point a and this is point b, looking at it from the top, when I flip it up and rotate it,2348

I’m going to get an isosceles right triangle.2355

This is going to be an isosceles right triangle.2358

This is a and this is b, taken the cross section, I flipped it up so I’m actually looking at it.2366

Let us see what we have got here.2377

Let us go ahead and call this h again.2382

Let us go ahead and call this b.2387

Again, we are looking at the area of the triangle.2390

We are going to find the area of the triangle.2394

This time, we are going to integrate along the y axis.2396

We are going to be integrating from -4 to 4.2400

Those are going to be our limits of integration.2402

Again, we have the area = the base × the height/ 2.2406

Because we are integrating along the y axis, we need a(y).2412

We need an area function that is a function of y because we are integrating along the y axis.2427

Let us go ahead and find a relationship.2432

If this is our x value, this is going to be our y value, the whole idea.2437

There is each point along the ellipse is in relationship between x and y.2447

The x right here, that is this.2462

This distance is our x.2468

The base is going to be twice the x.2471

This is the x, that is the x, that is our base.2474

We need to find x, in terms of y.2478

We need to multiply by 2, let us do that.2484

16x² + 4y² = 64, I hope that made sense.2487

The way we draw this triangle based on this thing that we have flipped up,2494

this half the base of the triangle is our x value, whatever x is.2499

We need to find an expression in y, we need to find the relationship between x and y.2503

The base is twice x.2511

We have got 16x² = 64 - 4y².2516

We have x² = 64 – 4y²/ 16.2526

Therefore, x = 1/4 √64 - 4y².2534

I know that you can simplify it more but I just want to probably leave it like this, does not really matter.2545

x is this, that is our x value and our y value is going to be this thing, whatever that happens to be.2551

The base is equal to twice x, that is equal to ½ × √64 – 4y².2560

We have our base.2577

Let us see what we can do about our height.2582

Let us draw our triangle again.2586

We have a right isosceles triangle, there we go.2588

We set this as the base, this is the height.2593

If this was our x value, this is 90° that means that is 45 and that is 45.2606

That is 45, that means this is 90.2617

h and x are the same.2622

h is actually equal to x and that is equal to b/2.2625

h is equal to x, we found x that is equal to 1/4 √64 – 4y².2634

Now we have our h, our area as a function of y is equal to ½ the base × the height = ½ of the base2645

which we said is ½ √64 – 4y² × the height which is ¼ × √64 - 4y².2659

Our function is equal to 2 × 2 is 4.2677

We are going to get 1/16, 64 - 4y².2681

You can simplify that out a little bit more, if you want.2689

That is going to be 4 - 1/4 y².2692

Therefore, the volume, we are going to integrate from, that was this.2701

We are going to add up all the triangles.2716

The area is this, we are going to go from -4 to 4.2719

The integral from -4 to 4 of 4 - ¼ y² dy.2723

Because this is symmetric, if you want, you can also write it as twice the integral from 0 to 4.2738

This area is the same as that area, you can do it this way.2748

Changing one of the limits to 0, if you want to, it is not a big deal.2751

4 - ¼ y² dy, there you go.2757

This is what we wanted, I hope that make sense.2765

Let us do one more problem.2773

Find the volume of the solid whose base is the region bounded by the function y = 3 - x² in the x axis,2776

and whose cross section is perpendicular to the x axis are squares.2784

Volume whose base is the region bounded by the function 3 – x² in the x axis.2792

Let us go ahead and draw this out.2796

This one should be reasonably straightforward.2800

3 - x², we go 1, 2, up to 3.2803

3 - x² in the x axis, this cross section is perpendicular to the x axis.2812

We are going to be taking slices perpendicular to the x axis.2818

This is the base, are squares.2825

When I take this slice out, turn it, it is a square.2828

s and s, these points of intersection by the way, -√3 and √3.2840

If you are wondering where that came from, set this 3 – x² to 0.2851

You are going to get x = + or -√3.2857

It is going to tell me what the 0’s are, what the roots are of this equation.2860

The area is equal to side².2864

If this is x, if this is my value of x, this is my value of y.2874

Therefore, that is actually equal to y.2884

Therefore, y is equal to 3 - x², it is a function.2889

The height of the square is 3 - x².2898

My area which is equal to s² is equal to 3 - x²².2908

Therefore, the area is equal to 9 - 6x² + x⁴.2920

Our volume, we took the slice this way, we are going to be adding up all the volumes this way.2932

It is just equal to -√3 to √3 9 – 6x² + x⁴ dx.2939

I will leave the integration to you, whether you want to use your calculator or do it by hand.2956

That is all, dealing with regions with a cross section is not a circle.2961

There is no rotation, but we can still deal with it.2969

Thank you so much for joining us here at www.educator.com.2973

We will see you next time, bye.2975