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

## Discussion

## Study Guides

## Practice Questions

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Polar Coordinates

**Main formula:**

Area =

Arclength =

**Hints and tips:**

Practice drawing lots of polar graphs − they do get easier with time.

Remember that when

*r*is negative, you go to the opposite side of the graph.Remember that each point has many different possible sets of polar coordinates. (This is different from rectangular coordinates, where each point has a unique (

*x*,*y*) pair.)To find the limits, you often have to draw the graph and find out what angle

*θ*makes*r*= 0, or when the graph comes back to meet itself.In finding areas, you often have to integrate

*sin*²*θ*or*cos*²*θ*. To integrate these, use the half-angle formulas from the section on trigonometric integrals.To find the area between the two graphs, subtract the two area formulas just as you would with rectangular coordinates. To find the limits, set the two functions equal to each other and solve for the angles

*θ*.To remember the arclength formula, it helps to recall that it comes from the distance formula between two points, which in turn comes from the Pythagorean Theorem.

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 itWhen it’s feasible, check that your answers make sense. Unlike area integrals in rectangular coordinates, which can be negative if a curve goes below the

*x*-axis, areas and arclengths in polar coordinates should always be positive. You might also be able to check geometrically that the area or length of your curve looks approximately right.

### Polar Coordinates

- Apply definitions for r and θ
- √{x
^{2}+ y^{2}} = 3tan(tan^{ − 1}([y/x]) - Reduce tangent and arctangent function
- √{x
^{2}+ y^{2}} = 3[y/x]

^{2}+ y

^{2}} = 3[y/x] (√{x

^{2}+ y

^{2}} )[x/y] = 3 [(x√{x

^{2}+ y

^{2}} )/y] − 3 = 0

^{3}= − 7cosθ

- Apply definitions for r and θ
- (√{x
^{2}+ y^{2}} )^{3}= − 7sin(tan^{ − 1}([y/x])) - Use trig identity
- (√{x
^{2}+ y^{2}} )^{3}= − 7([x/(√{x^{2}+ y^{2}} )]) - Isolate for Cartesian form
- (√{x
^{2}+ y^{2}} )^{3}= − 7([x/(√{x^{2}+ y^{2}} )]) - (√{x
^{2}+ y^{2}} )^{3}([(√{x^{2}+ y^{2}} )/y]) = − 7

^{2}+ y

^{2}} )

^{2}(√{x

^{2}+ y

^{2}} )

^{2})/x] = − 7[((x

^{2}+ y

^{2})(x

^{2}+ y

^{2}))/x] = − 7[((x

^{2}+ y

^{2})

^{2})/x] + 7 = 0

- Apply definitions for r and θ
- √{x
^{2}+ y^{2}} = cos(tan^{ − 1}([y/x]))sin(tan^{ − 1}([y/x])) - Apply trig identities
- √{x
^{2}+ y^{2}} = ([x/(√{x^{2}+ y^{2}} )])([y/(√{x^{2}+ y^{2}} )]) - √{x
^{2}+ y^{2}} = [xy/(x^{2}+ y^{2})] - √{x
^{2}+ y^{2}} ([(x^{2}+ y^{2})/xy]) = 1 - [(
^{3}√{x^{2}+ y^{2}})/xy] = 1

^{3}√{x

^{2}+ y

^{2}})/xy] − 1 = 0

- Apply definitions for r and θ
- √{x
^{2}+ y^{2}} = sec(tan^{ − 1}([y/x])) - Apply trig identities
- √{x
^{2}+ y^{2}} = [1/(cos(tan^{ − 1}([y/x])))] - √{x
^{2}+ y^{2}} = [1/([x/(√{x^{2}+ y^{2}} )])]

^{2}+ y

^{2}} ([x/(√{x

^{2}+ y

^{2}} )]) = 1x = 1x − 1 = 0

^{2})/25] + y

^{2}= 1

- Apply definitions
- [((rcosθ)
^{2})/25] + (rsinθ)^{2}= 1 - Expand
- [(r
^{2}cos^{2}θ)/25] + [(r^{2}sin^{2}θ)/1] = 1 - [(r
^{2}cos^{2}θ)/25] + [(25(r^{2}sin^{2}θ))/25] = 1 - [(r
^{2}cos^{2}θ+ 25(r^{2}sin^{2}θ))/25] = 1 - Simplify
- [(r
^{2}(cos^{2}θ+ 25r^{2}sin^{2}θ))/25] = 1 - r
^{2}= [25/(cos^{2}θ+ 25r^{2}sin^{2}θ)]

^{2}θ+ 25r

^{2}sin

^{2}θ)]}

- Apply definitions
- rcosθ+ [((rsinθ))/36] − 1 = 3
- Isolate and expand
- rcosθ+ [((rsinθ))/36] = 4
- [(36(rcosθ) + rsinθ)/36] = 4
- Simplify
- r(36cosθ+ sinθ) = 144

- Apply definitions
- 3(rcosθ) + 5(rsinθ) = 17
- Simplify
- r(3cosθ+ 5sinθ) = 17

- Apply definitions
- [(rsinθ)/(rcosθ)] − 16( rsinθ ) = 9
- Simplify

- Apply definitions
- [(rsinθ)/4] + rcosθ− 2 = 0
- Isolate and simplify
- [(rsinθ)/4] + rcosθ = 2
- r([(sinθ)/4] + cosθ) = 2
- r([(sinθ+ 4cosθ)/4]) = 2

- Make a table for plot points
θ r (θ,r) 0 0 (0,0) [(π)/4] [(3√2 )/2] ([(π)/4],2.12) [(π)/2] 3 ([(π)/2],3) [(3π)/4] [(3√2 )/2] ([(3π)/4],2.12) π 0 (π,0) [(5π)/4] − [(3√2 )/2] ([(5π)/4], − 2.12) [(3π)/2] − 3 ([(3π)/2], − 3) [(7π)/4] − [(3√2 )/2] ([(7π)/4], − 2.12 2π 0 (2π,0)

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Polar Coordinates

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 Equations 0:05
- Polar Coordinates in Calculus
- Area
- Arc length
- Lecture Example 1 2:14
- Lecture Example 2 4:12
- Lecture Example 3 10:06
- 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: Polar Coordinates

*OK, let us try another area problem.*0000

*We want to find the area inside the graph of r = 8sin(θ) and outside the graph of r = 4.*0003

*This one is a little tricky because we have not been given limits of integration,*0011

*And we have been given two different functions.*0016

*Somehow, we have to figure out what those functions look like.*0018

*Also, figure out the limits of integration ourselves.*0022

*r = 4 is probably the easier one, so I will start with that.*0027

*That is just all points at radius 4 from the origin.*0031

*That is just a circle.*0036

*r = 8sin(θ) is a little trickier.*0039

*If you graph r = 8sin(x), you get an exaggerated sin curve.*0042

*That goes up to 8 and down to -8.*0053

*Comes back to 0 at π, and again at 2π.*0061

*If you graph r = 8 sin(θ),*0066

*That starts out when θ = 0, it starts out at 0.*0072

*When θ = π/2, it goes up to 8.*0084

*That is 8.*0092

*When θ = π, it comes back to 0.*0094

*When θ is 3π/2, it is gone to -8.*0101

*It is just, let me show this in blue, retracing itself again.*0110

*When θ gets to 2π, it has come back to 0.*0116

*That is what the graph of r = 8sin(θ) is.*0131

*Its intercept on the π/2 axis is 8.*0140

*I want to show those 2 graphs together.*0144

*Remember r = 4 had a radius of 4.*0146

*Let me put that graph on there.*0150

*We are actually looking at two circles here.*0156

*We want to find the area inside the graph of r = 8sin(θ)*0162

*Outside the graph of 3=4.*0167

*Let me try to color that area in, in red.*0170

*That is the area that we are trying to find.*0175

*This means we have to figure out where those two graphs intersect.*0177

*Then integrate from 1 intersection point to the other.*0185

*To figure out where those two graphs intersect, I am going to set them equal to each other.*0193

*4 = 8sin(θ).*0197

*That means 1/2 = sin(θ).*0203

*There are two angles that have sin = 1/2.*0207

*That means θ = π/6, and 5π/6.*0210

*You can kind of see that in the picture.*0216

*This is π/6.*0218

*This is 5π/6.*0221

*What we are going to do is integrate using our area formula.*0225

*I think what I am going to do is just find 1/2 of this area.*0235

*I am just going to find that segment.*0240

*I will color that in black.*0243

*I will find that area and then multiply it by 2, then integrate it from π/6 to π/2.*0245

*Now I want to find the outside area - the inside area.*0256

*Well our formula for area is f(θ) ^{2}/2.*0262

*Here the outside area is 8sin(θ).*0267

*The outside curve is 8sin(θ),*0269

*So I am going to find 8sin(θ) ^{2}/2, that is the outside area,*0272

*- the inside area which is just 4 ^{2}/2.*0284

*That is what I am going to integrate.*0291

*That is nice because the 2's cancel right away.*0294

*So we get the integral from π/6 to π/2 of 8 sin(θ) ^{2},*0298

*So that is 64sin ^{2}(θ) - 4^{2} is 16.*0303

*Remember the 2's cancel on the outside d(θ).*0313

*That is the integral of sin ^{2}(θ),*0319

*We are going to use a half-angle formula to deal with that.*0323

*Remember sin ^{2} is 1 - cos(2θ)/2.*0324

*So, 64sin ^{2}(θ) is 32 × 1 - cos(2θ).*0330

*We still have - 16 dθ.*0340

*Here we have 32 - 16.*0347

*I will put those together and get a 16 - 32 cos(2θ), just 16(θ).*0352

*Now the integral of cos(2θ) is sin(2θ)/2.*0362

*The integral of 32cos(2θ) is 16sin(2θ).*0370

*We said we wanted to integrate this from θ = π/6 to θ = π/2.*0378

*We will plug those bounds in, and we get 16 × π/2 minus,*0388

*Now, if we plug π/2 into sin(2θ),*0397

*That is giving us sin(π), and that is just 0, - 16/π/6.*0401

*Which is 16 × π/6 for now + 16 × sin(2) × π/6.*0413

*Which is the sin(π/3).*0423

*We can simplify this.*0428

*π/2 - π/6 = π/3.*0430

*If we combine those two terms, we get 16 π/3.*0433

*Now, the sin(π/3) is sqrt(3)/2.*0441

*So 16 × sin(π/3) = 8×sqrt(3).*0447

*We get our answer is 16 π/3 + 8×sqrt(3).*0460

*This problem was pretty tricky but it was really the geometry that made it tricky and not the calculus.*0465

*We were given two curves and we were not really told how to handle them.*0470

*The good thing to do is to start out by graphing the two curves, putting them together on the same graph,*0475

*Then trying to find these intersection points.*0480

*We actually found the intersection point algebraically by setting the two equation equal to each other.*0486

*Once we found the intersection points, those gave us the bounds on the integrals.*0493

*Then we used this area formula.*0498

*Remember the area was f(θ) ^{2}/2 dθ.*0500

*Since we had to find the area between two curves, we looked at the area of the outside one - the area of the inside one.*0505

*Then we worked out the integration and we got our answer.*0514

*So our last example is another length of the polar curve r = 3+3cos(θ).*0000

*Again, this is one where we are not given limits of integration.*0008

*We have to figure it out by ourselves.*0011

*The way we are going to figure it out is by graphing the thing.*0014

*So, the way we can graph it is we can plug in some easy values of θ.*0028

*When θ = 0, cos(θ) = 1 so this is 6.*0033

*When θ = π/2, cos(θ) has gone down to 0, so we are just talking about r = 3.*0043

*By the time θ = π, cos(θ) = -1, so that has gone down to 0.*0054

*At 3π/2, cos(θ) is back up to 0, so r has come back up to 3.*0063

*At 2π, cos(θ) is back up to 1.*0073

*So 3 + 3cos(θ) is back up to 6.*0084

*That is the curve we are looking at.*0088

*We are trying to find the length of that curve.*0091

*I think it will be easier if we just find the length of the top part of this curve, the part I am showing in red right now.*0096

*So, we are actually going to integrate from 0 to π, and then we will multiply our final answer by 2.*0100

*We are going to do 2 × the integral from 0 to π.*0110

*Now, remember we have to use the arc length formula, which is this Pythagorean Formula.*0120

*We take sqrt(f'(θ) ^{2} +f(θ)^{2}).*0126

*f'(θ) is very easy.*0139

*The first 3 goes away because it is just a constant.*0143

*The derivative of 3cos(θ) is just -3sin(θ).*0147

*If we square that out we get 9sin ^{2}(θ).*0156

*f(θ) ^{2} is a little bit more messy.*0162

*We have to square out 3 + 3cos(θ).*0164

*That is 9 + 3 × 3 × 2 is 18 cos(θ) + 9cos ^{2}(θ) dθ.*0168

*That looks like it is going to be a pretty messy integral,*0187

*But good things happen here because we have 9sin ^{2}(θ) and 9cos^{2}(θ)*0190

*Those combine just to be 9.*0196

*We get 2 × integral from 0 to π of square root of,*0198

*Well the cos ^{2} and the sin^{2} give us 9,*0206

*And we have another 9 right there,*0209

*So that is just 18 + 18cos(θ) dθ.*0212

*Now, I can simplify this a little bit if I pull out the sqrt(18), so 2 sqrt(18) × integral from 0 to π.*0223

*Of 1 + cos(θ) dθ.*0236

*This is kind of a tricky integral.*0242

*There is a trick for doing integrals like this and it may not be completely obvious.*0246

*The trick is to remember the half angle formula.*0252

*Let me write that over here.*0255

*cos ^{2}(u), remember that that is 1 + cos(2u)/2.*0258

*If I solve that for cos(u), that says that cos(u) = sqrt(1+cos(2u)/2).*0268

*That is pretty nice because we have the sqrt of 1 + cos(θ).*0281

*We almost have this half angle formula set up for us.*0286

*All we have to do is let the θ be 2u.*0290

*dθ = 2du.*0297

*2 sqrt(18), what I would like to do is create this formula exactly.*0310

*I am going to put a sqrt(2) in here.*0322

*Then, multiply on a sqrt(2) on the outside.*0325

*So we get the sqrt(18) × sqrt(2).*0330

*Now the integral of 0 to π, of 1 + cos,*0336

*Now θ is 2u so this is 2u/2.*0340

*dθ is 2 du.*0347

*Let me simplify this a bit.*0355

*I am going to combine this 2 and this 2 and get a 4.*0357

*sqrt(18) × sqrt(2) is sqrt(36).*0363

*Now we have the integral from,*0365

*This 0 and π, those were values for θ*0374

*If θ = π, then u, since θ = 2u, u will be π/2.*0379

*If θ = 0, then 0 will also be 0.*0388

*In terms of u, this is the integral from u = 0 to u = π/2.*0394

*Now, sqrt(1+cos(2u)/2), the whole point of making this substitution was that that would convert into cos(u).*0400

*This is just cos(u).*0409

*This 2 I already moved to the outside du.*0413

*That took what looked like a very difficult integral, and converted it into a very easy one.*0419

*The sqrt(36) is just 6, so this is 24 × integral of cos(u) is sin(u).*0423

*We are evaluating this from u=0 to u=π/2.*0434

*That just gives 24 × sin(π/2) = 1 - sin(0) = 0.*0440

*Our final answer there is 24.*0452

*Again this qualifies as probably a pretty tricky polar coordinate problem.*0457

*What we are given there is just a polar curve.*0460

*And we were not given anything about any boundaries.*0465

*We had to first draw the graph of the curve and figure out what we were dealing with.*0470

*Once we draw the graph of the curve, it is not too hard to find out which part of the curve we want to integrate,*0479

*And multiply by an appropriate factor to set up our integral.*0485

*Then we used the formula for arc length, which is this Pythagorean Formula,*0492

*With sqrt(f'(θ) ^{2} + f(θ)^{2}).*0500

*We plugged that into the integral, simplified it a little,*0505

*And then we discovered that what looked like a very difficult integral, sqrt(1+cos(θ),*0511

*But in order to handle that, we remembered this half angle formula, cos ^{2}(u) = 1 + cos(2u)/2.*0516

*We did a little manipulation to get the integral to match the half angle formula,*0527

*And once we did, it turned into a fairly easy integral in terms of u.*0533

*Then it was easy to finish and get the entire length of the curve.*0540

*Hi, this is educator.com and we are here to talk about polar coordinates.*0000

*The idea of polar coordinates is that we are not going to keep track of things in terms of x's and y's anymore.*0006

*Instead, we are going to keep track of points in terms of the radius r and the angle θ.*0014

*Every point now will have coordinates in terms of r and θ.*0025

*We will talk about functions r = f(θ).*0030

*There are sort of two places that calculus comes in in polar coordinates.*0037

*If you have a function, here I am talking about plugging in different values of θ and getting different values of r as your output.*0042

*There are two things you might be interested in calculating.*0054

*One is the area, before we calculate under, here it makes more sense to talk about calculating the area inside a curve.*0060

*The equation we have for that is the integral from, these are values of θ, θ = a to θ = b.*0070

*Of f(θ ^{2}/2).*0085

*If you are given f(θ), this f(θ) is just whatever the r is, you square that and divide by 2.*0090

*Then you take the integral with respect to θ.*0098

*The second thing we are interested in calculating is the arc length.*0103

*What is the length of that curve?*0111

*The way you figure that out is you find f(θ ^{2}), f'(θ^{2}).*0114

*Then you use the Pythagorean theorem, you add them, take their square root,*0122

*And integrate from θ = a to θ = b.*0127

*Let us try those out using some examples.*0135

*The first example is the area inside the graph of r = θ, from θ going to 0 to pi/2.*0138

*If r = θ, then when θ = 0, r is just 0 but as θ increase to pi/2, r gradually increases.*0145

*We are trying to find that area there.*0160

*Let us work it out.*0165

*Our formula for the area is f(θ ^{2}/2) dθ*0170

*In this case, our f(θ) is just θ itself, so this is the integral of θ = 0 to pi/2.*0180

*Of θ/2 dθ.*0192

*The integral of θ ^{2} is θ^{3}/3*0194

*This whole thing is theta ^{3}/3/2, so over 6, evaluated from θ = 0 to θ = pi/2.*0196

*That is just pi/2 ^{3}/6.*0211

*Now, 2 ^{3} is 8 and 8 × 6 is 48 so our answer is pi^{3}/48.*0220

*That represents that area inside that curve.*0226

*To recap here, what we did was look at the function we were given,*0235

*R = f(θ), and then we just plugged it into this integral formula f(θ ^{2})/2.*0242

*Then we worked out the integral.*0249

*Let us try that out with a slightly harder example.*0252

*We want to find the area inside one loop of the graph of r = cos(2θ).*0255

*Perhaps the first thing that makes this example difficult is we have not been told what the boundaries of θ are.*0260

*We really need to look at a graph to figure this out.*0266

*What does a graph of cos(2θ) look like.*0273

*Well if we graph cos(x) to warm up, it starts at 1, it goes down to -1, then it comes back to 1 at 2pi.*0276

*If we graph y = cos(2x), we get a graph with the same basic shape, but it oscillates twice as fast.*0290

*If that is pi, and that is 2pi, it does a complete period in the space of pi, and then another complete period by 2pi.*0305

*If we graph r = cos(2θ), we take that graph and we wrap it around a circle.*0320

*In the sense that when θ = 0, we start out at 1 at radius 1.*0330

*By the time θ gets to be pi/4, it has gone to radius 0.*0341

*This goes down to radius 0.*0350

*At pi/2, r = -1 so you know that if pi/2 is up here, the radius has gone down to -1.*0353

*It comes back to 0.*0365

*Now we are graphing this part of the graph.*0368

*By the time we get up to pi it has come up to 1 again, so here is pi in this direction.*0370

*Then between pi and 5pi/4 it goes back down to 0.*0379

*That is graphing that part of the graph and if we graph this part of the graph going from pi/4 to 3pi/2,*0387

*It is negative so the graph ends up here and then it spirals back down to 0.*0400

*And back to 1 again when it comes back to 2pi.*0412

*We get this interesting 4-leaf clover and we are trying to find the area inside one of those loops.*0417

*A good way to do it might be to find that area inside there, inside half of one of those loops.*0425

*That is between θ = 0 and θ = pi/4.*0432

*Then we will multiply our answer by pi/2.*0441

*Our area is 2 × the integral from 0 to pi/4.*0445

*Remember f(θ ^{2})/2 so our f(θ) = cos(2θ) this is cos^{2}(2θ)/2 dθ*0454

*Our 2's will cancel, that is convenient, so we get the integral from 0 to pi/4.*0468

*Remember how to integrate cos ^{2} of something.*0475

*You write that as 1+cos(2 × that thing), but the thing is 2θ, so this is actually 1 + cos(4θ).*0478

*All of that over 2.*0484

*So this 2 over here was not the 2 above, that cancelled, but this came from the half angle formula.*0492

*Now we integrate this with respect to θ.*0497

*I am going to pull the 1/2 outside now, we get 1/2.*0500

*Now we have to integrate 1 + cos(4θ).*0505

*The integral of 1 is just θ + the integral of cos(4θ) is just sin(4θ)/4.*0509

*Then we evaluate that from θ = 0 to θ = pi/4.*0525

*Then we get 1/2 of pi/4.*0540

*Plugging in θ = pi/4 + 1/4 sin(4θ).*0545

*Sin(4θ) is sin(pi) which is just 0.*0553

*- plug in θ = 0, we get 0 - sin(4×0) is just 0 again.*0557

*Our final answer is just pi/8.*0566

*The tricky part there is that we were not given the boundaries of integration.*0570

*We really had to look at the graph of r = cos(2θ)*0580

*Then interpret from the graph what useful boundaries of integration we could use to find the area.*0584

*Once we found the boundaries of integration, we just plugged it into the formula f(θ ^{2})/2.*0590

*Then we worked out the integral and it was not too bad.*0600

*We will try some more examples later, this is educator.com.*0602

*Another example is the length of the polar curve, r = e ^{5θ} as θ goes from 0 to 2pi.*0610

*Fortunately we have been given the limits of integration.*0618

*We are going to set up our arc length formula.*0623

*Which remember is f'(θ ^{2}) + f(θ)^{2} dθ.*0629

*Our f'(θ) is well, r = e ^{5θ}.*0640

*f'(θ) would be (5e ^{5θ})^{2} + just f(θ^{2}) so that is (e^{5θ})^{2}.*0650

*We want to square root that and integrate it.*0665

*5 squared is 25, e ^{5θ})^{2} is e^{10θ}.*0670

*Square root that and integrate it.*0685

*This is sqrt(26e ^{10θ}).*0690

*I am going to pull the sqrt of 26 all the way out of the integral.*0704

*Then we have the sqrt(e ^{10θ}) which is just e^{5θ}.*0710

*We want to integrate this from θ = 0 to 2pi.*0718

*Now the integral of e ^{5θ} is just e^{5θ}/5*0728

*We want to evaluate that from θ = 0 to 2pi.*0742

*I can write this as sqrt(26/5) ×*0750

*If you plug 2π into e ^{5θ} you get e^{10π}.*0755

*If you plug 0 into e ^{5θ} you get e^{0}, which is just 1.*0763

*That is our answer for the arc length.*0770

*The key to that problem is recognizing that it is a length problem and then going to the arc length formula.*0776

*Which is this Pythagorean formula f'(θ) ^{2} + f(θ)^{2} square of that and then integrate it.*0784

*We use the f(θ) that we are given, work it through, and then plug in the limits that we are given.*0792

1 answer

Last reply by: Dr. William Murray

Mon Mar 20, 2017 4:57 PM

Post by Rohan Asthir on March 19 at 10:44:43 PM

Around 1:46-1:55 in additional example 4. You said at 3pi/2 it goes to -8, but you show it goes to +8?

3 answers

Last reply by: Dr. William Murray

Thu Jun 9, 2016 1:24 PM

Post by Silvia Gonzalez on June 5, 2016

Thank you again for your classes, it is clear they have been planned and they are very clear.

I have been doing the practice exercises and I am a little confused. On one side, the answers (steps) are not matched with the questions, but that can be solved paying attention. However I do not know if it is a repeated mistake or if it is something I did not understand, but when the area enclosed by a curve in polar form is asked, the first step shown is to find the derivative of the function and then they multiply it by the function squared and divided by two. It looks like a mixture of the area for parametric and polar forms. Could you please clarify this point for me?

1 answer

Last reply by: Dr. William Murray

Wed Aug 14, 2013 12:29 PM

Post by JASON WENZEL on August 3, 2013

Sir,

How did you determine that pi/4 would be the angle of measurement to use in example II. I am guessing that it is because that is where the two loops meet?

1 answer

Last reply by: Dr. William Murray

Mon Apr 8, 2013 8:07 PM

Post by Totaram Ramrattan on April 8, 2013

this video does not play..help

1 answer

Last reply by: Dr. William Murray

Tue Dec 18, 2012 2:25 PM

Post by Adrian Khaskin on December 17, 2012

In example 3 i believe its -1/5, not -1

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Last reply by: Dr. William Murray

Tue Dec 18, 2012 2:31 PM

Post by brandon dat on January 23, 2012

in example 4, isn't the limits for integration supposed to be (pie/6) to 0?

2 answers

Last reply by: Dr. William Murray

Wed Aug 22, 2012 1:35 PM

Post by VIncent Maguire on May 24, 2011

Where did the coefficient of 2 on the outside of the Integral come form in example 2? Its not from the formula for area.