For more information, please see full course syllabus of College Calculus: Level II

For more information, please see full course syllabus of College Calculus: Level II

### Surface Area of Revolution

**Main formula:**

Surface Area of Revolution =

**Hints and tips:**

To remember this formula, it helps to recall that the square root part comes from the arclength formula (which in turn comes from the Pythagorean Theorem). The 2

*πf (x*) part comes from the circumference of a circle, 2*πr*, where the radius*r*is the height of the curve*f (x*) that is revolving around the*x*-axis.Remember that you must integrate the square root formula above. A common mistake is to integrate the function itself, not the square root formula. Of course, this would give you the area under the curve and not the surface area of revolution.

A similar mistake is to mix this up with the arclength formula, which looks similar. Be careful which one you are asked for.

Don’t make the common algebraic mistake of thinking that reduces to

*a*+*b*! This is extremely wrong, and your teacher will likely be merciless if you do itA common technique in problems of this nature is to make a

*u*-substitution for whatever is under the square root sign. Then (hopefully) you can manipulate the expression outside the square root (which comes from*f (x*)*dx*) into being the*du*. However, you might have to do several steps of algebraic manipulation, pulling factors in or out of the square root sign, before this works.If the graph is being revolved around the

*y*-axis, simply switch the roles of*x*and*y*in the formula above. Be sure to check carefully which one the problem is asking for.When it’s feasible, check that your answer makes sense. Unlike area integrals, which can be negative if a curve goes below the

*x*-axis, surface area of revolution should always be positive!

### Surface Area 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 0:00
- Important Equation 0:05
- Surface Area
- Relation to Arclength
- Lecture Example 1 1:46
- Lecture Example 2 4:29
- Lecture Example 3 9:34
- Additional Example 4
- Additional Example 5

### College Calculus 2 Online Course

I. Advanced Integration Techniques | ||
---|---|---|

Integration by Parts | 24:52 | |

Integration of Trigonometric Functions | 25:30 | |

Trigonometric Substitutions | 30:09 | |

Partial Fractions | 41:22 | |

Integration Tables | 20:00 | |

Trapezoidal Rule, Midpoint Rule, Left/Right Endpoint Rule | 22:36 | |

Simpson's Rule | 21:08 | |

Improper Integration | 44:18 | |

II. Applications of Integrals, part 2 | ||

Arclength | 23:20 | |

Surface Area of Revolution | 28:53 | |

Hydrostatic Pressure | 24:37 | |

Center of Mass | 25:39 | |

III. Parametric Functions | ||

Parametric Curves | 22:26 | |

Polar Coordinates | 30:59 | |

IV. Sequences and Series | ||

Sequences | 31:13 | |

Series | 31:46 | |

Integral Test | 23:26 | |

Comparison Test | 22:44 | |

Alternating Series | 25:26 | |

Ratio Test and Root Test | 33:27 | |

Power Series | 38:36 | |

V. Taylor and Maclaurin Series | ||

Taylor Series and Maclaurin Series | 30:18 | |

Taylor Polynomial Applications | 50:50 |

### Transcription: Surface Area of Revolution

*I am going to do a couple more examples of surface area of revolution.*0000

*We are going to start out by rotating the graph of 2x ^{3} from x=0 to x=1 around the x axis.*0004

*So here, our f(x) is 2x ^{3}, we want to find f'(x), which is 6x^{2},*0012

*f'(x) ^{2} is 36 x^{4}.*0025

*1 + that is 36x ^{4} + 1.*0034

*The square root of 1 + f'(x) ^{2} is just the square root of 36x^{4} + 1.*0041

*The point of doing that was that you want to use your surface area formula. *0054

*That involves this big radical expression.*0057

*Our surface area is equal to the integral from x=0 to x=1, of 2pi × f(x), which is 2x ^{3}.*0060

*× the radical expression, 36x ^{4} + 1 × dx.*0079

*Now, I am going to pull the 4pi outside.*0092

*x=0 to x=1.*0097

*Key thing to notice here, this 36x ^{4} looks pretty nasty, but its derivative is exactly x^{3}.*0101

*Well not exactly x ^{3}, but x^{3} times a constant.*0110

*That suggests the substitution, u = 36x ^{4} + 1.*0115

*Then du is 4 × 36 x ^{3} dx.*0123

*We kind of have the du here when we have x ^{3} dx.*0131

*The only thing that is not quite right is the 4 × 36.*0134

*We can correct for that by dividing on the outside by 4 × 36.*0139

*Then we will have the x ^{3} dx gives you du.*0145

*We now have the square root of 36x ^{3} + 1, that is u.*0149

*So this is pi/36.*0156

*Integral of sqrt(u) is not too bad, you think of that as u ^{1/2}.*0161

*So the integral is u ^{3/2} and divide by 3/2, which is the same as multiplying by 2/3.*0169

*We want to evaluate this from x=0 to x=1.*0180

*We cannot do that directly because everything is still in terms of u.*0184

*I am going to write this as pi, I guess we can simplify the 2 and 36 to be 1 and 18.*0190

*Then 18 × 3 is 54.*0198

*Now u ^{3/2}, (36x^{4} + 1)^{3/2}, and we want to evaluate that from x=0 to x=1.*0202

*That is pi/54, now if we plug in x=1 to 36x ^{4} + 1, we will get 37 ^{3/2}.*0218

*- x=0 in there just gives you 1 ^{3/2}.*0235

*This can be slightly simplified to pi/54 ×, 37 ^{3/2} is the same as saying 37×sqrt(37) *0243

*- 1/2 ^{3/2} is just 1.*0253

*So we get our final answer there.*0258

*So, the key part of that problem is identifying y=f(x).*0266

*Then working through the formula, finding f'(x) and figuring out what the square root of 1 +f' ^{2} is.*0270

*Then plugging the whole thing into the surface area formula that we learned at the beginning of the lecture.*0280

*Then it looks like a tricky integral but the key thing is to notice that if we let u = 36 x ^{4} + 1,*0286

*We basically have our du set up for us with 2x ^{3}, we just need to correct for the constant there.*0294

*For our last example, I would like to find the surface area by rotating the graph of y=sqrt(2x) around the x axis.*0000

*Here, our f(x) is sqrt(2x), and f'(x),*0009

*If you think of f(x) is 2x ^{1/2}, then f'(x) is 1/2 × 2x^{-1/2} × the derivative of 2x,*0015

*So those 2's cancel and you get 1/sqrt(2x).*0031

*Now f'(x) ^{2} is just 1/2x, and if we add 1 to that, we get,*0038

*If we put those over a common denominator we get 2x+1/2x.*0053

*Finally the sqrt(1+f'(x) ^{2}) is sqrt(2x+1/2x).*0061

*Now we are ready to invoke our surface area of revolution formula.*0070

*The surface area of revolution is the integral on the bounds we are given are x=0 to x=1, of 2pi × f(x),*0075

*Which is the sqrt(2x).*0088

*× this big square root formula, 2x+1/2x.*0091

*All integrated with respect to dx.*0100

*So, this is actually pretty nice because the sqrt(2x) here, cancels with sqrt(2x) in the denominator there.*0106

*Now we have the integral of sqrt(2x+1), and a natural thing to do there is to let u=2x+1.*0115

*Then a du is just 2dx.*0130

*I am going to pull a 2pi outside, we get 2pi × integral from x=0 to x=1.*0135

*Of the sqrt(u) and dx converts into du, except that dx is 1/2 du.*0145

*I will put that 1/2 outside.*0154

*Sorry the integral from x=0 to x=1, of sqrt(u) du.*0157

*So, the point of that was that sqrt(u) is an easy integral.*0172

*You think of that as u ^{1/2}, and the integral of u^{1/2} is u^{3/2}/3/2 *0180

*Which is the same as multiplying by 2/3.*0193

*And we are going to evaluate that from x=0 to x=1.*0196

*But you cannot evaluate that yet because we have the thing in terms of u, and we need to convert back into x's.*0198

*We remember that u=2x+1, so we get 2pi/3, 2x+1 ^{3/2}, evaluated from x=0 to x=1.*0206

*So this is 2pi/3, now if you plug in x=1 to 2x+1, that turns into two times 1+1, so 3 ^{3/2},*0223

*- x=0 in there, into 2x+1, just gives you 1 ^{3/2}.*0240

*Again it simplifies down to 2pi/3.*0247

*3 ^{3/2} is the same as 3sqrt(3) - 1^{3/2} is just 1.*0250

*And we get our final answer.*0262

*To recap there, we identified f(x), that was y=sqrt(2x),*0265

*We plugged that into the formula to sqrt(1+f'(x) ^{2}),*0270

*Then we plugged our answer into the whole surface area formula right here.*0277

*Luckily that simplified down a little bit, then to get the integral we just needed to substitute u = 2x+1.*0283

*That finishes off that example and our lecture on surface area formulas is finished.*0292

*Today we are going to learn how to find the surface area of revolution.*0000

*What that means is that we are going to take a function y=f(x)*0006

*We are going to revolve this function around the y axis.*0016

*We are going to take this curve and spin it around the y axis and create a surface.*0022

*Then, what we are asking today is if you were going to paint that surface, how much paint would it require.*0039

*In other words, what is the surface area of the surface that you would obtain.*0045

*We have a nice formula that tells us the answer here.*0052

*The formula is the integral from x=a to x=b of 2pi × f(x) × sqrt(1 + f'(x) ^{2}) dx.*0054

*You might recognize part of this formula as something we saw in a previous lecture on arc length.*0072

*Indeed, that is not a coincidence.*0078

*The surface area formula comes from looking at a small piece of the curve,*0080

*Calculating its arc length, and then calculating what the surface area would be if we rotated that around the x axis.*0085

*Of course, that is where you get the 2pi/f(x) part of the formula.*0097

*Let us try this out with some examples.*0104

*The first example we are going to be finding the surface area of the cone*0107

*If we take the graph of y = 3x and we rotate that around the x axis from x=0 to x=2.*0112

*We would get that cone and we are trying to figure out what the surface area is.*0128

*Remember our formula is the integral of 2pi f(x) × sqrt(1 + f'(x) ^{2}) dx.*0134

*In this case, our f(x) is 3x.*0147

*So f'(x) is just 3.*0150

*1 + f'(x) ^{2} is 1 + 3^{2} which is 10.*0156

*So this part of the formula, the square root part, is sqrt of 10.*0162

*We are integrating from a to b is 0 and 2, from x=0 to x=2 of 2pi × f(x) is 3x.*0165

*× the square root of 10, dx.*0178

*That is our surface area formula.*0183

*I am going to pull the 2pi × 3 to the outside.*0188

*That gives us 6pi.*0192

*I will pull the sqrt(10) outside as well, that is the sqrt(10).*0197

*The integral of x is x ^{2}/2,*0203

*And we want to evaluate that from x=0 to x=2.*0205

*That gives us 6pi × sqrt(10) × if you plug in x=2 there, you get 4 - 0.*0212

*Sorry, 4/2 - 0.*0228

*That is 6pi × sqrt(10).*0232

*4/2 is just 2 so 2 - 0.*0236

*The whole answer is 12pi × sqrt(10).*0239

*Key points here, we are rotating something around the x axis.*0247

*We look at the f(x) that we are given and we calculate this square root formula by doing f'(x) and then 1 + f' ^{2}.*0253

*We plug it all in to the surface area formula and then do the integration to finish that off.*0262

*Let us try another example.*0270

*This time we are rotating the graph of y = x ^{3} around the x axis.*0272

*Our f'(x) is 3x ^{2}.*0279

*Our f' ^{2} is 9x^{4}*0288

*1 + f' ^{2} is 1 + 9x^{4}.*0297

*And, the square root of that is sqrt(9x ^{3} + 1).*0304

*So, our surface area is the integral from x=0 to x=1 of 2pi f(x).*0313

*So, I put 2 pi, now f(x) is x ^{3}, so 2pix^{3} × integral of 9x^{4} + 1 dx.*0326

*Now, this looks like a rough integral but it is actually not so bad.*0342

*If you notice, you have 9x ^{4} + 1 under the radical.*0348

*The derivative of that would 36x ^{3}, and we have an x^{3} outside.*0352

*What we can do is let u = 9x ^{4} + 1.*0358

*Then du is 36x ^{3} dx.*0368

*So, what this converts into is, I guess I can write dx or x ^{3} dx, is 1/36 du.*0376

*That takes care of the dx and the x ^{3}*0390

*I will put the 1/36 outside.*0394

*I am also going to put the 2pi outside.*0398

*We still have the integral from x=0 to x=1.*0403

*Now, 9x ^{4} + 1 just turned into u, and then we have du.*0408

*That is really a very simple integral now.*0416

*This is now pi/18, u ^{1/2} is the same as sqrt(u).*0420

*If we integrate that, that integrates to u ^{3/2}/3/2 or, 2/3u^{3/2}.*0429

*We are evaluating this not using u but using x=0, not x=1.*0442

*So let us keep going with that.*0450

*We still have a pi.*0452

*2/3 I can cancel that with the 18 to get 9 here and a 1 there.*0455

*Pi over 27.*0461

*Now u was 9x ^{4} + 1, and we are evaluating that from x=0 to x=1.*0468

*We get pi/27.*0480

*Now if you plug in x=1, I am sorry I left out my 3/2 there.*0485

*That 3/2 came from that right there.*0490

*Then we have 9x ^{4} + 1.*0495

*If we plug in 1 to 9x ^{4} + 1, that is 10^{3/2}*0500

*X=0, if you plug it in just gives you 1 - 1 ^{3/2}.*0507

*This gives us pi/27 × 10 ^{3/2}, is the same as saying 10 sqrt(10), - 1.*0516

*That is the answer for the surface area of this graph we rotated around the x axis.*0535

*Again, the key point there is identifying your f(x)*0543

*Running it through this formula, plugging it into the general surface area formula*0548

*Then, we had a pretty tricky integral there.*0554

*The key thing there was observing that the derivative of 9x ^{4} + 1 was more or less x^{3},*0556

*So we could make this substitution that gave us a very nice integral to solve.*0568

*Let us try another example*0575

*This time we are rotating the graph of y = 2x ^{2} + 1 from x=0 to x=1.*0578

*There is a very important difference in this example which is we are rotating around the y axis.*0584

*Remember that our surface area formula that we learned before used the x axis.*0591

*We had the x axis before, that means we need to convert our surface area formula*0605

*To adapt to the fact that we are now rotating around the y-axis.*0611

*That means we need to look at the surface area formula and sort of switch everything from x's to y's.*0615

*Let me write that down.*0621

*The surface area is now the integral from y =, I will not write a and b, c to y = d, of 2pi f(y).*0625

*× sqrt(1+f'(y)) dy.*0640

*That means we need to convert everything here into functions and terms of y.*0648

*Including our function and the limits.*0654

*We were given this as if y were a function of x.*0659

*Instead we need to convert everything to x as a function of y.*0663

*Let us go ahead and convert that.*0667

*Looking at the function first, we get y-1 = 2x ^{2} *0669

*We can divide both sides by 2 there, so x = sqrt(y-1/2).*0679

*If you plug in x=0, into the function,*0688

*That would convert into y = 1.*0693

*If you plug x=1 again into the function, *0700

*That would convert into y=3.*0704

*Now we have everything in terms of y.*0710

*Again, let us try to figure out what f'(y) and what this square root formula turns into.*0713

*f'(y), well we have got a square root of something*0722

*So its derivative is 1/2*0728

*Because we think of its square root as all of that stuff to the 1/2.*0732

*1/2 times all of that stuff to the -1/2*0736

*I will write it down here in the denominator, y-1/2*0739

*× the derivative of that inside stuff by the chain rule*0745

*The derivative of y-1/2 is 1/2.*0750

*Let me try to simplify this.*0754

*I am going to put the two 1/2's together and put 1/4.*0757

*Now this denominator, I am going to flip it and bring it up to the numerator*0760

*That is 2/y-1.*0765

*That is f'*0770

*We need to find f' ^{2}.*0772

*f' ^{2} is 1/16 × 2/y-1.*0776

*We can simplify that into 1/8 × y-1.*0782

*1 + f'(y) ^{2} is, 8 × y-1, 8y-8+1/8y-1.*0790

*What I did there is I wrote that 1 as 8y-8/the denominator.*0809

*8y-8, because I wanted to combine everything over a common denominator.*0818

*I can simplify that down a little bit into 8y-7/8(y-1).*0826

*Now we need to take the square root of that.*0837

*What we have is the square root of 1 + f'(y) ^{2}.*0840

*Is the sqrt(8y-7/8(y-1)).*0853

*I am going to put all of these pieces together into a big integral. *0863

*The integral is the surface area from y=1 to y=3.*0867

*Got those from here and here.*0874

*2pi f(y), I got that from here, that is the sqrt(y-1/2) × sqrt(1+f'(y ^{2})),*0878

*That is 8-7/8(y-1) × dy.*0894

*That looks pretty messy but it does simplify.*0905

*The y-1's cancel and then we are left with a 2 and an 8 in the denominator, that is 16.*0906

*We can pull that out of the square root and that is a 4 in the denominator,*0913

*And so this simplifies down a little bit.*0919

*2/4 gives us 1/2, I will pull the pi outside.*0922

*So, we get the integral from y=1 to y=3 of square root of, I think the only thing that is left there is 8y-7 dy.*0930

*Now we can use u = 8y-7 so du = 8 dy.*0945

*dy is 1/8 du.*0955

*What we get now is pi/2 × the 1/8 that we got from the du here.*0962

*The integral from y=1 to y=3 of u ^{1/2} du.*0974

*That is pi/2 × 1/8 u ^{1/2}*0986

*The integral of u ^{1/2} is u^{3/2}/3/2, which is the same as multiplying by 2/3.*0993

*And, we want to evaluate that from y=1 to y=3.*1004

*I can simplify a little bit.*1008

*My two's cancel, and we get pi, combine the 3 and the 8 to give 24.*1010

*That is pi/24*1019

*Now, u was 8y-7 ^{3/2}*1022

*Finally, we get pi/24 × *1034

*Now, if we plug in y=3 to 8y-7.*1044

*That is 24-7, which is 17 ^{3/2}*1047

*If we plug in y=1 to 8y-7 which is 8-7, so -1 ^{3/2}*1055

*We can clean that up a little bit into pi/24 × 17 sqrt(17).*1063

*1 ^{3/2} is just 1.*1075

*Our final answer is pi/24 × 17 sqrt(17) -1.*1080

*Probably what made that problem a little difficult besides it being a little bit complicated on the algebraic side.*1085

*Was the fact that we were given the y axis instead of the x axis*1093

*Which means you have to take your original x formula and translate everything into terms of y.*1098

*In turn, you have to translate things from y as a function of x to x as a function of y.*1106

*The x values that you are given have to be converted into y values.*1112

*Once you do that, you walk through the process of calculating this radical 1 + x'(y) ^{2}.*1122

*Then it is still kind of a messy integral but the square roots sort of cancel nicely *1133

*And, you end up with something that is not too complicated at the end.*1136

1 answer

Last reply by: Dr. William Murray

Fri Apr 15, 2016 5:20 PM

Post by Gautham Padmakumar on April 13 at 05:20:28 PM

At 10:43, I noticed that you made a mistake when you wrote down the integral formula about the y axis. You left it as f'(y) instead of (f'(y))^2

Thanks for the lecture series by the way! It really helped me review for my calc finals

1 answer

Last reply by: Dr. William Murray

Tue Feb 11, 2014 4:18 PM

Post by Katrina Forrest on February 8, 2014

Professor Murray,

I just want to comment on the brilliancy of your lectures. You present the material in a way that is structured, concise, and comprehendible. I wish more of my instructors followed your style of teaching!

1 answer

Last reply by: Dr. William Murray

Fri Sep 6, 2013 12:35 PM

Post by A De Lama on September 2, 2013

WOnderful lectures, by the way. This really is the future of education!!

1 answer

Last reply by: Dr. William Murray

Fri Sep 6, 2013 12:34 PM

Post by Anubis De Lamali on September 2, 2013

when solving for x in terms of y, why is it not "+ or -" in front of the square root?

THanks!!