2020 Spring SMS Mathematics Lecture 10

Añadido: 2026-05-19

[00:00:08]hey guys welcome uh thanks for joining me today we're going to re continue on with our lecture n and then if there's enough time we'll we'll get to uh lecture 10 as well but we've still got time so uh there's nothing for us to be concerned about at all well but we still we got time okay so uh what we're looking at today is our polar coordinates okay so with our polar coordinates so we think about uh changing frames um and changing frames basically represents a change from uh our normal cartisian coordinates okay to our oops going to draw on that to our uh so there's our one there's our origin there's our zero here's our z0 origin okay so we use a transformation to go from cartisian coordinates in terms of X and Y okay to our polar coordinates which have a representation in terms of R and Theta so instead of being in terms of X and Y the are in Theta so there's a slight difference in the way that um we think about these coordinates and so we've got to get used to them now it's going to involve a bit of exact values trig and all of that stuff that you've kind of used in the past uh if we do a resolution for R we'll get a value of R = < TK of X2 + Y2 so that kind of thing also from these guys here it's not hard to work out the Tan x uh tan Theta is rise over run y/x okay and so what we're able to do is make using these basic translation ideas make a change from a point XY uh to R Theta and back again if we need to so um the recognition of how these things go together is sometimes a little bit uh not obvious so we've got to spend a little bit of time um uh just getting getting to know some of those ideas so for our theaters uh we could go all the way around so we could have a domain uh that's um zero to Theta 2 pi uh or we could have uh minus Pi but there's usually a a constraint or restriction on the interval and that would be then just plus pi okay okay so all right so what we'll do is we'll have a look at the next part in terms of our integration over this polar plot uh this is our po plot here instead of doing our integration uh DX Dy uh we'll be doing it uh in the new frame involving R Theta so I I haven't written it directly because there's a little bit of development that needs to happen so this one's fairly straight forward we get that idea okay but here there's a little bit of development that we have to build up if you like to get our da okay so we'll have a look at that all right so if we take a look at our polar sub part and we're having a look at this rectangle that's been pulled out of the Polar part so if we were looking at this part we might say it's this little square here okay so we want to have a look at the increments in the polar part so there's our little square and we've got an area da okay so we can consider that in our rectangular coordinates quite easily Okay so we've got our dxdy for our polar coordinates though we need to consider the impact of the new environment that we find ourselves in okay so I don't know why they're quarters length of a small element is approximating to Dr and the width of a small element is Rd Theta so we need to have a look at that so if I take this uh and draw this down to its origin and this one as well we can see that they're going to converge down here okay and so we'll have at the Theta so an increment in the angle Theta Okay so we've got our full Theta and then we've got our increment adding so there's our Theta and there's our the Theta so it's not easy to pull together without a bit of support there okay and then we've got our um d l so if we think about our radius so that's our R and then we've got a little Dr here so this is the Dr for this one and add the Theta so on top of our R we're going to have an A Dr and our Theta will have a d Theta all right so what does the area represent in here what does this area here represent and so what we've got is our r with an increment yeah that's Theta thank you with our increment r so our area is r d Theta d r okay so that's the area of this particular part here okay uh likewise if we were to have the L here R Theta but this is the one that we're interested in up here okay so in terms of uh DL V sign sorry r r sin Theta D Theta D Theta okay so the area that should be equ sorry I shouldn't have let those quarters stay on there uh that's just the way it's printing so what this is that dxdy is being translated to r d r D Theta so we've got a new frame of reference that we have to add to our integrals when we do those okay so a new frame of reference that we have to put together for those and when you first encounter this uh it's not so straightforward um students like to think of it just in terms of Dr Theta but we have to take into account the r uh of our length if you want you can put uh if you want a little okay so if we go from X to R so from cartisian to Polar so we've got cartisian too um I guess it's a good time given that it is a little bit complicated why do we bother why do we bother doing uh cartisian to Polar transformation why would we do that what's the reason that we do any of the integration techniques that we use it's to make life simpler okay so it's to make life simpler so the only reason that we would do this is because in the cartisian coordinates this is just a horrible s but if we do the transformation uh using our r d d Theta then we're likely to be able to have an equivalent that's quite a bit simpler okay so what we've got here is f functions of theta um transposed over our cartisian coordinates and the increments in our Theta and R all right so we make a change that's the first thing we do to RDR D Theta now what you'll notice is that this here is Dr so that's regular and now thinking of incremental change D Theta is incremental change but notice this R is a new variable added in so it'll be a product with anything that's preceding it so it can be taken over and placed in front if this has already got an R then that becomes an r s because of this guy existing here so it just catches students una aware initially we need to be aware that we've got to have uh this r on top all right so if we have a look at our function so here's our integral uh for some function over dydx and then we do the transformation so the first thing you'll notice is dydx is transformed to r d r d Theta now what about X and Y what happened to them well what's the equivalent of X we did that previously if you look at the definitions that you learned back in high school X is R cos Theta and Y is R sin Theta so we just make those sub substitutions R cos Theta that's your x r sin Theta that's your y okay so we do this trans this substitution if you like of equivalent forms so the transformation involves okay so the forms of the equations the forms of the variables okay so X is R cos Theta and so so we've got our increments shown between the two regions all right so let's have a look at our uh let's just do this this way and think about integrating this particular segment this shape a given this way okay and we're given a cartisian coordinate of function X DX Dy so we've got to go through and substitute for both of those so we're going to have to substitute an equivalent form for X substitute an equivalent form for dxdy okay so where we have an integral sorry I want to draw on this one so where we have an integral for our a x DxD y I'm kind of spacing them a little bit okay we're going to have an equivalent now what's X what's the equivalent form of X based on the transformation from cartisian to Polar well previously we see it so we've got our integral X is going to be replaced with yeah thanks very much that's it R cos Theta what are we replacing there's no y so we're only doing x what are we replacing dxdy with okay and we've already got some idea of what the limits of integration are going to be okay and so we've got our R in terms of the radius okay so 0 to 2 okay and so that'll be uh fairly straightforward for our inner and then Theta will go from 0 to Pi and 4 0 to Pi and 4 okay so we're doing that's our transformation so we've done our transformation from cartisian to polar now that one's fairly straightforward some of them uh get out of hand okay some of them get out of hand now one of the things you'll notice in this particular example is that we've got numbers numbers for our limits of integration so we can multiply them by fabinis and group them okay so we can write and we don't have to do the inner outer there's nothing nothing wrong with us doing the inner outer by the way we can do the inner outer but to simplify it it might be just a forthright to invoke for bonies and in that case what we'll do is we'll do the RS together now R * R will be R 2 d r so that you can cross the M if you like just to make sure you got them all out of the way and our radius is 0 two multiplied with uh 0 to P4 cos Theta to Theta so for V is been invoked and we're doing the product of the Integrations we could have done Ina and there'll be other occasions where we can do Ina but I just wanted to show you that uh particular example okay so you can can make life a bit easier if you can okay so uh for our part here then um what are we going to do Up and Under uh now the the differential of s is cos the integral of cos is minus s Okay so we've got minus sign remember that this is a product so just be careful here this is the not a let's just put a big product line there - sin Theta between 0 and piun on 4 so we have to do our uh substitutions accordingly so we got 2 Cubed on 3 - 0 Cubed on 3 time - and 4- minus s0 so just take your time and make sure you get each of those parts okay and make sure you get your groupings so for the algebraic part that'll be St quite straightforward okay so that's gone that that was a sum so we're left with 2 2 4 2 8 okay and we're multiplying it now this guy is asking from exact values trig what's the Y at Pi on 4 what's the Y at Pi on 4 so Pi on 4 is here so it's the one that's got the same length there and there so the the Y is 1 /un 2 so it's -1 / < tk2 and what's the Y at zero so here's our Z and what's the Y there the Y there is zero so we can kind of forget about everything out of here now because s minus 0 is just minus 0 so what we've got here is 1/ < tk2 now you're going to find doing the polar rectangular stuff if your exact Valu trig is not up to standard um that you're going to find it a little bit tricky okay I wish I could make a shortcut or give you something that would help you but if your exact value trig uh hasn't been worked through uh then unfortunately it's a bit problematic so what we end up with is um now uh differential of s is C so the integral of C is sign and I put minus there that's very naughty of me so that's a plus so we get 8 on 3 > 2 uh the usual convention is to take the third off the bottom so we'll multiply that by 3 < tk2 okay that would give us uh 24 8 tk2 might I should have done that perhaps back here oh always make things more complicated don't I trying to tell you to make it simpler so what I might do is I might just do the root two here so 1/ < tk2 is going to be < tk2 over 2 so it's the same thing < tk2 over2 so that's 8 on 3 * < tk2 over2 that's probably easier to bear okay so we get um a little bit of cancelling going on so we'll get four on uh 4 < tk2 on three okay so a little bit of extra work there to work through just watch your integrals make sure you take your time okay all right so it is possible uh and I've put all of that on the computer so I'll post this afterwards but on the computer the printout obviously doesn't explain anything so you jump to those areas straight away all right uh now we could get some cartisian coordinates that uh have some very horrific style uh problems in them so we would be grateful to be able to change and transform the system so that we can move to um so I'll just bring it to here and we'll think about this integral in the bottom here wow what a horrible integration Okay so we've got a circle we've got the limit of integration so we've got for X which is on the inner okay so let's have a look at the Domain we've got - < TK 1 - y^ 2 1 - y^ 2 uh x andun 1 - y^ 2 wow so that's an interval okay and Y is simply one so these intervals represent um the half circles so x^2 + y^2 is a circle uh with a radius of one okay and these limits are the p m parts so in other words they're the bottom half and the top half okay the bottom half and the top half so we need to recognize that this region is in fact a circle of radius one and we go around radius one okay so from that we can see that integrating this uh putting these into our by FTC Parts wow who would that that's rather difficult isn't it okay that's rather difficult so what we want to do do is try and do a transformation of this into our components so we've got our DX Dy so that's going to become r d r d Theta and then we've got our X2 and y^2 components now who can tell me what X2 + y^2 would look like okay so X2 + y^2 is going to be in its alternate form here we go change forms now so we've got R cos Theta squ + r sin Theta squ we know that that's going to be R 2 cos² Theta + r 2 sin s Theta we can take the r out together R 2 cos² theta plus sin s Theta and so who can tell me what this is who can tell me what that is please who can tell me what that is please cos s theta plus r² th so if we've got our rectangle yes thank you one so an actual fact by doing the transformation we have simplified this okay so this guy here is going to get translated to R 2 wow that's really made it simple hasn't it Okay so we've got our 0: one and we've got our zero all the way around 0 to 2 pi okay so we can build up our integral using this information so we're going to have r d r d Theta following our routine x^2 + y^2 is substituted for x squ our R is 0 to 1 and our Theta is 0 to 2 piun how easy is this and again you can do this thing now notice there's no Theta terms here so if we do for ven it's going to look a little bit odd but that's okay uh we'll do 0 to 1 now notice that R * R is R cubed d r multipli by the integral from 0 to 2 pi Theta now you know what that means I pointed this out to you last time you know what this is here okay you know that that is effectively going to conure a Theta where there isn't a Theta there will be a Theta if that was T so on and so forth okay so in this regard uh We've now done our transformation and moved all of our parts over and how good are these okay so now it's up and under so get uh four over four I forgot to do that last time okay and then we've got our what are we Conjuring here so we've got Theta haven't we between 0 and 2 pi okay so that's just a straight substitution in numbers Okay so we've got 1 over 4 subtract Z four Theta which is 2 piun - 0 so that's obvious so this is 1/4 that's gone that's gone so we've got 1 over4 * 2 pi and so we get a cancel so we get Pi on 2 okay and that's our final result p 2 so the transformation process is a bit tricky in picking your geometry doing your equivalence and making sure that you redraw so that you can get new limits okay now I've done all of that on the computer side so this is where it just jumps out at you now wait what where who what what there's my RDR D Theta that's the substitution for R2 according to this other work up here okay that was all of that so those things aren't always obvious so if you find yourself struggling a little bit how do we get to that point it's probably not as complicated as you think so they are interesting and they are doable um but it's uh you've got to be patient with yourself okay so let's have a look uh at another application and let's think about that so use polar coordinates to eval valate or an improper integral now why are we able to do uh an improper integral because we're going to um we're going to rely on the fact that the function has this decaying property okay this decan property okay so we're going to rely on that now what can we do to make things a bit easier for us well it turns out that now you're going to say to where did you get that problem well okay so it turns out that this also has an equivalent form and if I put in y that they are the same okay that's helpful why is it helpful okay because we can uh and this is so I also so if I just kind of put a bit of a dotted line across there so that's also I so one of the techniques we'll use here is to see these properties and use them together okay okay so what we're going to do is come over here I'll just start a new page I think so to make life easier we're going to I 2 multiplied so remember that there's a multiply CU I'm doing for V here okay so now we've got something where we can think about uh our index rules okay so what we can do is use our index rules what's the rule we can sum them and we can put the DX and Y together okay so all we've got to do is take the square root at the end and we can make this into uh a pretty straightforward integral so the integral we're looking at doing is replacing a single integral with a double integral by squaring okay now it's not obvious why we're doing it I get that uh in terms of experience but we we're using two integrals a single integral and a double integral to overcome the difficult ulty of dealing with this improper integration okay so we're able now to make a transformation from our double integral in D xdy to our integral in r d Theta now what is this guy up here let's look at our parts we've got r d r d Theta and then we've got this bit up here so - x^2 - y^ 2 is that not the same Asus X2 + y^ 2 is that not the same as cos 2 thet + y^ 2 sin 2 Theta ah is that not the same as uh R out the front squ - R 2 is that not one again so this whole thing here is going to become minus r 2 for these reasons okay so it's uh our transformation so let's do our transformation remembering that we're still doing I2 for the purpose of dealing with the uh indefinite integrals sorry yeah okay so what we're going to have here are the new integrals e- R 2 r d r d Theta okay so our RDR is going to take its Lea from X and Y with are going to Infinity whoops okay and we're still going to go 0 to 2 pi so 0 to Infinity so that's one we don't know what that is so we're just going to have 0 to Infinity okay and we'll also have 0 to 2 pi okay so that's uh uh reduce the complexity of what is ostensibly a quite complicated integral okay so what do we got to do now well we've got a little bit of work to do this is not straightforward by any means because of this chain rule this is a u substitution okay so we'll use the U substitution idea to help us uh forward with that so in this case we'll do the inner even though because we've got we've got to account for this uh part here okay so the first thing we want to do is uh integrate 0 to Infinity e- R 2 Rd so that's what we're looking at Okay so we've got to do a u substitution okay so what's going to end up happening is we're going to swap these around and we'll get minus a half du equal R Dr R Okay so there are our substitutions we're substituting these guys okay so we're going to have minus a half integrate to U that's our U substitution I know it's not straightforward but the beauty of it is e to the U aha that was the goal the goal was to make that simpler Okay so we've made it simpler uh We've also got to just check our Integrations using function notation we've changed our limits zero so we've got minus infinity squar and - 0 2 so we've got 0 and we're just going to have uh so it's just minus infinity isn't it so where before we had a zero and now we've got a minus infinity now that makes sense because we've got this decaying part okay so it's going to Decay to zero all right the next thing we want to look at is um uh we want to have a look at is something that you would have known from school uh show the area of a circle is < r s so I guess here we get to use a transformation to show that the area of a circle is PK R 2 so we've got an integration over a domain okay and we are going to do a conversion from our X Y cartisian coordinates to our polar coordinates okay so there's our region now the way that we'll do that is that we'll transform the X d y into r d r d Theta so it's fairly straightforward our R is 0 to R so there's our 0 to R and Theta is going to be 0 to 2 pi okay so you can see that it's a fairly straightforward now because it's all in numbers all we've got to do is split them okay so we'll split them and uh operate on both of those so that's up and under it's R2 R where we substitute algebraically and this one will be simply 2 pi minus 0 so we'll get 2 pi so that's your product okay that one is just 2 pi so we get the substitution for R so we get 2 piun / R 2 the twos cancel and you're left with the area of a circle is p piun r 2 so I don't know uh it's now you know how to do the Integrations and the transformation it's a very straightforward uh integration no big deal there but uh a very interesting idea all right so um the next thing we want to do is have a look at the general changing coordinates now what I'm going to say to you is that this is a bit of a wide knuckle ride you know when you hang on tight whoa hang on tight you might have thought you've been hanging on tight at the moment but no that's not the case at all so in this case here we can do the Matrix transformation idea where we have our uh X and Y so there's our origin and X okay and we find some value 1 one okay and then we look at making a transformation now you can see what's going to happen we're going to be using a chain rule okay to get this transformation to to this point here 1 one okay uh so we would like a way of writing da equals the what in terms of new coordinates okay so that's an understatement so basically what we've got in our transformation of uh one one coordinate so I'll just put that back up okay and a one one is this one come on where are you now one one and so the are the area element from XY so that's our coordinates frame of reference to St we use this thing called the Jacobian and you guessed it this is the debt okay this is the debt form and you'll notice that you've got dxds so there'll be a an equivalent dxds dxdt dyds dydt and we've been able to do that so that transforms this coordinate system using the Jacobian okay so where we'd have our uh domain in fxy to dxdy okay we'll now have a double and the double will look like this and we'll have x s of x X and then we'll have y sorry I want to uh change that to a comma so then we'll have y s a t that's better y d sdt Okay so we've got this uh transformed integral so it is a bit of a White Knuckle ride it's something going to be uh no I wouldn't do that to you okay I won't put a jacobia you in this course because we don't have enough time to spend a lot of time on this stuff we would do it in a normal full course and Analysis is a subject that does that but because half hour course is statistics we won't be putting that into your examination so you can go okay that's uh that's fine all right so just an example uh cartisian coordinates okay so if we go ahead and do uh all of this we'll end up with uh R cos² this is this multiple minus- R sin s okay and then we can take out the minus of minus is a plus Okay so we've got R squ plus r sin s we can take R out as a common function yeah it's very tricky to recognize that as one okay so we get okay all right so we can expand this idea further all right so the next and we can do uh cartisian to um elliptic coordinates and we can use uh an elliptic transformation and here's our transformation alternate forms I know you're smiling aren't you you're going wonderful okay but um we can uh believe it or not we can manipulate all of this to our advantage so we can go for uh Theta usually 0 to 2 pi also V which is our uh positive okay so we can um in terms of uh our formula we've got x^2 over a 2 cos 2 U + y^2 / a 2 sh squ so that would be our base equation that we would be able to manipulate in order to do substitution of similar forms okay so we would we would be able to do that uh so X over a cos U is cos d and Y a shine okay so then we can use these as uh our transformative our transformative information so building up the uh the ydx equivalent and so we get this dub so this is our new transformation pairing okay and uh you'll notice that the slide stop there okay so we can do more advanced coordinate swaps uh in order to make Transformations between different systems not just cartisian to Polar but to elliptic coordinates as well or and really the skies the limit so there are subsequent causes in mass that you can do that don't have the tie upward stats the way we have this one one tied up uh Advanced maths and calculus would be Advanced calculus Advanced maths and physics would be subjects that uh would cover more of this in depth okay so uh we're at 4:30 so what we're going to do is um that's taking us to the end of nine uh so we're going to finish there with nine and then we're going to jump on to 10 okay and we'll have a look at triple integrals okay so for triple integrals uh we want to uh integrate over volume over region in three dimensions so integrals in 1 D are defined as a limit over Sun so this is a revision again that you're familiar with this in terms of its ran sum in 2D our double integral has an equivalent IJ room excuse me room and sun and now we go one step further and we Define our integral in 3D as a limit of sum not just as areas underneath okay and doing areas but boxes the boxes are the elements of integration for a triple integral so our change in volume is not just X and Y it's also Z and so the re integral looks a bit more complicated but I think you're getting the gist of how these things are put together so we've got uh X there this one was Y and the vertical one Z okay so as the number of boxes increase the limit of the slices goes to infinity and so we can see this grow and subtract with smaller and smaller infinitely small boxes um to get our coordinates so we probably won't get through all of this today but it's worthwhile making a start and getting it out of the way for us in case we've got uh particular things we want to focus on for revision all right another way of writing this is to write it as a triple integral and notice the change it's no longer a region it's a volume okay so that's often what you'll get and then we've got the limit for our uh volumes approaching Z for all of the boxes that is the boxes are going from a bigger size to a smaller size infinitely small okay and that's this idea here um again for we've seen this before for our uh our double integrals again we just look out for that okay there it is notice the difference this has got your fun function in it for these guys what's going in here okay not this but we understand what that means now okay and so we can uh look at it that way all right so for some application it lends itself to a volume of one all right as in 2D 2D the integral can be computed iteratively a rectangular box in 3D is described so these will be simp ones we wouldn't give you a nightmarish transformation for example polar transformation or other but yes you can do all of that with triple integrals so what we would do is look at keeping it simple uh probably where you'd be able to invoke fa beanis okay um the order isn't important okay if uh there's numbers in the limits of integration all right um so the reason that you swap them around is to make it simpler okay so for X that's our A and B for our Y which is out the back here we've got c and d and then for our back corner and our top Corner we will have p and Q okay so we're using those uh out the back so just that c should be out the back here okay so um probably the best thing we could do is have a got an example okay integrate the following so we've got xy^ s z is our function okay DX Dy d z where is a rectangular box with the following and you'll notice they're simple number equations so they've been allocated their respective XY Z integral uh limits of integration so x uh is being put at the back one and two because for beinging says the order is not important okay uh Y is being put in the middle 1 to three and 0 to 1 for Zed is going in the middle okay so if we do the integral 0 to 1 um then is there a zed yes there's a zed okay 0 to one so that's pretty straightforward uh Dy okay so provided you uh recognize that you can invoke fies first to change the order and then you can invoke the multiplication of the um terms you can simplify something that looks pretty horrific and we don't ask you to go much beyond that that's actually about as difficult as it gets okay so for this guy here Up and Under we'll end up with uh z^ 2 / 2 between 0 and 1 we'll get y 3 over 3 between 1 and 3 we'll get x^2 / 2 between 1 and 2 so we'll get 1^ 2 / 2 - 0^ 2 / 2 that does that guy so that's a half this is 3 Cub over 3 - 1 Cub over 3 so this is a product so this is going to be 3 3 are 93 are 27 uh 27 / 3 is 9 that's a thir so that product is 9 - A3 for that one uh for this one here uh we've got 2^2 / 2 - 1^ 2 / 2 4 over 2 is just 2 2 - a half so just be careful with all of your fractions and you what are you adding and summing and what are you multiplying so this is a half so that's got to be 27 all 3 - A3 that's got to be 4 -1 / 2 so we've got uh 4 3 12 oh we can cancel some of that I guess why don't I do that I always go for it don't know sometimes a better off pausing uh so two will go 13 so I'll be left with 30 two and that's your answer all right so it's possible to do them but use these use these techniques to help you okay uh where they get harder and harder uh and we'll do that now but I'm just making a bit of a point if we did do something like for an exam this would be the type that we would use okay we would use it so that would only have numbers in the limits and so then you'd be able to uh do that so I put the computer print out there okay so we kind of draw a line there with that Aster and say that's about as hard as it get now could have e to the whatever in it you'll see that in past if you look but um at this point it won't be like one of these guys okay so describing uh more General regions in 3D in a 1D region we describe it by an inequality in 2D we looked at a pair an outer and an inner set of inequalities okay and in 3D we need an extra pair of in inequalities to describe the domain so we've got our A to B our G and the convention is we use uh U with zed and you'll notice the order is going up constant function of one function of two okay just note that as it goes on Okay so we've got our three parts we've got our constants for A and B then we've got our G of X and G of Y okay a g of X and G of Y and then on top of that we've got our surface which is this guy so the upper part is this bit and the lower part of the surface is this bit so it's a bit more complicated it's a bit of an understatement isn't it all right so then if we integrate over 3D domain look at the form of the equation we've got a three value function XY Z we've got the XD yd Z and then we put in our uh domains so that we've got our complete set all right so it would be reasonable for us then to uh to do an example that would be reasonable for us to do that okay so let's have a look at an example oops we'll stop there okay so in this particular example we've got to do a little bit of geometry so we have to interpret our part so we've got xal 0 we've got planes xal 0 y = 0 and zal 0 and we have another plane give it in this form now you can plot that using technology if you want to you can plot the plane to see what that looks like but extensively what it means is we're going to have our X our Y and our Zed from the origin we'll go X will be 1 0 to 1 y will be 1 - x so 1 - x will be this Line 1 - x is this line 0 to 1 - x that way if you like and then we've also got Z is 1 - x - Y which will be up here so we've got a tetrahedon which is effectively a rectangular pyramid that's sitting on its point or sitting on one of its faces okay so that's the 1 - X part and so we've got a few little manipulations we need to do in order to get each of these guys out okay so so when Z is z so when Z is z uh you'll get x + y = 1 that is y is = to x -1 there it is okay uh sorry I've done that wrong I've written x - one as be 1 - x did that okay 1 - x all right what else have we got we've got when Y is zero we're going to have x + z = 1 so Z will be equal to 1 - x and we can otherwise arrange this as Z equal 1- y all right so we're getting our limits from the basic geometry of our planes and that's very difficult I want to acknowledge up front that for you guys to put that out and organize that in that way um is very very difficult you'd have to go through and think very carefully about your planes your points of intersection what your lines are and all of that stuff together to generate this set of domain integrantes once you've got that uh then you can go ahead and put your integral together okay so integral together for this one uh it looks like this so there's our uh Z dydx is our Zed function and then we've got our integration limits have been added Okay so z y and x okay so all you've got to do then is follow your rules these are now filled with functions so you've got to do your inner and then you'll do your third one and finally you'll do your outer so you must do it this way can't invoke for bonies there's no trick around this you've just got to do the hard yards so the inner one will be uh an integral between Z and 1 - x - y for Zed d z so just be careful how you put this on the page all right so the integral is going to be up and under between 0 and 1 - x - y by the fundamental theorem and so we'll get Z will be 1 - x - y all 2 - sorry on 2us 0^ 2 on 2 okay so the second term is uh algebraic so it's gone and so then we get our next part now is it easier uh to do something with this or to leave it as it is because it's going to go inside the next integral and the next integral is 1 - x so sometimes it's worth trying it we can pull the half out the front so that'll simplify it a little bit we've got 0 to 1 - x that's this guy and we're doing D so we'll just do the Y and we've got 1 - x - Y 2 now you can do the expansion if you want to okay um but we'll see what that brings us okay so we'll end up with a half if we do yeah I put the two over here I just brought the half out which is fine I can do that okay no that's all right that's fine I don't get everything right Lucas that's good um so what I want to do here is think about this 1 - x - y and I multiply that 3 by -1 i' get y + x -1 well in either case I might just do it as it is we're going to end up with a negative I think anyway so the minus would be okay so let's just do that all right so for this one here that would be a u substitution okay and the U is going to be all of this with respect to Y and you're just going to come back with a minus one okay so I'm going to take the minus one out and then we'll end up with 1 - x - y cubed over 3 between 0 and 1 - x so that's the integration completed for that middle set okay so then we've got to put our uh terms in so we'll get minus a half now what are we doing Dy so everywhere there's a y we're going to substitute that so we'll get 1 - x minus bracket 1 - x / 3 so you got take you take care okay and then minus 1 - x - Z that should have been cubed and that can be cubed and the three can be taken out too can't it I think the three can be taken out okay so I'm going to have to continue on the next page because I think I've run out of space Okay so we've got uh so six I'll put that as six okay so inside here got 1 - x -1 + x cubed over 3 - 1 - x cubed over no take the three away Jason you've done that already you counted for that okay so what do we got here - x + x 1 -1 okay so that's becoming zero and so all we're left with is -6 - Time 1 - x cubed so Theus time Aus makes for a plus so we'll just get 6 1 - x cubed okay okay now that goes into the final integral okay and the final integral so we're going to have 1 16 integral 0 to 1 DX and it's 1 - x 3 okay so if I convert that around I'll get a minus back okay so it's a little bit easier to process all right so we're going to get a negative again so we're going to end up with just a positive U substitution U equal so that'll be just one this will become four well so this is zero so that term's gone and this is -1 the^ 4 over 4 so it's an even so it's just one quarter oh0 minus one yeah and so I've got 124th okay that's what I've got but we can check that using wolf okay okay so we're going to explore another version uh I'm just wondering about the time so I'm just checking I'm going to leave oh yeah I'm going to leave cylindrical coordinates until next week so we'll have a look at cylindrical coordinates next week and that'll finish off that'll finish off the course for us we've still got uh quite a few to get through about 20 slides so that'll be fine okay there's a bit of problem solving that goes with that so I want to make sure we get that right okay I know your heads will be full just try and take note of where I've suggested where it's pertinent and useful don't get too hung up on the complexities um because if you spend a lot of time at this end of the course in this last week and a half's bit uh it'll probably rob you of where you need to develop other stuff as well okay all right so just stick to your plan stick to your platform and uh you'll be fine all right I'll see you in the labs thanks everyone for coming along bye now for