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

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

## Discussion

## Download Lecture Slides

## Table of Contents

## Transcription

### Volumes III: Solids That Are Not Solids-of-Revolution

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
- Solids That Are Not Solids-of-Revolution
- Example I: Find the Volume of a Pyramid Whose Base is a Square of Side-length S, and Whose Height is H
- Example II: Find the Volume of a Solid Whose Cross-sectional Areas Perpendicular to the Base are Equilateral Triangles
- Example III: Find the Volume of a Pyramid Whose Base is an Equilateral Triangle of Side-Length A, and Whose Height is H
- Example IV: Find the Volume of a Solid Whose Base is Given by the Equation 16x² + 4y² = 64
- Example V: Find the Volume of a Solid Whose Base is the Region Bounded by the Functions y=3-x² and the x-axis

- Intro 0:00
- Solids That Are Not Solids-of-Revolution 0:11
- Cross-Section Area Review
- Cross-Sections That Are Not Solids-of-Revolution
- Example I: Find the Volume of a Pyramid Whose Base is a Square of Side-length S, and Whose Height is H 10:54
- Example II: Find the Volume of a Solid Whose Cross-sectional Areas Perpendicular to the Base are Equilateral Triangles 20:39
- Example III: Find the Volume of a Pyramid Whose Base is an Equilateral Triangle of Side-Length A, and Whose Height is H 29:27
- Example IV: Find the Volume of a Solid Whose Base is Given by the Equation 16x² + 4y² = 64 36:47
- Example V: Find the Volume of a Solid Whose Base is the Region Bounded by the Functions y=3-x² and the x-axis 46:13

### AP Calculus AB Online Prep Course

### Transcription: 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 formula*0622

*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/h*2080

*× 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 base*2645

*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

2 answers

Last reply by: Professor Hovasapian

Wed Dec 30, 2015 1:08 AM

Post by Joseph Berk on December 29, 2015

As always - great lecture, but there is a mistake here. The formula for a circle of radius 2 is not x^2 + y^2 = 2, but x^2 + y^2 = 2^2 and the answer should be equal to 1/3 Bh - which it clearly isn't.