For more information, please see full course syllabus of Multivariable Calculus

For more information, please see full course syllabus of Multivariable Calculus

### Polar Coordinates

- The conversion from rectangular coordinates to polar coordinates uses θ = tan
^{ − 1}[y/x] and r = √{x^{2}+ y^{2}} . - For x = 0 and y = 1 we have r = √{0
^{2}+ 1^{2}} = 1 and θ = tan^{ − 1}[1/0] = 0 (radians).

- The conversion from rectangular coordinates to polar coordinates uses θ = tan
^{ − 1}[y/x] and r = √{x^{2}+ y^{2}} . - For x = 1 and y = √3 we have r = √{1
^{2}+ ( √3 )^{2}} = 2 and θ = tan^{ − 1}[(√3 )/1] = [(π)/3] (radians).

- The conversion from rectangular coordinates to polar coordinates uses θ = tan
^{ − 1}[y/x] and r = √{x^{2}+ y^{2}} . - For x = [1/2] and y = − [(√3 )/2] we have r = √{( [1/2] )
^{2}+ ( − [(√3 )/2] )^{2}} = 1 and θ = tan^{ − 1}[( − √3 \mathord/ phantom √3 2 2)/(1 \mathord/ phantom 1 2 2)] = [(5π)/3] (radians).

- The conversion from polar coordinates to rectangular coordinates uses x = rcosθ and y = rsinθ .
- For r = 1 and θ = [(π)/4] we have x = 1cos( [(π)/4] ) = [(√2 )/2] and y = 1sin( [(π)/4] ) = [(√2 )/2].

- The conversion from polar coordinates to rectangular coordinates uses x = rcosθ and y = rsinθ .
- For r = 5 and θ = − [(π)/2] we have x = 5cos( − [(π)/2] ) = 0 and y = 5sin( − [(π)/2] ) = − 5.

- The conversion from polar coordinates to rectangular coordinates uses x = rcosθ and y = rsinθ .
- For r = 2 and θ = [(7π)/6] we have x = 2cos( [(7π)/6] ) = − √3 and y = 2sin( [(7π)/6] ) = − 1.

^{2})/4] + [(y

^{2})/9] in terns of r(θ).

- Recall that in polar coordinates x = rcosθ and y = rsinθ .
- Substitution yields 1 = [((rcosθ)
^{2})/4] + [((rsinθ)^{2})/9]. We now solve for r so 36 = 9r^{2}cos^{2}θ+ 4r^{2}sin^{2}θ = r^{2}(9cos^{2}θ+ 4sin^{2}θ).

^{2}= [36/((9cos

^{2}θ+ 4sin

^{2}θ))] and so r(θ) = [6/(√{9cos

^{2}θ+ 4sin

^{2}θ} )]. Note that in polar coordinates the pi m affects the orientation of the curve.

^{2}+ y

^{2}} )] in terns of r(θ).

- Recall that in polar coordinates x = rcosθ and y = rsinθ .
- Substitution yields 1 = [1/(√{(rcosθ)
^{2}+ (rsinθ)^{2}} )] = [1/(√{r^{2}(cos^{2}θ+ sin^{2}θ)} )] = [1/r]. Solving for r yields r = 1.

_{0}

^{π}∫

_{0}

^{2cosθ+ θ}drdθ

- We first integrate in respect to r so that ∫
_{0}^{π}∫_{0}^{2cosθ+ θ}drdθ = ∫_{0}^{π}r |_{0}^{2cosθ+ θ}dθ = ∫_{0}^{π}( 2cosθ+ θ ) dθ .

_{0}

^{π}( 2cosθ+ θ ) dθ = 2sinθ |

_{0}

^{π}+ [(θ

^{2})/2] |

_{0}

^{π}= [(π

^{2})/2]

- We can use a polar respresentation to obtain dA by having dA = rdθdr and f in terms of r(θ).
- Note that we want the region inside the ellipse 1 = x
^{2}+ [(y^{2})/4] (our interval of integration for r) but between y = x and y = − x (our interval of integration for θ ). - Since x = rcosθ and y = rsinθ substitution yields 1 = x
^{2}+ [(y^{2})/4] = ( rcosθ )^{2}+ [(( rsinθ )^{2})/4] or 4 = 4r^{2}cos^{2}θ+ r^{2}sin^{2}θ = r^{2}(4cos^{2}θ+ sin^{2}θ). - Solving for r gives r(θ) = [2/(√{4cos
^{2}θ+ sin^{2}θ} )] and so r ∈ [ 0,[2/(√{4cos^{2}θ+ sin^{2}θ} )] ]. - To find the interval for θ we find the point of intersection between 1 = x
^{2}+ [(y^{2})/4] and the lines y = x and y = − x. - For y = x we have 1 = x
^{2}+ [(y^{2})/4] = x^{2}+ [(x^{2})/4] = [(5x^{2})/4] or 1 = [(5x^{2})/4] solving for x yields the points ( [2/(√5 )],[2/(√5 )] ), ( − [2/(√5 )], − [2/(√5 )] ). - Similarly for y = − x we have ( [2/(√5 )], − [2/(√5 )] ), ( − [2/(√5 )],[2/(√5 )] ). Note that our points of intersection are ( [2/(√5 )],[2/(√5 )] ) and ( − [2/(√5 )],[2/(√5 )] ).
- We can find the angle θ from these points by using θ = tan
^{ − 1}[y/x] and the location of the coordinate plane. - So ( [2/(√5 )],[2/(√5 )] ) is at θ = tan
^{ − 1}(1) = [(π)/4] and ( − [2/(√5 )],[2/(√5 )] ) is at θ = tan^{ − 1}( − 1) = [(3π)/4]. Hence θ ∈ [ [(π)/4],[(3π)/4] ]. - Since f(x,y) = 5xy then in polar form f = 5(rcosθ)(rsinθ) = 5r
^{2}cosθsinθ.

_{π\mathord/ phantom π4 4}

^{3π \mathord/ phantom 3π 4 4}∫

_{0}

^{[2/(√{4cos2θ+ sin2θ} )]}5r

^{2}cosθsinθrdrdθ or dA = ∫

_{π\mathord/ phantom π4 4}

^{3π \mathord/ phantom 3π 4 4}∫

_{0}

^{[2/(√{4cos2θ+ sin2θ} )]}5r

^{3}cosθsinθdrdθ

*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
- Polar Coordinates
- Polar Coordinates
- Example 1: Let (x,y) = (6,√6), Convert to Polar Coordinates
- Example 2: Express the Circle (x-2)² + y² = 4 in Polar Form.
- Graphing Function in Polar Form.
- Converting a Region in the xy-plane to Polar Coordinates
- Example 3: Find the Integral over the Region Bounded by the Semicircle
- Example 4: Find the Integral over the Region
- Example 5: Find the Integral of f(x,y) = x² over the Region Contained by r= 1 - cosθ

- Intro 0:00
- Polar Coordinates 0:50
- Polar Coordinates
- Example 1: Let (x,y) = (6,√6), Convert to Polar Coordinates
- Example 2: Express the Circle (x-2)² + y² = 4 in Polar Form.
- Graphing Function in Polar Form.
- Converting a Region in the xy-plane to Polar Coordinates
- Example 3: Find the Integral over the Region Bounded by the Semicircle
- Example 4: Find the Integral over the Region
- Example 5: Find the Integral of f(x,y) = x² over the Region Contained by r= 1 - cosθ

### Multivariable Calculus

### Transcription: Polar Coordinates

*Hello and welcome back to educator.com and multivariable calculus.*0000

*Today we are going to talk about polar coordinates. Now, polar coordinates, most of you have studied before, it is just a different way of representing a point in the (x,y) plane using a length and an angle as opposed ot a length and a length, the x and the y.*0004

*So, if you studied it before, great. This will be a quick review. If not, then this will be something really, really great.*0020

*Polar coordinates are actually very important and although we will be discussing polar coordinates and talking about some of the conversions that we are going to be making between rectangular and polar, really what we are concerned with is doing double integrals over regions in the (x,y) plane that are a little bit more easily expressible in terms of polar coordinates.*0028

*So, let us just jump right on in. Okay. Let us go ahead and make a couple of copies of the coordinates... the Cartesian coordinate system.*0047

*We have got one like this, and let us just take a point here... this is the point (x,y), and there is another way of expressing -- so we have 2 numbers to express a point in 2-space.*0057

*Well, there is another way of expressing this particular point. Instead of this distance x and this distance y -- so, let us go ahead and do this here -- what we can do is we can express it as a length R and an angle θ measured from this point.*0074

*That is it. That is what a polar coordinate is. It is just another way of expressing where a particular point is.*0092

*You are just using different numbers. This is a transformation. You are literally taking this representation of a point and converting it into this representation of a point. They represent the same point, but you are actually doing a little bit of a transformation.*0099

*Now, given... if you are given R and θ... and you want to convert to rectangular coordinates, well, the transformation is x = Rcos(θ) and y = Rsin(θ), you know this already.*0117

*Again, you are talking about this little triangle here. If this is θ, well, this is going to be x, if this is R this is Rcos(θ, this is Rsin(θ). This is how you convert back and forth.*0137

*Now, going the other way... if you are given the point (x,y) and you need to convert to polar coordinates, well, R = x ^{2} + y^{2}, also written as R^{2} + x^{2} + y^{2}.*0151

*Then θ itself is equal to the inverse tangent of y/x. So, let me write this a little bit better here. Of y/x.*0171

*That is it. So, if you are given one coordinate system, you can move to another coordinate system and this is very, very important in mathematics. Being able to move from one coordinate system to another, for any number of reasons.*0185

*Mostly because certain problems are actually expressed easier, or other times it is because it is easier to sort of visualize them physically... it could be for any number of reasons.*0195

*Let us just do a quick example.*0206

*Let the point (x,y) = 6sqrt(6)... find in terms of... ah, why say find in terms of, we will just say convert to polar coordinates, that is it... so we want to converse these to polar coordinates.*0214

*Okay. So, we are given (x,y), this is x, this is y, so we are going to use this set of transformations here and we want to find R and we want to find θ.*0248

*Again, notice, it is 2 numbers, 2 numbers, you still need 2 numbers to represent the point in 2-space.*0256

*Well, R = x ^{2} + y^{2}, so we have 6^{2} + sqrt(6)^{2}, all under the radical, so 36 + 6... 42, so sqrt(42) is R... nice and easy.*0263

*θ, well we said it is the inverse tangent of 1/x so y = sqrt(6), this is 6 and when I put it into a calculator, I get 22.2 degrees or if we can avoid some of these crazy lines here... 0.39 radians, that is it, either one is fine.*0282

*Hopefully you are reasonably familiar... um, not familiar, certainly you are familiar with radian measure... hopefully you are comfortable with radian measure. You definitely want to start thinking less in terms of degrees and more in terms of radian measure.*0314

*Again, radian measure is an actual number that you can do something with. This degree business, it is a geometric idea. You cannot really do math with 22.2 degrees.*0329

*When you enter 45 degrees in your calculator, your calculator converts it into radian measure and then does the math with it.*0339

*Let us just do another example here. This time we are going to express a particular equation in terms of polar coordinates.*0347

*So, example... express the circle x-2 ^{2} + y^{2} = 4 in polar form.*0359

*Okay. Well, that is fine. You know we have the transformation. Wherever we say x we put in Rcos(θ), wherever we see y, we put in Rsin(θ), and we just work it out.*0378

*As far as simplification is concerned, you know we simplify it as far as we can within reason. So, let us go ahead and do... so x is Rcos(θ) -2 ^{2} + Rsin(θ)^{2} = 4.*0391

*So, we get R ^{2}, cos^{2}(θ) - 4Rcos(θ). We get + 4 and we get + R^{2}, sin^{2}(θ) = 4, so the fours go away, and we are left with... we can factor out an R^{2} and let us go ahead and take cos^{2}(θ) + sin^{2}(θ) - 4Rcos(θ) = 0.*0413

*Well, this is just 1, so we get R ^{2} = 4Rcos(θ) and we can go ahead and cancel out an R. We get R - 4cos(θ), there you go.*0449

*So, this polar representation is, well, this particular representation is the polar representation of the Cartesian equation.*0475

*This is the equation of a circle, it is the circle who's center is (2,0), who's radius is 2, so this is going to be one point, that is another point, that is another point, then we are going to have 2 units over that way, this is a circle centered at the point (2,0), its polar equation is this: 4cos(θ).*0485

*Now, let me specify what value that θ is going to take in this particular case. Here the equation of the circle was given in explicit form, where x and y show up on one side of the equation. It is not saying that y is explicitly this function of x.*0511

*I mean, it is, but this is given in implicit form. In this particular case, it is saying that as θ is the independent variable and R is the dependent variable, so in this particular case, θ is going to vary over certain values.*0528

*Well, we know that the cos(θ), well R is a distance, right? It is a distance and a distance has to be positive.*0544

*So, in this particular case, θ, the values of θ are greater than -pi/2 and less than or equal to pi/2, so when θ takes on the values from -90 degrees to + 90 degrees, cos(θ) is always positive, therefore R is always positive and that is what is going on.*0551

*So, we specify the equation and we also specify the range of the independent variable, because we are expressing it in terms of -- you know, this explicit form -- so hopefully that makes sense.*0572

*All of these things are absolutely important, you have to specify what θ is going to be. Let us say that you were only taking half of a circle, well then θ is not going to go from -pi/2 to pi/2.*0588

*It might go from -pi/2 to 0, so you have to specify what θ is going to be.*0596

*So, now, let us demonstrate how we graph a function that is actually given to us in polar form.*0603

*What is the best way to do it? Well, use your mathematical software, or use your calculator set to polar form. That is the best way to do it. The quickest way.*0634

*You can do it by hand just by making a table of values of R and θ and running θ through the particular range of values, calculating what R is and connecting the dots, just like you did for (x,y).*0643

*Okay, so let us go ahead and take R = 4cos(θ) so we already know what it looks like, so let us see what it is like when we are given the equation, and then we graph it.*0658

*θ and then θ is > or = to -pi/2, and < or = to pi/2. Okay, let us just make a table of values here.*0674

*I am going to go ahead and make the table of values over on this side, so we have θ and we have R.*0686

*So, θ, when θ = -pi/2, well cos(-pi/2) is 0, 4 × 0 is 0, so R is 0. Let me go ahead and draw what it is that I am doing here... so 0... a point of length 0 is -- we are always starting from the origin, that is what we are doing -- a length of 0, so that is the point.*0697

*Okay. So let us go to -pi/3. -pi/3, when we put -pi/3 into this, cos(-pi/3) is 1/2, so 4 × 1/2 is 2.*0723

*Then we do -pi/4, we do the same thing, we just run through from -pi/2, -pi/3, -pi/4, -pi/6, 0, and then pi/6, pi/4, pi/3, all the way to pi/2, and we get a bunch of values.*0738

*Well, here, for this one it is going to be 2.8 and for here -- let us see -- when we keep going we are going to get 0 = 4.*0757

*okay, so here is what is happening. At pi/2, the length is 0. At an angle of -pi/3, which is just about here, we get a distance of 2.*0768

*Let us just put that there. At -pi/4, we are going to get a distance of 2.8, okay? At 0, we are going to get an R of 4, so literally what you get is when you trace out all of the angles from here all the way to here, what you are doing is you are going from -pi/2 to pi/2.*0783

*You are going to get a bunch of points. Well, those points are going to be precisely this circle. So, what you are going to end up getting is this.*0810

*This is what you are measuring. As you move from -pi/2 to pi/2, you are going to get values of R, and then when you connect all of the dots, you are going to get this circle, this particular circle, which is x-2 ^{2} + y^{2} = 4.*0820

*That is all you are doing. You are just making a table of values, you are going through values of θ calculating values of R, putting dots there, and connecting the dots. That is it.*0841

*Ok. Now. So, what we are concerned with is taking double integrals of functions over regions in 2-space, which we have done already.*0853

*We want to do it in polar form. We want to do a change of coordinates. Now this whole change of coordinates thing, now-a-days in the era of very, very sophisticated mathematical software, it is probably unnecessary.*0864

*Ultimately what you are going to do is actually write out the integral.*0878

*You can just stick it into your mathematical software and it will solve it for you. You do not necessarily have to solve it by changing coordinates to polar form and then running the polar... so it does not really matter anymore.*0882

*In the old days, converting something to polar form made handling the integral a little bit easier, but again, now it does not really matter. It does not matter how complicated an integral you have. You just stick it into your software.*0896

*So, what is important is being able to actually set up the integral. You can let the software solve the integral, you do not have to worry about that.*0907

*Again, this is historically important. The coordinate system is important, and then -- you know, chances are that the problems that you have on your quizzes and your tests are going to be such that they are not overly complicated, but you are going to have to be able to actually convert from rectangular to polar and be able to integrate it.*0915

*It is important from a practical standpoint, but ultimately it is not that valuable anymore. Not like it used to be, because now you can just use math software to solve any integral, no matter how complicated.*0935

*So, let us see. So what we are concerned with, well, I do not need to write that down. What we are concerned with is integration.*0950

*So, when we are converting -- actually write these words down, I tend to write a little fast, my apologies... muddle up my writing -- when we are converting a region, in the x, y plane to polar coordinates, we have to make changes to the integral and the relationship is as follows...*0961

*Ok, so, let -- I am going to do this one in blue, okay... there we go -- okay, so, let t(Rθ) = the actual transformation from polar coordinates to rectangular coordinates that's Rcos(θ), Rsin(θ) and I wrote it as a column vector.*1022

*You can also write it as Rcos(θ), Rsin(θ), this is a transformation, so this is equal to xy, that is what we said.*1055

*If you are given polar coordinates, if you want to convert them to rectangular coordinates, this is the transformation that you use.*1066

*So, let t be the transformation Rcos(θ), Rsin(θ).*1073

*Now, the integral over a particular region a, in the x, y plane of f(x,y). This is just the standard double integral, dy dx, when I make the conversion into polar coordinates, it is the integral over the region a expressed in polar form of f(t(Rθ).*1079

*In other words I form the composition f(t) and then I multiply by R dR dθ.*1110

*This dy dx, this area element that we know of as dy dx, differential y, differential x, dy dx is actually equal to not dR dθ, it is actually equal dR dθ × this factor, which is the radius.*1116

*In order for this integral to work out when I am making the conversion, from polar to rectangular, rectangular to polar, I need to do this. I need to include it.*1137

*So that is it. Any time you have this integral, and you make a change, what you do is you take f(t), the transformation and then you solve this integral which will often times be easier -- we hope.*1146

*Okay. Now we are going to discuss what all of this means, where all of this comes from, this R dR dθ this factor here.*1159

*We are going to discuss where this comes from a little bit later on when we talk about the change of variables theorem. When we talk about a change of coordinates, not just from rectangular to polar, but a general change of coordinates, how we actually change the integral.*1167

*We are going to be talking about determinates, Jacobean matrix, Jacobean determinates, things like that and it will make more sense, but for right now we just want to develop some technical facility.*1181

*We want to be able to be given a particular function or a region in the x, y plane, we want to be able to integrate over that region using polar coordinates.*1192

*We want to develop technique first, get comfortable with that, and then we will talk about what this means.*1201

*So, for now, we just want to integrate, so let us just do an example. Example 3.*1207

*Let f(x,y) = xy.*1220

*Find the integral of f over the region bounded by the semi-circle who's center is the origin (0,0) and has radius = 5.*1233

*So, basically, so x ^{2} + y^{2} = 5^{2}. That is the equation.*1286

*We want to do -- so let us go ahead and draw this region, oops, we do not want that, let us try this again -- so, we have a semi-circle... so this is our x ^{2} + y^{2} = 5, we want to integrate this function xy over that region.*1291

*That is it. That is all we are doing, and the radius is 5 -- oops, sorry, 5 ^{2}... whoo these lines are really, really causing a lot of difficulty here... let us see x^{2} + y^{2} = 5^{2}.*1328

*Ok. I think I just need to write a little slower. Let us go ahead and do this.*1346

*Well, let us go ahead and convert this particular region into polar coordinate form, so let me draw the region again, so that we know what we are looking at.*1351

*We have this semi-circle, and the equation is x ^{2} + y^{2} = 5^{2}.*1363

*So, when I put my transformation in, I get R ^{2}, cos^{2}(θ) + R^{2}sin^{2}(θ) = 5^{2}, = 25.*1370

*Well, this just becomes R ^{2} = 25, and it becomes R = 5.*1395

*So, R = 5 is the polar representation of this circle x ^{2} + y^{2} = 5^{2}, this is the polar representation... and θ notice θ does not show up in this particular thing.*1401

*θ goes from 0 to 2pi. In other words, 5 stays and now all I am doing is taking this line which is 5 and just swinging it around to θ and I am going to trace out this whole circle.*1417

*In this particular case, it is not 2pi, since we are talking about a semi-circle, it is just pi. That is it. So, that gives us... this right here is the polar representation of this region.*1433

*Now, let us go ahead and do the integral. The integral is equal to -- we said it is the integral over a expressed in polar coordinates of f(t), right? R dR dθ.*1447

*Well, that is going to equal, well d(θ), θ goes from 0 to pi, so that is that one... and R goes from 0 to 5, right?*1467

*What we are doing is we are taking -- here is -- we are taking R from 0 all the way to 5 and then we are swinging it around.*1485

*We are integrating this way and then we are integrating that way, the θ. Then, f(t)... well, f was equal to xy.*1493

*Well, f(t), t was (Rcos(θ),Rsin(θ)), so xy is Rcos(θ) × Rsin(θ), which is R ^{2}cos(θ)sin(θ), then × R dR dθ.*1505

*That is it, so when we actually -- let us bring this back down here -- now, that is going to equal... we can separate this out.*1532

*This R ^{2} and this R becomes R^{3}, so this integrand actually ends up becoming -- so, we will rewrite it as 0 to pi, the integral from 0 to 5, R^{3},cos(θ),sin(θ), dR dθ.*1547

*If you want, you can separate this out from 0 to pi, sin(θ) cos(θ) dθ, 0 to 5 R ^{3} dR -- I mean, you can solve it that way.*1572

*It does not matter how you actually write it out. Again, ultimately you are just going to put this into your mathematical software. When you go ahead and solve this, I am not going to be concerned with actually solving it at this point.*1590

*From now on I am just going to go ahead and leave it to you to take care of actually solving the integral itself. That is secondary.*1604

*What is important is this right here. Being able to set up the integral, making sure that these upper and lower limits of integration are correct, making sure the integrand is correct, and making sure that your area element is there. That is what is important. Being able to set this up. The rest, a computer can take care of.*1610

*When we do this, we end up with 0 as it turns out. That is it.*1628

*We integrated the function xy over the semi-circular region. We could have just left it alone and done it in terms of x and y, but again we are trying to gain some practice in polar coordinate form.*1634

*We found out that the polar representation in this particular region is R = 5, as θ runs from 0 to pi, so our upper and lower limit of integration is 0 to pi. It is 0 to 5 with respect to R.*1648

*We did f(t), so we put in the transformation x = Rcos(θ), y = Rsin(θ), into this... that gives us the integrand. We have our area element, and the rest is just math, technique, that is it.*1661

*So, let us go ahead and do another example. So, example 4.*1679

*Let us see. We want to find the integral of... excuse me... f(x,y) is equal to xy/x ^{2} + y^{2} over the region bounded by... well, we want y > than or = to x, we want x^{2} + y^{2} > or = to 1, and x^{2} + y^{2} > or = to 2.*1691

*These are the boundaries of our particular region. Let us go ahead and draw what this region looks like.*1737

*Okay. y > or = to x, so let me go ahead and draw the line y = x, that is that line right there. This is that.*1749

*x ^{2} + y^{2} > or = to 1, so let me go ahead and draw the unit circle... that is the unit circle -- let me make this a little bit longer.*1759

*And, x ^{2} + y^{2} > or = to 2, so -- yes, that is going to be a circle of radius 2, rad 2. This is R^{2}, let us just say we have this circle, a circle of radius sqrt(2).*1773

*Sorry about that, this is a little bit odd but you see what is going on so we want the area that is... all the x's and y's such that y > or = to x,.*1792

*That is bigger than the unit circle, less than the circle of radius sqrt(2), so what we want is that region right there.*1802

*We are going to integrate this function over that region.*1816

*Well, let us go ahead and find what this region is expressed in terms of polar coordinates. We know that the equation for a circle in polar coordinates, a circle of radius a is R = a, that is it.*1820

*In this particular case, this circle... R = 1... this circle is R = sqrt(2)... R = 1, R = sqrt(2)... and this region right here, from here onward, this is just pi over -- 45 degrees all the way 225 degrees.*1837

*So, R goes from 1 to sqrt(2) and θ > or = to pi/4, less than or equal to 1, 2, 3, 4, 5pi/4.*1862

*In this particular case, the polar representation is R goes from 1 to 2 and we are going to sweep out an angle from 45 degrees all the way to 225 degrees. That is going to fill in this region. That is it.*1878

*This is the polar representation of this region. Now we can go ahead and do our integral.*1894

*Well, let us go ahead and find f(t) first, so f(t), when I put in Rcos(θ), Rsin(θ) over all of this squared... you are going to get R ^{2} cos(θ) sin(θ), that is the numerator, and for x^{2} + y^{2} you are going to end up... that is R^{2}.*1899

*The R ^{2}'s cancel, so we just get cos(θ)sin(θ), that is f(t).*1920

*So, our integral is equal to well, θ runs from pi/4 to 5pi/4. R runs from 1 to sqrt(2). cos(θ)sin(θ) is the integrand, and R dR dθ is the element.*1927

*When we go ahead and put this into mathematical software, we end up with 0 again. Okay. This is a bit of a coincidence. I know.*1955

*I just happened to pick this intervals and these regions that always end up 0, it is not always going to be 0 as you will see in just a minute, we are going to get something that is not 0. This is just coincident.*1963

*Let us do one more. This time we will go ahead and actually express the equation already in polar form.*1976

*So, give us a little practice in graphing also. So, example 5. Okay.*1987

*Find the integral of f(x,y) = x ^{2} over the region contained by R = 1 - cos(θ).*1995

*So, in this particular case, they gave us the equation of the particular region in polar coordinate form already.*2021

*Let us go ahead and draw out what this region is.*2026

*When you do a table of values, or when you put it into your calculator, you are going to end up with something that looks like this.*2031

*That is called a cardioid. It is called a cardioid because it looks like a heart, that is it.*2042

*In this particular case... this is the region that we are concerned with. We are going to be integrating over this region and we are going to be integrating this function.*2048

*Let us go ahead and find what R is. R is going to run... it is going to be > or = 0, < or = to well, 1 - cos(θ)... so in this case, this is an actual function not specific values just like before.*2057

*You know, we can have a function in the inner integral. It goes from 0 to 1-cos(θ), whatever that happens to be.*2075

*θ is going to run from 0... we are going to start with 0 and go all the way around, sweep all the around to 2pi. That is it.*2084

*These are our upper and lower limits of integration. Now let us go ahead and let us do f(t).*2094

*Well, x ^{2} is R^{2}, cos^{2}(θ), so the integral equals... well θ is going to run from 0 to 2pi, x is going to run from 0 to 1 - cos(θ).*2102

*Our integrand is R ^{2}, cos^{2}(θ), and our area element is R dR dθ.*2123

*It is my recommendation to always put the R dR dθ there. Then if you want to you can go ahead and rewrite this as R ^{2}, cos^{3}(θ).*2133

*It is just good practice because a lot of times you will just sort of forget the R. Make sure that it is there.*2140

*Let me go ahead and put those little parentheses around it so that you know they are actually separate things. This is the integrand, this is the conversion factor, if you will.*2148

*Then when you put this into mathematical software, you end up with the following number, 49pi/32, so you see, it was not 0.*2155

*That was just a coincidence that we kept getting integrals that were 0.*2166

*Okay. That was polar coordinates and that was integration in polar coordinates.*2170

*Thank you for joining us here at educator.com, we will see you next time. Bye-bye.*2174

1 answer

Last reply by: Professor Hovasapian

Mon Dec 8, 2014 4:10 AM

Post by William Dawson on December 7, 2014

can you always break apart the product in the integrand and integrate the r part over dr , then integrate the theta part over d(theta)(as long as there is no r with the theta and vise versa), then multiply the results of the evaluated integrals?

1 answer

Last reply by: Professor Hovasapian

Thu Oct 10, 2013 1:29 AM

Post by Heinz Krug on October 9, 2013

Hello Raffi,

aren't there an infinite number of semicircles with the condition center(0,0) and radius 5 in the example 3? Wouldn't there need to be an additional condition of y>=0 to arrive at the one semicircle that you have chosen to calculate?

Heinz