Lecture 15. Triple Integrals in Cylindrical and Spherical Coordinates

okay how are you guys good you have a good week I'm good yeah so but it looks like in a few of you I hope we haven't lost anyone or maybe it's just the subject is too easy for some people so well we also have this wonderful resources so for those watching online welcome but do come to visit us sometime alright so we are talking about integrals right and I know that last week we discussed double integrals and triple integrals and four double integrals you talked about integration using polar coordinates and so this is actually very it's a very useful tool for integration that sometimes things simplify when you use some interesting stuff on now on the floor when you use a different coordinate system we saw that when we talked about as early as like a second first or second week of the class hello you guys are lacking in discipline after you know not having me for a week I hope not so you know remember that I'm really hot you know strict but fair so all right so already at the very beginning we saw that many our solutions simplify when we use polar coordinates for example in describing curves on the plane and you saw last week that integration also simplified double integrals simplify if you use polar coordinates so now the next question is what about triple integrals is there an analogue of a polar coordinate system for triple integrals and how can we use such a coordinate system to effectively evaluate triple integrals and in fact because now we are in three dimensions there is more than one choice to make there are more options and there are two particular coordinate systems which are very convenient in three dimensions which are called cylindrical and spherical coordinate system so we will discuss them in turn first we'll talk about cylindrical cylindrical coordinate system and cylindrical coordinate system is really just an off shot of the polar coordinate system in two dimensions in a way we don't introduce anything new in triple integrals we use the same tool which will ready use effectively in two dimensions so the way it works is like this the usual coordinate system XYZ is replaced by another coordinate system where on the XY plane you use polar coordinates on the XY plane and we just add the Z variable the way we just added Z to X and y so the result is that instead of the XYZ coordinate system norton which we normally use we pass through the coordinate system which will have R theta and Z so x and y i get get replaced by r and theta in the same way as for polar coordinates and z is just thrown in as an extra variable so the formula is expressing these coordinates in terms of these new coordinates are exactly the same as for Polar's they are R cosine theta and R sine theta and then we have this Z and so Z we have on both sides and it's the same variable okay so what are they what is this coordinate system whose for this coordinate system is good for describing objects which are cylindrical well hence the name right so what do I mean by cylindrical a cylinder in general is something which comes from an object on the XY plane by kind of sweeping a surface by by using that let's say a curve on XY plane so a cylinder itself a cylinder itself is like that we take we take a circle and we if we if we just move it up and down vertically we sweep a surface and that surface is what we normally call a cylinder cylinder all right so a circle the equation of the circle on the plane simplifies in polar coordinates R equals R capital where this is a number and this is the one of the variables one of the coordinates in the polar coordinate system and this number is just the radius of the circle so in other words it's a circle of radius radius R so when we when we move it upside down move it up and down like this along the Z where this z axis we create the surface in which the equation is the same in 3d the same equation means that we also have an arbitrary value of Z and therefore we get this the cylinder and the cylinder of radius R and so just like in integrating on on the plane in double integrals it is beneficial to use polar coordinates and describing things related to the circle or to the disk or sectors in of the disk or annuli and things like that it is beneficial to use the cylindrical coordinate system for triple integrals in describing things in integrating things related to cylinders and various things that you can obtain by you from the cylinders okay so so what does the triple integral look like in cylindrical coordinates there is an important point which we have to which you should not forget which is that when we pass to a new coordinate system there is some advantage which is that some of our not simplified but there is a sort of a price to pay for which is that we have to insert an additional factor in the integral okay so for polar coordinates coordinates something that was discussed last week if you have an integral of a function f da you can write it as a double integral in polar coordinate a f of r and theta but then instead of simply putting the Rd theta which you would normally put you insert an additional factor which is R so you get rdrd theta we instead of the usual dxdy which we would have in a polar coordinate in a normal coordinate system what is the reason for this just to go over it one more time because we will now see how it works for cylindrical and then for spherical coordinates everything boils down to the area of an elementary object with respect to this coordinate system and elementary object is is a rectangle which you obtain by saying that X is say in this case two coordinates x and y X between some fixed value say X 0 and X 0 plus Delta X and Y is between some fixed value and that value plus some increment Delta Y for the XY coordinates for the rectangular coordinate system this inequalities describe a rectangle in which one of the sides of Delta X and the other side is Delta Y right and so [Applause] the area of this rectangle is simply Delta x times Delta Y and this is what gives us the expression for da at DX dy and so it leads to a description of the double integral as an iterated integral in X and then Y or 1st in Y and then X Y fubini's theorem but the key point is the area of the elementary rectangle which is given by this formula now when we pass to polar coordinates the analog of this elementary rectangle is the following we have to look at all points which are which which are given by these inequalities say R between R 0 and R 0 press plus Delta R and theta in between some fixed value of theta zero and theta zero plus Delta Theta and the picture will be different right because because this gives us a sector of angle Delta Theta and so and then the condition for the four are to be between R 0 and R 0 plus Delta R let's assume that this is R 0 right just like I am assuming that this is Theta 0 of course in English in in this calculation eventually you would like Delta R and Delta Theta to become very small whereas R 0 and theta 0 are fixed right but I'm just drawing this picture I kind of magnify everything so it looks like Delta Theta looks like sort of the same magnitude as theta 0 but in fact it should be much more and so then if this is Delta R this and this is confined within Delta Theta and the elementary object instead of this one we get a kind of it sort of looks like a rectangle but not exactly because first of all there is a certain angle here and these are not straight lines but these are these are segments of a circle right so what's the area of this the area of this one actually we cannot I mean it will be we can but it will be different it's not so it's simple to calculate give exact answer but because we are going to take the limit anyway eventually we only need that sort of a good approximation to this and what's a good approximation to this we can think of this we can think that when Delta R and Delta Theta are very small this will look like a rectangle with the size Delta R and what's the other side well this actually has length R 0 times Delta Theta right the segment of a circle of radius R 0 confined within the angle Delta Theta is given by this formula R 0 times Delta Theta and so we approximate this by the area of the rectangle with the size Delta R and R 0 times Delta Theta and the result of this is just the product of this and we get R 0 Delta Theta Delta R and so that's where that they are that's the R which shows up in the in the formula for the integral here I denoted by R 0 because because I wanted to emphasize that in for calculating this particular area I fix R 0 I fix this length but when we do a general calculation this will just become R and Delta Theta will get replaced by D theta Delta R by D R in the limit right and so that's how we end up with this formula so this is the most this is the only sort of non-trivial thing to remember about polar coordinates don't forget to put this factor R which actually could be a good thing I made it sound like a surprise to pay that we are being taxed for sort of for the convenience of using these coordinates but actually sometimes and often times this could be a good thing and the typical example is let's say you have an integral like this something like 1 minus x squared minus y squared DX dy so you see this is complicated right this is actually something we discussed it's kind of similar to something we discussed earlier this is this will be very difficult to take just as an integral in x and y but when you pass two polar coordinates this will become square I'm not writing the limits but I'm just giving a rough sketch here this becomes 1 minus R squared right and now we get this extra factor R and then we get the Rd theta and so you see this actually becomes a much better integral because I couldn't reduce a new variable let's say T which is R squared and if I do that then DT becomes 2 times RDR and and then this actually gives me one-half of DT and this becomes this becomes a square root of 1 minus T instead of 1 minus R squared and this is much easier to calculate rate because the antiderivative of this is just a 1 minus T to the power 3 cos times 2/3 minus 2/3 whereas for this one it's much more complicated well and it's imagine if it were it was like exponential function or something so so then it becomes even worse right so sometimes actually having this factors if it actually works to our advantage but in any case you have to remember to insert it to get to get the right answer now so this was all about the polar coordinates but cylindrical are not that far away far apart because for cylindrical the only thing that changes is that we now throw in an extra variable namely variable V so we add the variable T and now instead of the double integral we have a triple integral over some solid some region in a three-dimensional space and so we have some FG a where this da usually info in normal in a Cartesian coordinate system we simply write as DX dy DZ right but now if we want to use polar coordinate system we'll have to write this function as a function of R theta and Z and then we'll have to write dr g theta DZ and the question is what should we put in front that has to be some factor and that factor is the fact responsible for the area of the elementary domain like this in the usual coordinate system in the 3-dimensional space in the case when I say usually mean Cartesian coordinate system the elementary object is a box of sides Delta X Delta Y and Delta Z right so it's actually no one correct me what is it what what is it a mistake here DV that's right see I'm a little rusty after the break but Luke will do correct me when I do when I make this mistake so DV because we emphasize that this is the elementary volume it's volume it's not area anymore the a of course was for for area and V is for volume so the elementary volume opposite for this guy is just just Delta X Delta Y Delta Z which gives a gives rise to DX dy DZ in the integral and allows us to calculate this integral as an iterated integral first x and y and z or any other any other order that you choose but now we choose this coordinate system and so we have to be careful but we have to now look at the elementary object in this coordinate system and what is this elementary object well we just define it in the same way as before by simply I shouldn't have done it this way I'm actually jumping ahead let's just do it like this it really should be cylindrical so it should be it should be like this right so it is it is complete in a way it's a kind of it's what you get by combining the two things in this see in this this place this side of the of this bit of this object looks like looks like this it's yes why is it called cylindrical because this is a cylinder right okay he saved you by the way because I was kind of approaching to count how many seconds it will take for you to notice me but he saved you so you should take him so the cyllage cylindrical is called because you you see the you make a change on the XY plane and then you just throw in the Z variable in other words that you change the coordinates just on the XY plane and then you kind of let the the third variable the same as before so it's the same idea if the idea for say taking the circle and kind of just moving it up and down to get this object know the circle is fixed right it doesn't grow it doesn't the the the radius of the circle is fixed it's this R capital right just think of a ring which is made of metal or something you just move it up and down vertically along the parallel to the z axis right so what's the result you sweep a surface which is the surface of the cylinder right you can only use footage no you don't use it's only four cylinders but well what's in a name you know it's it's a it's a name so it is it is derived from this right but it's not supposed to explain everything we can do with this coordinate system right any other questions all right so what what's the elementary volume of this well we already know the approximate area of this that's RDR our Delta R Delta Theta and we know that this is Delta G right and so the volume is RDR Delta R Delta R Delta Theta Delta Z so it's the same R which we had or maybe if you want R 0 if you want this to be R 0 the same R which we had in the calculation of the polar coordinates so the bottom line is this factor is exactly the same as in the polar coordinate system it's not no not more not more complicated ok so that's the formula going to use and now let's do an example let's do an example [Applause] so evaluate triple integral over e of the function e to the Z where e is enclosed by the parabola with Z equals 1 plus x squared plus y squared the cylinder the cylinder x squared plus y squared equals 5 and the xy-plane okay so actually let me get let me draw a bigger so the paraboloid looks what is the probability look like it it starts at the point 1 on the z axis and it so it opens up like this right and the cylinder is okay I just erased it so that's assuming that let's assume that this is radius this is going to be radius square root of 5 right because we have 5 with the square of the radius as always so the radius itself is square root of 5 and so that that's a circle at the base of the cylinder and we want the figure which is a solid which is confined by by this by the paraboloid by the cylinder and by the xy-plane so it is the the inside of this so it's kind of concave the surface kind of concave it goes inside right like like a fancy fancy glass so what we need to do is to calculate the integral over this this is our region e of the function e to the Z so when we do triple integrals you have to so then you have to write it as an iterated integral okay so you have to choose in which order you're going to integrate so so you you're going to end up with an integral so you have e to the Z here actually these one of the variables so we just write it like this if this is good right because he's one of the variables in this news in cylindrical system and then you're going to have it are let me emphasize this R again that's the same are back before and then we have rdrd theta and DZ so that now we want to write it as an iterated integral so you have to choose in which order to integrate okay so what's the what's the best way to do it for this excellent for the cylindrical coordinate system is always you see the point is that you have to choose what the base of this actually this is a general general approach to when you have to choose the order you have to see whether what's a good projection of your object is there a good projection on to say XY plane or Y Z plane or ZX plane that's the first question and once you have a good projection everything projection to something nicely right so in this particular case it sort of does project nicely onto the XY plane it projects just onto onto the disk onto V which is our less than or equal to square root of 5 the disc of radius square root of 5 right so that's this that is yellow let's let's draw it with yellow and then for each point in this disk we have to we'll have to integrate so first of all we'll have to integrate over the disk and this will be the this is I'm talk about the outer integration I'm starting from outside when you project onto this you are choosing the first two variables which normally if you were if we were using x and y and z-coordinates that would be integrating first x and y its then you decide you decide between x and y later but first we choose the first two and then you then you choose which one goes first right so that's the strategy which I proposed that you're going to end up with three different integrations right so the last one will be the integration over the remaining variable with respect to its projection so here I project onto the XY plane so this will be D Z but let's let's now see what happened here so this is going to be the usual coordinate system and the Cartesian coordinate system this would have been just DX dy which we wouldn't have to decide the f dy or dy DX and once you decide then then in the in the last integration you have the freedom to choose what what x and y are so that means that you have a point here X Y inside and then you the limits will be the limit you have to sort of draw the segment starting from the bottom of this of this object to the top of this object which indicates from the disk which is part of the XY plane to this paraboloid and you would have to integrate in the last integration you would have to integrate from the bottom value to the top value which in this particular case what does it mean the Z goes from 0 right two to this value but what is this value this value is going to be one plus this is going to be the value of Z on the paraboloid so that is in the Cartesian coordinate system this would have been 1 plus x squared plus y squared but now we're replacing this with R squared so because we are doing the integral in the in the polar coordinate system we will actually write it as 1 plus R squared where it is understood that R squared comprises x squared and y the sum of x squared + y squared yes that's right well up and down the cause I have chosen the projection on to the XY plane right so the the way I suggest to do to the integral is the following that first let's choose the two outer integrations the person the second right so geometrically it means that you are projecting your three-dimensional region three dimensional solid onto one of the planes one of the coordinate planes I'm talking now about the really the Cartesian coordinate system but cylindrical core this is not that far away from the Cartesian so the same analysis sort of applies here so so the projection here is going to be the disc right but in general it could be square it could be something else so that will be taken care of by these two integrations by the limit here that's right the projection goes in those two and the inner integral is the remaining variable which in this case is Z so to put the limit here you have to take one of the points in the projection in the image of the projection which is this point say X Y which actually because we are doing polar coordinates or cylindrical coordinates this would have to be recorded as R theta right so this point would be R Theta and we'll have to take the segment along the third variable which is the variable going from the bottom to the top from the sort of the bottom lid to the top lid of this of this figure of this region which as I explained the bottom is on the XY plane so the G is equal to zero at the bottom and at the top it's what it's we have to look at the equation of the parabola it's 1 plus x squared plus y squared which because we are working with cylindrical coordinates will be written as 1 plus R squared okay let me finish the integral and you ask me if you have more questions so so then what what remains to be done what remains to be done is that we have to put also here R and theta so I have to choose some order in which we will do R and theta so let's say we put here they are and put here D theta because now we are actually see once we are here in this to out outer into integrals we are actually doing a double integral we are deciding how to how to split the double integral into iterated integral so and what we are working with is already the image of this three-dimensional object onto the XY plane which is nothing but the disc right so for the disc we know what the limits are first so we know that it doesn't matter in which order to take RN theta and second of all we know that R goes from zero to the radius which is in this case square root of 5 and theta goes from zero to two pi right that's because the disc this disc is described by well actually I don't even have to write one more time I just have to add here theta from 0 to 2pi and finally we should we should not forget to put the function which is e to the Z and we should not forget to put this factor R so I'm I'm introducing a slightly different notation from before right normally we would write this as e to the Z times R times dr d-theta and then we put sort of brackets and we think about integrating first with respect to R and then there's respect of theta and sorry dividing this respect to Z then this respect to theta and this respect to R but I'm writing it in this way which i think is a little bit more suggestive that you write the deal the gr the G theta DZ the D of the variable you're integrating right next to the integral so you remember which one it is you see because in the old way you would have to put the are the last one right so the last one will get paired with the LA the first thing to grow the next-to-last will get shared with the second integral and so on this way it's a little bit more intuitive so I prefer to write it this way but you can choose whichever way you like but I hope that it's clear what I mean well this certain is not clear because it says on 2 and then it says 0 so let me write it here and arrive on 2 like this I think now it should be clear okay any questions about this no yes that's right so in fact I could have put our all the way all the way here right oh yeah yeah yeah yeah you're right you're right you're right you're right right so I actually I actually did it in a wrong way in the wrong order so this so did I do it in the wrong order now hold on no the older of course matters because see the point is that no no I did it correctly who thinks that I did correctly who thinks that I didn't do it correctly okay you guys who failed of course now we are here to find the truth you know so I but see here's here's the way I thought about it this afternoon actually and I convinced myself this is right but here's the point the point is right so the point is that we are let me this is actually very kind it could be very confusing so let's actually do it slowly you should not worry so much about the fact that here there is a dependence on R in the inner integral what you should worry about is the fact that there is a dependence on R in the limit so you certainly don't want to put this outside you see because you want to end up with the number okay in other words we can disagree on many things but there is one thing we should certainly agree with which is that the integral is a number okay it should not depend on R it should not depend on theta it should not depend on anything it's a number you cannot possibly get a number if the last integration has the limit which depends on a variable okay so it's not going to work but more conceptually the way I'd like to think about this is as follows and I would like to think maybe in slightly you know it could be some work more complicate domain so what I'm doing is I'm projecting so there is some E and let's say I want to project it onto the XY plane that's right now let's talk about just the Cartesian coordinate system let's not worry for now about cylindrical and spherical even in the Cartesian coordinate system the way I would like to think about integration is as follows that I have a region on a three dimensional space right and what I do is I project it on to the XY variable so now if I have a triple integral over e when I have half GV first of all I split it like this it's going to be a double integral over D over this D all the single integral with respect to the remains of the remaining variable easy you see where for each for each X&Y in D for each x and y in Indi you will have some lower limit let's say it L of X Y and you'll have some upper limit U of X Y right so that's what's going to look like yes so then I would just write like oh and here yeah that's right so then here I would put da like this yes that's right in the all the way right so you want me to write this an old way okay so this in the old way would be zero to square root of 5 0 to 2pi 0 to 1 plus R squared right e to the Z D D D theta dr oh I'm sorry RDR yes very good thank you see because it's red so I don't I don't agency are you are well usually with the right are the argot yeah are gr okay fine whatever we can put our inside here maybe it's better to kind of keep track okay so then normally we would put brackets like this is that okay this make sense so what I'm doing is just I'm trying to avoid using the bracket and instead just putting the differentials right next to the integral so it becomes a little bit more clear to me anyway but you are free to use whichever way you like so you know so this is sort of a new way to write and all the way to write would be fgz well it's all with the same da over G so but the the main point which I would like to make is that the first to integrate integrations in the formula C it's very confusing because when I say the first integration although I mean the first integral although I mean the action the creation you perform its oppinion right the first integral you write is the last one it's the last thing to go your ID is the first you calculate okay that's right that's right so this is the outer the outer goes for the outer is written first so that's why it goes those last you see so you have to figure this out on your own I guess some point you have to think about it and have a clearer picture because because like I said it's very confusing what do you call first would equal second who is on first right so the the last integration the last integration is outer one and the last integration corresponds to the projection of your three-dimensional region onto the plane okay that's the point all right so having established that let's let's evaluate finally this integral so what do we get [Applause] we get zero squared 5 to 5 and then here we get R times e to the 1 plus R squared minus 1 right and you get dr this integration just gives you 2 pi because the the integrand does not depend on theta so you just get 2 pi times our e to the 1 plus r squared minus 1 right and finally you integrate over R and so so here you want to you want to use you want to change variables so just I just write the answer so you get 2 pi is an overall factor and then you get 1/2 e to the 1 plus R squared minus 1/2 R squared between 0 and square root of 5 right and so the answer is 2 pi times e to the 6 minus e minus 5 one half but because one half and to get cancelled so you get e to the 6 minus e and this guy gives you 5 right so that's correct ok yes it's pi that's right so it like this Thanks ok cool so we're good with this yes it is universally accepted that are you talking about this formula or this form this you mean this integral yes but I I'm not sure I understand what the question is which line first of all this one is it this one yes here a little bit the left-hand side or the right-hand side sorry the question is is it okay to put the R before the function of R okay the answer is yes it is okay I am suggesting with notation in the book it's not used right but it is the pricing mathematician I can assure you that a lot of people use that you know it's it's certain interchangeable they do commute in other words it doesn't matter which order you write them it's a it's a more interesting point whether these guys commute you know whether it's whether it is important how you write the Rd fate or g theta dr this is a much more subtle point which i'm going to going to get into too much but certainly dr come you can put on the left or on the right of the function of r so that's not no worries here okay let's move on so the next one is called spherical coordinates and it's a little bit more interesting sorry spiracle coordinates and of course the question is right away why are they called the spherical coordinate because they should remind you of a sphere okay I do as you will see so let me actually keep this let me keep this picture so spherical coordinates work in the following way suppose you want to invent a coordinate system in which the sphere it is a sphere which gets the simplest possible equation you see so in the in the cylindrical coordinate system it is a circle that gets to say in a polar coordinate system it is a circle which gets the simplest equation R equals a constant that's a circle over a of a given radius in the cylindrical coordinate system the simplest equation is the equation for the cylinder okay R equals R is a circle in choler coordinates because R is a cylinder in the cylindrical coordinate system so cylindrical conceding to G and now I want to have some variable row such that this equation will give me a sphere and that's called a spherical coordinate system and that's some 3d so that means that this row should really be the distance from the origin of my coordinate system to my point because the skier was here with the center at the origin the sphere with the center at the origin of radius R capital this one is going to be the set of all points such that the distance from the origin to my point is given by this by this number R let's say it's square root of 5 I can a previous example okay so know that this is not the same as R in the cylindrical coordinate system because R in the cylindrical coordinate system is not the distance from the origin to this but rather it's a distance from the origin to the projection of my point onto the XY plane so that's not the same thing right so you see right away that this is not the same coordinate system and sure enough this simplest equation Rho equals a constant describe is not a cylinder but a sphere okay but now so now I got the first variable of my spherical coordinate system but that's not enough I need three variables right because I'm a three dimensional space so I need to complete this row by two additional variables two additional degrees of freedom if you will so that I could give each point a unique address by using those three coordinates what are they so the standard conveys a different way it's actually to do it by the standard convention which we are using in this class and this book is the following that we we measure two angles in addition to rule we measure two angles and the first angle is exactly the same as before theta it's the state of the cylindrical or polar coordinate system okay and so we need one more and we find this one more by measuring the this angle and the code is Phi so the spherical coordinate three coordinates are all Phi and theta where only one of them is part of the of a cylindrical coordinate system which data is part of the cylindrical coordinate system but the others are you okay so by the way the notationally phi sometimes is also is written like this it's the same thing so I find it easier to write Phi like this but when you type for example and then in the book it's written like this so you can use whichever notation you like so Rho Phi and theta the first step of course is to express the user coordinates the Cartesian coordinates in terms of Rho Phi and theta and to see that indeed these coordinates are determined the points on in space and also to see what the ranges are so what is the formula well what we can do is we can use the old formula which is R cosine theta and then note that R is equal to Rho times which this is equal to Rho times the sine of this angle see the point is that this has the right angle this is a triangle this is a triangle which has a right angle and this is Phi so you can find this distance which is our of the cylindrical coordinate system by taking R or multiplying by the sine of this angle so this is a Rho times sine of Phi and now I substitute this in here and I get Rho sine Phi cosine theta yes which access is five measured from this method from Z right so I did not put labels but I should yes XYZ in standard Y another question Rho is not it does not belong to any plane a priori it's in space somewhere right it's just I have a point it kind of flows around it like this right Wiggles around it's not it doesn't belong to any particular plane it's free to move right because our point is free to move in through space and so the segment connecting our point to the origin it does not necessarily belong to any plane which one this triangle it has the right angle here because this is the projection I don't know if it looks like a project it looks a little bit it's a little bit crooked I guess it should be more fair it should be a little bit more parallel to the Jeep's line than what I made it look like better slightly okay so this is the perpendicular which we draw up from this point where our point this is our point we drop it on to the XY plane so I guess the rest this should be clear any other questions yes that's right if they'd always taken from the x axis for the projection everything is a little bit not straight on this picture so this is our this is the old R which is not this this R is not part of this vertical coordinate system it's part of the cylindrical coordinate system the reason I have drawn it I've written it here or put this label is to connect to make a connection between the spherical and cylindrical coordinates and also to simplify the derivation of the formula for XY & z because you I do it in two steps I first recall the formula for the cylindrical coordinates which is really the same as for polar coordinates and then I just substitute R equals Rho sine Phi and then I do the same for the walk or the Y variable and then Z is what is D to find thee we have to complete this to the rectangle I'm sorry it should be a sine theta that's right thank you all right and to find these I have to use another right triangle which is this one yes again I didn't do it very well maybe more like this okay so this is a right triangle this is Phi so Z is Rho times cosine Phi so if you visualize if you visualize this picture you see that you don't have to memorize or write on your cheat sheet the formulas for the strip of course very easy you have to remember the polar coordinates of course but I mean you don't remember polar coordinates then it's like remembering your phone number in some sense and I think so if you remember polar coordinates then then the spherical coordinates are very easy just by using by memorizing this picture visual or visualized in this picture okay what are the ranges so are always like R Rho is a non-negative number although for R if you remember we had certain convention in polar coordinates which sometimes allowed our to be negative but in polar instead of a coordinate systems system that would be make things too complicated so we don't do that so so Rho is non negative theta is just like in a polar coordinate system or in the cylindrical it's between zero and two pi and what about Phi what are the ranges for Phi zero and pi exactly because you can go so if this is the vertical this is a vertical on the z axis you can go all the way down to PI but if you go more than PI then you can sort of approach it from a different direction and it will be less from that direction with less than pi so that's what we do so it's going to be from zero to five you see what I mean yes that's right from the vertical so in other words what I'm saying is let's say so this is the axis and this is the origin so you have a let's suppose that our point actually lives on the on the Y heap on the Y Z plane so and let's let's start rotating it so this is quite for this point this is Phi for this point this is five for this point still less than pi right so we get here it's still I mean I'm trying to say is this this is this right we measure angle from here but finally when we go like this we should not measure it like this but you should measure it this way you see them in so this is wrong and this is right which is fine so you choose whichever is shorter which is smaller and that's going to always be between 0 and pi so pi would be this this is pi which is an point on the negative Phi equals PI response to point on the negative part of the G axis did you have a question all right now what are what are these coordinates good for well the first answer is has already been given in this coordinate system it's much easier to describe the sphere so the sphere is now given by the equation Rho equals R instead of instead of x squared plus y squared plus Z squared equals R squared of the rest of the Cartesian coordinate system or R squared plus V squared equals R square r-squared of the cylindrical coordinates it's just a simple as possible equation Rho equals a constant so that's in a way justification for the use English coordinate system and for the name but that's not the only object which has a nice representation in in this coordinate system we can also represent night and a cone by the equation Phi equals some fixed angle Phi 0 so that's the cone right let's let's look at this picture again so let's suppose I fix I fix some Phi i65 and I look at all all the points which have a given angle like this Phi 0 6 to the z-axis so then it's like this is the z-axis and my point is like this I don't know I don't have enough visual aids so I'm I have to trust your imagination and your intuition may be like this is better I don't know so the point is that this is the angle so the angle is fixed so it has to be the same angle between your points and in the same z-axis so the result is the cone and because I don't specify what role is I don't specify the distance it could it can go as far away as I want so it's an infinite cone kind of upper cone going in a to infinity right so it's not bounded unbounded cone so don't think that I just drew one level curve for it but it goes to infinity so the cone is given by this formula which is again much simpler than v equals square root of x squared plus y squared of the cartesian coordinate system or Z equals R of the cylindrical coordinate center which is not so bad but this is little better this is to handle so this are the things you should remember the cone and the sphere have a very nice expression in terms of in terms of this coordinate system now what about the integration what about integration in circle coordinates in spherical coordinate system so once again we have to figure out what is the volume of the elementary object with respect to the spherical coordinate system so these are the elementary object for the Cartesian and the cylindrical coordinate systems which give us the usual form of DF dy DV and r vr D theta DT so now we have to do exactly the same for for the spherical coordinate system so what is the elementary object now and again by elementary object I mean simply that you impose the condition that Rho is between some Rho 0 and Rho 0 class Delta Rho and what was my secretary about Phi Phi is between Phi 0 and Phi 0 plus Delta Phi and theta is between some theta 0 and theta zero plus Delta so let's draw this object which corresponds to this inequalities like this tygris Thank You Vince well this whole thing should be kind of noted line okay and like this and like this okay you see if you I mean this well in fact this I can do good like this so maybe it's better so here what happens here first of all this is Delta Theta Delta Phi right this is my this is Rho zero so this thing is actually between two spheres convention two spheres which are very close radii very similar idea so this is something which is between those spheres and it's first of all we cut it by saying that the angle goes between you know it can find within Delta Phi and also when we project this down onto the XY plane and we project it down onto the XY plane various angle I'm not going to draw it all the way for the projection going to be Delta Theta so this is the projection onto XY plane you see so now I have to evaluate the volume of this just like I have to evaluate the volume of this and in this in my other coordinate systems so what is it like what is this volume equal to well we don't calculate exactly we approximate it and we approximated by the volume of the cube so it not the cube but a box a box whose sides are given by the lengths of this this kind of circular segments that are going to be all circular segment okay so the this one is very easy to find well first of all this one is very easy to find this is just Delta Delta Rho this is judge cultura that's one side okay now this one this one is also easy to find this is easy this is Rho 0 and the angle here is Delta Phi so this is Rho times Delta Phi it's similar to the way we found that to be R 0 x times Delta Delta Theta so it remains to find the this one which I guess you cannot see this color actually is better than the other ones kind of greenish can you see this one yeah oh good ok so this one is a tricky one because you would be tempted to take Rho 0 times Delta Theta and this is not correct so this is this is only one there is only one set of points here the point is that this is the same as this in other words this actually is not this is not part of a circle of radius Rho zero and confined within the angle Delta Theta on the contrary it is actually part of a circle on the on the ply on the XY plane but the radius is not the radius of that circle is if this which is not rosy ro but it is it is r it is Rho zero times sine Phi so that's the only point that you have to realize to get the right answer and so and then you've got you've got Delta Theta I'm trying to draw this like this this is a project this is on the XY plane and so this segment is actually each length is Rho 0 times sine Phi Delta Theta ok so the answer is for the volume it's approximated by Rho 0 sine Phi Delta Theta times Delta Rho times Delta times Rho 0 again this is 0 0 3 and times Delta Phi yes I'm I am I'm trying to calculate the volume of this right and I am approximating it by the volume of a box like this which has the falling sides this side is Rho 0 sine Phi Delta Theta that's this times Delta Rho which is this side times Rho 0 times Delta Phi is the yellow side so the end result is Rho 0 squared sine Phi times Delta Rho Delta Phi Delta Theta ok so the the the upside of this is that when you compute a triple integral in spherical coordinates you have to introduce the fact that all squared times sine Phi it's a little bit more complicated than for than for cylindrical polar coordinates the triple integral over some solid is Theta Rho times Rho squared sine Phi well since I wrote it in red before let me write in red again a Rho squared sine Phi that's the factor to remember V Rho D Phi D theta ok 1 so we do this calculation once and for all then we just memorize it Rho squared times sine Phi and and now after that we can do this triple integrals by using iterated integrals for these three variables Rho Phi and theta let's see how it works in practice so here's an example compute the integral of x squared plus y squared plus Z squared DV where e the region is bounded by the XZ plane and the spheres x squared plus y squared plus Z squared equals 9 and x squared plus y squared plus Z squared equals 16 so between two spheres and also the XZ plane so first of all I have to say the general comment on OpenType say only one next thing next week will help me turn on the midterm I will not tell you which coordinate system to use right that you have to decide yourself so the question is you have to look for clues which coordinate system spherical cylindrical you have to remember one thing that this expression x squared plus y squared plus Z squared simplifies in the spherical coordinate system so if you see x squared plus y squared plus Z squared how many times 1 2 3 times chances are it's best to use in this case the spherical coordinate system if you die mean because this is Rho squared right this is Rho squared if you see x squared plus y squared that's R squared R being the R of the polar or cylindrical coordinates that's how you know that you probably should do use polar coordinates or cylindrical coordinate depending whether you use dual are doing a two-dimensional double integral or three-dimensional triple integrals but in this case it's it's very here you'd be better off using spherical coordinates now we convince you like an try to visualize this we could try to visualize this think of us here first of all let me draw let me just draw just a quarter on this picture actually yeah about a quarter so that's that's the part of the sphere this is a sphere of radius three because nine is three squared right and this is fear of radius 4 because 16 is 4 squared so let's say this is a this is a sphere of radius 3 and then this is the inner sphere and then there is an outer sphere which has radius 4 so see this one is inside the other and what you're looking for is what you want to look at is the 3-dimensional shell which is in between the two in between these two spheres it's like an analyst by 3 dimensional analysis between two circles and this is between two spheres and also you have to bind a better bound is is the XZ plane so this is the XZ plane so actually that's not the whole thing the whole this is 1/4 of the whole thing because you will also have something in the in this quadrant and in those two quadrants which are behind the zy plane so this is 1/4 so to speak of the picture if I try to draw the whole picture it will be more confusing so you just have to draw the same thing but three more times and that's what you need to cut it now actually here the way it's phrased it sounds almost as though it is not it is ambiguously defined because it doesn't say in which side of the XD play it could be on this I draw it on this side where Y is positive but it doesn't say that so in principle I could also look at the other half but the point is that the answer does not depend on which path you choose because the function which you use the function that you use it does not change if you if you flip Y if you replace Y by negative Y so if that were not the case the problem would not be well-defined would not have been well-defined but actually this is well-defined because the answer is the same no matter no matter on which side of the XZ plane you look okay now to compute this you write C this is one case this is one a coordinate system where you cannot so easily visualize the way I explained in a previous discussion for cylindrical and and Cartesian coordinates where I was talking about projection because these coordinates do not involve any of the variables of the Cartesian coordinate system does not involve X different one of Y does not involve Z so you have to be a little bit user imagination and you have to just try to describe this by inequalities in the coordinate of the spherical coordinate system so the inequalities will be in this case as Rho is between 3 & 4 we know this right then what about what about what about 5 well 5 actually can be anything from 0 to PI it takes the maximum range and then you have to here the condition is on theta if there was no condition about the XD plane if we didn't say that if the problem didn't say that one of the bounds was the ZX plane we would write that theta is between 0 & 2 pi which is the maximum possible value right this is a total range of theta but because we are bounded by XZ plane that means that theta can go from 0 to only up to here but it cannot go further because we are on this side you see I mean that means that state actually goes from zero to PI and that's what that's what this extra bar about gives you right so these are the ranges and so actually in the spherical coordinate system this very complicated very complicated domain looks very simple right it looks a like a box it looks like a box and so the integral that you get describing this is triple integral is actually going to be the simplest possible triple integral with respect to this coordinate system and you can choose any order you like for for the integration because the bounds do not depend on other variables you see they are fixed so if it's like the basic integral yes very good question is it is it going to be like this for in all coordinates in all for all problems for spherical coordinates or just for this one just for this one of course because you can imagine more complicate domains in which the bounds will depend on other variables right so I'm just getting the simplest possible example okay so so here you're going to have say row from 3 to 4 and then 5 from 0 to PI and theta from 0 to 5 and then you don't forget your function which is Rho squared and don't forget the extra factor which is Rho squared this one Rho squared sine Phi so times Rho squared sine Phi D theta D Phi 0 Rho squared sine Phi huh sorry why is it Rho square times Rho squared it's called Rho Rho so that's actually a good question on the midterm would be what is this variable called and so if you write for pt4 there for the explained that you get to that point deducted this is a roll right anyway so this Rho squared comes from the integrand which is x squared plus y squared by d squared which is Rho squared and this Rho squared comes from the extra factor so the net effect is that you got rows row to the force sine Phi so that you get a very simple manageable integral okay I'm sorry hold on hold on this one one minute left yes letting it roll squared sine Rho squared sine Phi is always there and then you have the function okay so we'll continue on Thursday

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Length: 79min 36sec (4776 seconds)

Published: Thu Aug 03 2017