Initial Sxw Made Easy

Initial sxw

so hopefully we Scout on MyMathLab if you haven't please try to do that as soon as you can again any issues that occur the faster we can get it fixed the better off you're gonna be so any issues let me know there's that many hundred or one eight eight eight on the syllabus so just try to get everything in public I guess we haven't quite finished 6/1 but does anyone have any questions on the pill I think we have two more examples left in 6.12 we might just briefly mention this it's it's example eights building off of what we've already done with position and velocity just another step removed so they're actually given the acceleration initial velocity and initial position and we want to find the velocity function and the position function so remember we have two ways to do this there was the antiderivative method where we could find the antiderivative and then use the initial conditions to find the constant C or we could use the fundamental theorem of calculus method which was here in this box so if we look at it through that lens it's you know it's the function equaling to the initial value of that function plus the integral and then this B here of course is the derivative of s right so we could just make a similar statement with regard to velocity and say that V of T is the initial velocity plus the integral from zero to T on this this functions derivative right so V Prime there's a of X DX and in the same way get the velocity function from acceleration so hopefully that's not that's not too far of a stretch [Music] this is a spot of tea for us - espera that's whatever did but that's that's kind of the pattern and you know if there are other derivatives involved we can successively find the other functions so we could also use this method here to find both the velocity and position and example I think it was 8 so what's example Aid and it doesn't specify a particular strategy so in that case it would be up to you as to which one you won in two years so I don't know if you want to work that and check to see if it's right we're just gonna skip that cuz we're a little bit behind but it's almost identical to the ones we've done so far that's alright and so this last example is going to illustrate more of a general general change so here at the bottom of the page with example 8 on it we we did briefly look at this table and said you know there's position velocity acceleration type stuff but more generally we can think of it as net change future value rate of change etc so those those can all be applied to you know broader topics like population for instance so when records were first kept t equals zero so that's saying the initial population was 250 people and during the following years we were given this growth rate so that's P prime of T is the population growth rate not the population so they want to know what is the population after 20 years and the second part is to find the population at a time again I guess there's a couple of ways to do this you know we wouldn't necessarily have to start out with Part A the population after 20 years entails that we know the population function and then then we plug a T value of 20 into that so you could start off just by finding the population function P of T and then plug 20 and to answer a so we could do B first and then a or we could do a and then it doesn't really matter but let's try to just fit this in with what we're doing if we're trying to find the population function P of 20 what would that how would we construct that you 2250 Merck didn't want P up the initial okay yeah either one we have the initial value P of 0 plus the integral of 30-plus to start be right right of peace Ron get 31 plus do you'll agree with that yes so we're using that I mean and this is essentially like finding the finding the position function given the velocity using the fundamental theorem of calculus method or if you go back a previous page that is the future value function so the quantity is the initial quantity plus the change of the sum of the change of that quantity and remember in this generic statement here this T value [Music] this T value here matches this upper limit of integration so that's since we want to know the population at 20 our upper limit of integration is 20 and of course our initial was America measure at zero but if it's something different it's going to match the lower valid there so we just need to compute that and you know again thinking ahead in Part B we have to find the population function P of T it doesn't specify how to do so so we could we could say all right well I'm given a derivative I want to find the function so let me find the antiderivative of P prime that will give me the population function plus C and I'll just need to figure out what C is and that give me Part B at which point then I can plug 20 to that all right some of them we're going to sew the antiderivative let's just take this 30 out T plus 3 halves 19 of the 3 house times two thirds evaluated for each one 250 plus 30 DFO or the antiderivative at the upper bound 20 + 2 xx to the three halves of course the second one went to zero all right so you know as we expect this is going to output a number and that number represents the population after twenty years should come out to be a two thousand six hundred and thirty-eight I usually I really don't care I mean in there yeah we're measuring the number of people and yeah we would probably say I don't know probably say six 2638 because we wouldn't count that 0.85 that makes sense it's nothing too terribly off from what we're doing and then the population of anytime does does anyone have a preferred method would you rather do the antiderivative and consult foresee or do you prefer to this method here anybody have a preference this one you're talking so you know there's a couple like this on your homework it doesn't matter how you start so I think you know looking at it one way and then the other is helpful but it's totally up to you and won't make sense I guess so capital P of T would be equal to what how would we find it through the integral this one plus where people yeah this is what is that broad we go from where we were just leave it okay so this would be the this would be the antiderivative method then right yeah because we're just going to find the antiderivative this is this is going to give us the same antiderivative we have up there and then from there we use the fact that P of 0 is equal to 252 5 see yeah so that would be the antiderivative method use that time I see all that the other method P of T with the fundamental theorem method what would that look like 215 value from the zero function you would output it to the variable s okay there's a dummy variable for animation purposes or everyone quest yes okay place the all right good yeah so you know feel free to use whatever variable here the thing is is you just we want this since this is a function of T you want this upper limit to be T whatever variable you use here sxw just make sure it's not a team so we've got our initial value again this is going to turn into a function of T so our antiderivative it's very similar to what we had about above and then and go ahead and distribute the 30 so we can get rid of the 2/3 I guess it really doesn't matter but however you prefer and then this is evaluated remember the zero to T in this instance since we chose s those are s values we got 215 plus over this quantity 30 t plus 20 times 2 u to the three-halves the upper limit - all of that the lower limit will just give us here so there's our there's our function of T it's 250 13 and 22 the same result you get by solving for C etc [Music] I guess so again now that we have a population function for any T if you go back into Part A plug 20 in and we could also do that now just to verify that they are they to coincide that's the basic idea though do y'all have any questions on that when the last page was just a summary so 6.2 we're not we're not going to really do a lot in 6.2 you know you do have homework over it just I'm just curious how many people took calc 1 here so if you're taking a couple in here you really ended kept longer with this section or should have so we might do a couple of examples just to get an idea I know I have like seven or eight examples we're definitely not going to do well with them and the idea is kind of a more generalization of the standard Riemann sums that or areas that you looked at so remember those were for the Riemann sums when we initially looked at when we had you know some function a to be and we wanted to know the area what we had to do was just cut this up into a bunch of little pieces and then we looked at the the areas of the individual rectangles and so that's what's represented here when you see this this is think about this as the height and the width so that's the area of the first rectangle and we do plus the area of the next rectangle plus right but that only gave us a finite you know even if we went to a thousand we've got a thousand rectangles approximate area so then we said you know in order to get an exact area we had to let the number of rectangles approach infinity so that's that's the same as this limit Delta X represents you know let's just say that's one of our rectangles Delta X is just the width of each rectangle so we put in on infinitely many rectangles in there then their width has to go to zero eventually to get infinitely so that was fine for that particular situation what we're gonna extend that to is whatever what if the what if we're not looking for the area that's bound between a function and the axis what if we want the area between two functions and that's what six point two is all about we had these two functions and we wanted to know what the area was and again the same strategy is applied that we're going to cut that region up fill it up with a bunch of rectangles compute the area of each rectangle and then add all that up so then and this idea does extend into six three and six four so here when we said what's the height of this rectangle well it was just this distance from the function to the axis right so it was it was the distance f of X minus zero so we just call that f of X because it was redundant too right we measure the height of each one of these rectangles like what with the height of that one be the one and thanks - to you they write f of X minus G of X so the height of that particular rectangle you know if you knew this Y value was 10 and this Y value was live it's the height of that rectangle is 5 right you just take the top minus the bottom so that's that that idea might you know might be easily overlooked in this type of situation but I think it's more important in 6 3 & 6 for when we start looking at volumes to kind of get your bearings and compute volumes with with a similar notion ok so it's here it's f of X minus G of X for the height of each rectangle more generally when you're when you're slicing things in a vertical direction you know here since the thickness if you will of each slice is in the X direction if it told us a lot of information it said well since I'm slicing it this way we always ended up integrating those with respect to X but in this situation as well when we're integrating with respect to X the the height of each rectangle is also determined you know just more simply by remembering it's the top minus the bottom same thing here is just the bottom function is zero top minus bottom bottom function being white is zero so what do you think then the area looks like the area of this region balint by those two curves would be what integral from A to B minus 2x there's the height of each rectangle here's the width of your table so that integrand is the area of each rectangle remember a definite integral is just an infinite sum so we're summing all the areas of the rectangles so that you know you can look at this as the the more generic Riemann sum what we started off with was the very specific three-month Psalm where this function G of X was always 0 so you know if we can look at it this way you know we can tilt our heads 90 degrees and look at it another way and it works exactly the same so if we have this function let's call it h of y in this function that's W why it's called the intersection point here see indeed it's the same idea what is the area about between those two curves let's fill this up with a bunch of rectangles now run it this way you know I'm gonna keep in mind that we're gonna exaggerate one of these rectangles it has it has a thickness now in this direction which is the Y direction so if we're going to look at it from this viewpoint then we have to integrate with respect to Y so that means just like here a and B were both x values or integrating with respect to X two functions of X if we're going to look at it this way our bounds have to be Y values our functions have to be in terms of Y right so it's it's the exact same concept just rotate it if you will Zelda somewhat clear to have any questions so you know why would we what would we have two ways to do that what does it matter why not just want why wouldn't we just integrate with respect to acts all the time okay yeah so like if you had multiple values you mean you would have multiple integrals to compute yeah so certainly one of these might be a little bit simpler to compute than the other yeah that's that's pretty much the standard of why we have multi ways to do things not only that though if you remember you implicit differentiation does anybody ever so like the reason we differentiated things implicitly was because why what do you remember it's really difficult to do it explicitly and yeah sometimes it's even impossible to solve a function for y for X so we may not do this example but I do one of the point out this this example number three for instance we're given two functions of Y they're cubic functions of Y right so we could say X is y cubed minus 4y and say I don't want to do that with respect to Y because I'm just used to doing that with respect to X you're gonna have a whole lot of trouble solving that for y because you've got to solve a cubic equation right and those aren't easy to solve so in a situation like this we would say well you know it's it's clearly easier to do this with respect to why it's already the functions are already in terms of why and even if they weren't like they're difficult to solve for like if that were just like X equals to Y you know that would be a pretty even trade-off as far as solving for X or Y so sometimes it's not saying that this one's impossible to solve for y but you know each one of these curves here like this red curve would represent you would have three different functions of X representing that red curve so you know extremely difficult to integrate with respect to X so we've got these two ways to look at it again they're telling us the same thing it's just the area bound between two curves I'm going to skip that first example twos more straightforward and yeah I just want to mention this one here sometimes you have to split up your region a lot of times the difficult part here is you know looking again at the more generic situation the difficult part here is figuring out what this a and B are in other words where do those two curves intersect this is oftentimes that's not given to you so if we need to find the region bound between eight over x squared Phoenix and X sometimes there's confusion as to you know like which region do I look at its the region bound by all three of those so sometimes it's a little bit sometimes it's totally obvious sometimes it's not totally obvious but the area we're finding is not this piece right here for instance because that piece is only bound by X and 8 over x squared we want the area that's bound by all three of those so we want to find this area here okay so kind of similar that's the area we're finding so notice it's it's really beneficial to draw these out or to at least try to visualize these regions especially in the next two sections so if we're if we're looking at this thinking about filling this up with rectangles I think that's the first step to find the area and volume for anything is to just visualize which direction are the rectangles running because that's going to tell that's gonna dictate the rest of the problem since my rectangles are vertical up and down again will exaggerate one of these but it does have a thickness in the x-direction so this thickness right here is DX which tells us that I need to integrate with respect to X my bounds need to be in terms of X and my functions need to be as f of X where G of X okay so that I think that's always important as a step one how do you know okay you can do it in terms of Y as well yeah so that's why I would recommend drawing a picture so you can keep your bearing like if I wanted to take this region but have my rectangles run this way then the thing I set up my bounds are going to be in terms of Y the function is going to be in terms of Y order order you know so whichever one you should you're going with it stick with it here we could do either or these functions are already in terms of X so I'm just going to integrate with respect to X so I don't have to solve but we could certainly do it this way when you find the intersection of points we need to look at the y coordinates you know so just be super mindful but I think this is definitely the most important thing to establish first and I think it's a little bit more evident as to why when we start looking at volumes so we're gonna stick with this one here if you want to try it with respect to why you should get the same area same area whether we look at it this way or this way any other questions about over out right now so we see here though that alright we've got this intersection point where 8x intersects X this intersection point where 8x intersects 8 over x squared and this intersection point I mean from the graph we might say we might guess that that zero is print that one's sort of trivial so I guess there's a couple of things to do at this point we've established kind of a setup we want to find the points of intersection so we know where we're integrating from but remember when are when our slices are rectangles our elements whatever you want to call those things whenever they're running in the vertical direction the height of those things are measured by one by how how do I measure this height the one yeah the witch swish function bottom - top or top - bottom without top - the bottom so having this picture we can see all right we'll all of these rectangles here the top minus the bottom it's always 8x minus X right until you hit this sort of transition point where now these rectangles are the top eight over x squared minus X you see the top in the bottom well actually the bottom function here stays the same but the top function is different depending on where we're at so really you've got to break this up into two different pieces we're gonna put this we're gonna put this dotted line here because that's sort of the transition when the rectangles how we define their height when it changes is everybody okay with that so we have this one and then we have that's making that area wants us to draw one of our rectangles here can you not do the right function as a loss function what do you mean the right - the left - the bottom four for what we're doing here no because if we're doing right - left we're measuring something that that direction and those rectangles all have thicknesses in the y-direction if you will so that would be a DUI situation if we're doing right - left so again we need to find these points of intersection and just highlight these this this point of intersection here as far as the x-value is the same that dotted line is the same so those are our points of intersection which give us the the upper and the lower limits of integration all right so this this point of intersection here we said looks like it's at the origin but we can just verify its when X is equal to X or when 7x is equal to 0 or when X is equal to 0 so we know this this x value is zero and maybe we would have guessed that from the graph but some of the examples the graphs aren't provided so you know you want to make sure that you're drawing a picture is sort of get your get your bearings so that's got an x value of zero the second point of intersection here let me just that points of intersection is when eight over x squared equals 8x so we'll multiply here by x squared and then X cubed equals 2 8 x cube equals 8 when X is 2 it's 2 a negative 2 but again from the graph you know clearly we're not intersecting at negative 2 so that's got an x value of 2 [Music] sorry what this one intersection is about the heart rate yeah I think so yeah when the red intersects the blue when eight over exes so that one's that one's an x value to my bad so I'll just fix it this way this is this is this one is it too this one's our purple so that's about that's an x-value and then right the blue one is where the green and red intersect when a tax okay so again multiplied by multiply both sides by x squared so when X cubed equals one when X is two off [Music] you know and don't you know don't rely on the graph unless you can like look at this and say oh yeah well it looks like looks like from the graph that it's one you know you need to plug one into both of these to see that it's the same sometimes those things might be just a little bit off it could be 1.02 or something like that so you really want to make sure that they are intersecting where they appear to be intersecting so that's our x value of 1 so so is this one [Music] all right so you've got two different integrals two computers to calculate that area we'll go ahead and put it down here but notice to find this area exes are exes ranging from 0 to 1 and and it's always the top minus the bottom which never changes D X minus X as for the first piece plus the area of the second piece and that was or whatever at that any questions let's go draw this here all right so we could just call that first one 7x how do we integrate 8 over x squared make it right we just turned it into a power so this is 8 X to the negative 2 X DX and that's our setup I think I'll just stop there everything else is just at our rules there's straightforward yeah if you're going to work that you should get to 26 you'll have any questions on that example so this this one here we mentioned earlier in terms of why you just highlight these two functions this function here on the I guess on a lot the red one is the y cubed minus one whatever this color is [Music] and that's the why cute found so we've got those two curves and it says for y in between negative 2 and positive 2 so you probably can't see it very well there's y equals to 2 negative 2 rather and here's y equals to 2 so a given uh says find the area between these graphs for this particular y-value you know in effect what that y-value does from negative two to two just gives us sort of the top and the bottom of that region but it's it's the area that's shared by all of that stuff so we're really finding all of this area and we'll just set this up but just be mindful now that our cross sections our slices are now running vertically which means these these rectangles have a thickness in the y-direction so we need everything in terms of Y so the area is the integral anyway I want to take a stab at it the right - the left yes so what are the bounds of our integral negative 2 to positive 2 right it's a Y value of negative 2 y value of positive 2 and again the the air of the bounds of integration are determined by wherever you're filling your rectangles so we start with this rectangle here we end with this rectangle up here so is everybody able to do that all right and then what functional reintegrating motional hy - right h- g in other words the right - the left be careful since we've got to subtract off the function on the left and the function involves more than one term make sure you're subtracting off the whole thing okay so that's the setup again you're just integrating powers of Y everybody okay with the setup whether any questions don't roll over so if you're not if you know none of any of these setups and you have questions on please ask because what we're gonna do in the next section is we're going to take something like that and revolve it around an axis to create a solid and we're gonna calculate the volume of the solid so it's it's calculating volume in tells us to only focus on area and so this this idea of how we're calculating this area is perfectly relevant to the volumes that we calculate all right I'm gonna let's go it's a good example for and then we'll look at example 7 to find the area of the region bounded by natural log of X by equals 2 y equals to 0 and x equals to 0 by setting up an integral with respect to Y so here they do tell us integrate with respect to Y we'll see why limit so no graph provided hopefully you remember the natural log graph super common so what is the natural log graph look like this way relative to the to the board yeah okay yeah gol agree is international law what is this point here the one zero it's a natural log function y equals to two right here why equals to zero right here and x equals two zeros right so we're finding the area of this region so notice it tells us to integrate with respect to Y so which which direction are the cross sections running vertical or horizontal vertical so remember if they're running if they're running vertical right then they have a thickness in the x-direction right so that would be perfectly fine if we needed to integrate this with respect to X it says with respect to y so we need our rectangles to run this way so they have a fitness dy so we don't care with that okay so then what would our over the area of this region look like take a shot besides change y equals tax it's containing integral 2:02 Hawaii and dy all right just look at that for a minute what do y'all think so we can look at the balance if we're going to integrate with respect to Y as we're told this Y value here of course is zero we're gonna fill the region up with rectangles that stop at that y value there which is a y-value of two so we've got to see where the two makes sense y e to the Y and not Ln of X it's harder to yeah yeah all right one how do you take the integral of the natural log of X what do what functions derivative is the natural log of X yeah so how do you know that integration by parts which we haven't talked about so that's kind of off the table the other reason and the main reason is because it says with respect to Y so this needs to be a function of Y not a function of X if we were integrating with respect to X then natural log of X as it is is perfectly fine so then since we want to integrate with respect to Y this function needs to be in terms of Y or sulfur X so if Y is the natural log of X e to the y is equal to X so that function of Y which is X that function of Y goes years does that make sense so you're gonna have a real hard time if you say all right well and with respect to why my bounds are from zero to two which is why and then you try to integrate ln of x DX first of all you're integrating with the wrong with respect to the wrong variable you can't integrate ln of x yet and zero and two aren't x-values they're Y values so you know there is potential to mix up stuff here like I just keep harping on the volume stuff because it is it is very very parallel to what we're doing here but if you if you have to find the volume using a certain method and you don't know how to set that method up you're gonna use a different method which may give you the same volume but you're going to end up losing points because you didn't do the correct method and most of the time when people do the incorrect method I think this set up is incorrect - so it's gonna make sure that we're you know all of our bearings are straight before we start tackling that stuff any questions on this yeah do you want an actual testimony yeah I would leave it as if sacked the antiderivative is the other why so yeah we did a squared minus one so that's exact that's right any questions on that honestly if you're in physics you're probably drawing pictures at least some diagrams to solve some of your problems no shame in drawing diagrams here either I would I would definitely recommend it if you're not given a graph and you have to find these areas in these volumes the first thing you should do is draw the graph okay we're gonna look at example seven use the most efficient method to set up the appropriate integral to find the area between the so it's just a we'll just set this up and we'll say to find the area between the curves of sine X sine of 2x on the interval from zero because this is PI here all right so we've got these two curves here which I don't know if you can see that faintly this red to me and then we'll do this other one so which ones which which one which one's the sign of X the red does everyone agree you're in sign it was like so let's let's try to establish the intersections and then we'll take a break a little bit of thought required to find where these intersect again maybe this point here is trivial and this point here is trivial we know that the interval is from zero to five but we need to find this point of intersection too we've got two options right we can integrate this with respect to X which means my rectangles would be running in this direction if I integrate it with respect to X again the the height of each rectangle top minus the bottom that changes as we pass over that point so we're going to have two different integrals describing this area if we do if we do this with respect to why would that be easier you can see even right there that's kind of a problem right if we did integrate this with respect to Y then I have to say this function here is solve for X so X is sine inverse of Y do want to integrate that so I mean that would be enough to just stop and say well I don't want to integrate inverse sine another integrative side it's in terms of X already so let's do DX dy do this most efficient method with respect to X okay so the end points they're fine this intersection point when does the sine of 2x sine of 2x equal to the sine of X how do we solve that equation I used to try anything sine X sine 2x two cosine X sine X right right so you gotta remember the identity for sine of 2x this this identity is really common for us so this sine of 2x we're gonna write as 2 sine times cosine why why are we doing that find a point of intersection between side is the first one in this section yeah who AMS not necessarily you can't cancel out the Sinex I mean I guess you can we could divide through by sine of X if what they're sine of X is not zero right we can't divide by zero if the sine of X is zero we can't divide both sides of the equation by sine X we see here sine of X is zero but the reason we're writing it this way is because now we have basically two terms this term in this term both of them involve sign so we're just going to get both of those terms on the same side and connector that sign out so I mean yeah you if you divide both sides of this equation by sign you do get the the equation 2 cosine 2 cosine X equals to 1 but you would totally miss out on which side is equal to 0 for this particular equation it's not that important because we do have the graph with the endpoints 0 and pi but leaving out the X values when sine is equal to 0 it's it's just not good right you're leaving off half the problem even though leaving it off here is not that big of a deal but you never want to divide through by a function of X when that function of X is zero see here you instead you want to factor it out if possible okay so one when this is zero when sine is zero and when two cosine X minus four we know sine of zero or sine of X rather is zero when X is zero hi and so forth but again we're not concerned with any values after pod that's outside of our this one we've got another zero here when cosine is 1/2 so cosine is 1/2 when X is 1 PI over 3 and the next one's outside of the interval from 0 to PI right is 5 PI over 3 so again we're not concerned with these so those are our points of intersection at zero at PI and at PI over three so now we know that point in the middle here happens at PI over three so everything about a see my same like 2x equal to the scene case that we can that you X equal to X I mean excellent only way that we you know well and after that we can lady by signature suitable side so you just sign you couldn't divide just sign no I mean it even seemed to X equal to CX and no one is that extra in what you define to sciocracy X and we don't miss any well kind of if if we know like with the sine of 2x is equal to sine of X when 2x is equal to X gives us that C value of 0 yes and we at least that right so but wouldn't we miss this one here because that's also when that's also when it's 0 or also win in their equal right when we miss that pie so yeah you've got to be careful with trig functions there a little bit you know that I think the frustrating thing with trig functions is there's so many identities for them right when you look at solving an equation it's not always intuitive like what why should I even do in two if I'm gonna use an identity which which one of the 35 identities is going to help me out right so I will say that you know when you have a situation like this you want the same trig functions on both sides so you can factor that out there's more than two terms like this then typically it's it's a situation where it might it might be some sort of quadratic type of thing but again they're all kind of unity we don't the things you did a pre count to prove identities and stuff I promise the things here and count to don't really get that involved it's really just if you can remember basic stuff you'll be alright does anyone have any questions on this will them after break we'll set up the integral any questions on that at all okay we'll take ten minutes affection like I said typically the points of intersection are usually the most difficult part unless you have to split it up into a couple of different integrals so then our area would be at the first region found by 0 and PI over 3 and again in here we want the top function minus a lot of functions since our rectangles are running vertically top minus bottom DX plus the area from PI over 3 to PI again top minus bottom okay so what so to sideline to sign either way around here it's let's our setup so that we were just asked to set it up so I didn't actually compute this Katie questions on that are there specific instances where you'd want us to write out to us or would it be better like that yeah like what do you mean as far as like you integrating it or like the two sine X cosine X would you prefer to write that out or leave it detective X okay so it really that depends on what you're doing so that's for like solving the equation you know we wanted to write that out as 2 sine cosine but if we have to integrate if we have to integrate this what do we know is the same as this well I mean okay which which one looks more appropriate to anti-fun antiderivative of first one all right can we find the antiderivative of this one maybe not yet at least but with what else I can try to write the only other technique we would have is a u-substitution so if if we if we did write it like this depending on their identity and the trig functions involved you may or may not be able to integrate a product because we haven't technically talked about how to integrate a product of two functions of X but if you did write this one here you know hopefully you're at the point where you don't have to do au substitution for this one but if you are then that's fine it's just going to be you're gonna be a lot less efficient with a lot of things we do so it's easier just to know that that the antiderivative of that okay so what what could we possibly try as a u-substitution here it seems okay so sine X is equal to what say the first part okay so T or you okay so DT s cosine X DX right so this part is now our new differential and this is just to you we can certainly integrate to you so yeah either either way here is fine but sometimes like I said you got to be careful as to it depends on what you're doing if we were like trying to integrate some other trig function that we would is an identity one we might be able to do pretty easily the other one like okay any other questions okay so there's summary here and then I'll go ahead and add another page for a couple of things in the homework to remember six people's work come on when we use my comma already the other one was for cosine of 2x can you let it remember the identity for cosine 2x yeah cosine squared minus sine squared in the number for cosine of 2x there was actually three different identities all based off of this one right here so if you remember this one the Pythagorean identities and I guess you know backing up even further if you remember the like that the sine of 2x is the sine of X plus X and you remember that identity for sine and that's where this identity comes from did you just get sine of X cosine of X plus sine of X cosine of X to get two of those right same thing with cosine of 2x cosine of 2x comes from the sum identity some a difference but some here for cosine but if you can remember these rather this specifically this one for cosine and the Pythagorean identities sine squared plus cosine squared is equal to one and the other two identities for cosine of 2x come through so the other versions of them 2 cosine squared minus 1 and 1 minus 2 sine squared again it you know which one you work with depends on what else is present in the problem you're trying to solve like if there's a if there's a cosine squared and a cosine then that's a that's a quadratic type of equation or solving so you'd probably want to write cosine of 2x s this one here in the middle if there's if there's a sine squared and a sine present then that's again a quadratic so it just really depends on what else is present in the equation you're trying to solve see what else yeah the integrals for cosine squared and sine squared absolutely so integrating these is quite common and to integrate those we rewrite those in terms of their identity 1/2 cosine squared X is 1/2 times 1 plus cosine sine squared 1/2 times 1 minus cosine so those only worthwhile remembering for this sections homework and going forward in general all right everyone okay any questions belong to 63 so the next two sections deal with volumes the general slicing method and the disk and washer method is in sixth one degree at six point three and then the shell method is in six one four does this just sister reminder when we find an area what we did was we filled this region up with rectangles when we computed the area of each rectangle and added that alright so that when we see this that's the sum of these rectangles height x eats rectangles thickness and height times the thickness is an area so we're just adding up a bunch of areas so when we're talking about now here finding a volume we're gonna look at a couple of like geometric figures just to kind of get the sense of how that notion is going to be extended and then look at more like exotic regions here so this prism really the key is the consistency of the figure throughout right like if we imagine cutting this prism or this cylinder or this prism like no matter where you cut it as long as your cuts are consistent you're not cutting one at an angle one horizontal or whatever then they're gonna generate the same cross section if I slice into this prism this way they're all going to be these right triangles so the way we can look at the volume of this thing is we can say that let's just consider this a thickness we'll call it t for thickness and we'll calculate the area of the face and then multiply it by in stigmas to get the volume so we know this this area is 1/2 its base times its height so then the volume of this region it's just that one half the base times the height times the thickness same thing here if we were to if we were to slice this way let's say horizontally all of our cross-sections would be circles so we find the area of that cross-section and then we multiply by its thickness so our volume is PI R squared times two and cross-sections here are all rectangles call that Lincoln wins and missed again thickness is everybody okay with that notion that's that's the notion that does apply to these volumes that we calculate so the first thing we're going to look at is this general slicing method that then sort of parlays into the disk and washer method and the shell method in the next section so like if you have to find the volume of a loaf of bread you know each each cross-section is pretty much uniform and you know those those weird little crusty in pieces are always kind of funky but we have the cross-section we slice that up into a bunch of different cross sections of the same shape if you will as they are well what's the area of this cross section times its thickness we've got uniform thickness for each piece of bread we just need to know the area of the face times the thickness of this piece the area times the face of the second the third the fourth the fifth and just add all of those volumes up so that's that's what the general slicing meant that refers to I think it's you have to presented here on the next page so you know some sort of the volume of this gold figure here or we're still gonna slice this up into a bunch of different regions but instead of like in this in this setting where we're refining an area we just found a length times the width for the volume we want to look at an area in terms of thickness so we're gonna we're gonna slice this region up and notice that all the the areas would be sort of this parabolic shape so we would say the area of each cross-section is a parabola times its thickness and the more of those things we insert or the more cross-sections we perform the more accurate that is so once again we're letting the number of those things go to infinity which involves an infinite sum which again an infinite sum is what an integral is okay so the the key at the to finding these these volumes here is we're gonna we're going to integrate the function a of X where that a of X is the area of the cross-sections for the same doesn't really sell you so the cross-sectional areas is that function that we integrate the differential DX here is now is playing the role of thickness as as over here with Harry it was just playing kind of the same role but it was it was a one-dimensional thing so we have so I see the parallels here this is an area length times width this is a volume area times thickness so when we're given a figure like this we're gonna say the first thing we really need to the only thing we really need to identify that requires effort is what is the area of the cross section so if we're finding an area like we were in 6.2 would just say well what's the height it's at the top minus the bottom here it's a little bit more effort because we have to find the area so this example we're going to look at it says use the slicing that the defined the volume of the solid with a circular base of radius 5 so here's our circulation for circular base and it says whose cross sections that are perpendicular to the base in parallel to the x-axis are equilateral triangles this is kind of so we've got this circular base with this cross sectional equilateral triangle and you know depending on where we are we've got to imagine filling that entire region up with all of these equilateral triangles and when we do that we generate this figure here so we're finding the volume of this thing right here so I mean yeah the the the figure looks complicated but fortunately for us every cross section is the same just like in those pure prisms and then the night cylinder so we can use this particular method to determine the volume which entails all we need to do is determine what what's what's the area function and again the area function is attributed to the cross sections so we know that the cross sections are equilateral triangles and the area of an equilateral triangle is just the 1/2 of times its base times its height but of course that's not a function of X or Y so now the problem is or the trick is we've got to rewrite B and H in terms of X or Y depending on what variable were integrating with respect to so the way this problem is set up for us which what are we integrating with respect to X or Y it's kind of skewed here right respect so it says the the cross-sections are perpendicular to the base and they're parallel to the x-axis so here's our x-axis our cross-sections running in that direction so there's one cross-section the apex of that triangle of course is coming out at you so if dxdy d wide right that's that's a rectangle right it's like this thing here its thickness is in the Y direction so that's a dy we can exaggerate this if we need to so we have to we have to have our bounds our functions in terms of why is everyone clear on that okay so we know the cross-sectional areas are 1/2 base times height so for I'm just going to write it here for this we need B and H in terms of why so remember this thing right here is the base of your equilateral triangle so the length from here to here or the length from here to here is the length of the equilateral triangles so in the base in terms of why is to find hell sorry two times the square root of 25 minus x squared yeah nice right and how are we getting that well I'm measuring something that's running in this direction if this were an area we would say that in order to measure that height I just do the right function minus the left function correct it's the same thing nothing's changed here how long is this green line how long is that green line it's this boundary minus this boundary or the right - the left there will always be the case when you're slices run horizontally okay so what is the right - the left well it's it's dependent upon the equation of that boundary of that circle right so the equation of our circle we're told that the circle has a radius of five so the equation of the circle is x squared plus y squared equals 25 but remember we want everything in terms of Y so we want to rewrite the equation of our circle in terms of Y in other words we want that circle to be x equals 2 so Louis right here our circle just solve for x so this is remember we've got two representations for X the positive X of course is this piece over here and the negative x over here to make sense anybody have any questions and so then put our rectangle here again now we know exactly what the right - the left is in terms of Y because we're it's with respect to Y so there's the right - the left sorry that's a negative as well what - the - which gives us two of those so that's that's B that's what B is equal to in this particular setting any questions and just looking at that figure we saw that those things of course those things varied as we move throughout so that that triangle has much different base length than this triangle has a much different base length than that triangle so everybody good with that okaythat's be in terms of Y now we need H in terms of Y you are finding the area of the cross section so it is the height of each cross section it's the height of this triangle so that's what we've got a we've got to measure that in terms of hopefully in terms of Y if we have any hope of integrating this to find the volume any ideas here yeah yeah right and why would we want to do that right there was there one able to hear that we've already defined the base in terms of why if we can relate h-2b we've got B in terms of Y and we've related H to B which means we can relate age to Y as well and it like in this two-dimensional setting hopefully it's pretty obvious that you know this this thing coming out at us there's there's no way to measure that in the XY coordinate system right we need an Xyz coordinate system to measure that so our only hope is to relate this age back to this being which relates it back to life so fortunately for any any equilateral triangle there is always a specific relationship between the base and the height of an equilateral triangle so there's our age there's our bee and all we need to do is as was mentioned once I wasn't here Ryan well Ryan was saying was we just couldn't look at this can cut it in half [Music] this is still age this this you know is is what what is this link here B because it's an equilateral triangle what is this length here yeah half of B we cut it in half so we've got this right triangle now that we can use the Pythagorean theorem to relate H&B so pathetic come this use the Pythagorean theorem solve for H or solve for B whichever one and then what you'll get is the relationship that's true for any equilateral triangle can always relate the base in the height you may get a yet there's sulfur this is H squared we solve for H squared so for any equilateral triangle the height of it is always root three over two times in space okay that's good for us because we already know what we already know IP is in terms of why just now that be in terms of why is there to route 25 minus y squared okay so that turns into just the root of three times the other all right so we've got the base so the base we've got the height we know now know the area of the cross-sections in terms of Y does anyone have any questions these are rather involved when we look at the other volumes with the the two methods mentioned shale and washer shale washer method they're they're a little bit more straightforward it's because it's kind of like the formula is always the same here with these general regions it depends on the boundary and the cross-sections so usually these problems are a whole lot more involved than the rest of six point three and six point four but if everyone's okay then wall will go ahead and continue because we're pretty much done you've got the area in terms of why because we have dy from the very beginning is one-half the base times the height so then we'll just clean that up slightly that's gone we've got root 3 times 25 minus y squared so that's the area function we need to integrate that function volume is the integral of the sum of the area times the thickness that makes sense questions how did you get to that messy singing to that from me from here to here you mean yet so the what happened the two are gone this route times this route is this thing squared so it takes something takes away that square root so yeah it's good for us because it's a lot easier to integrate now to what are the balance [Music] yeah why - why earning of it so we're integrating with respect to okay so from negative yeah it would be negative why do I write yeah but right remember the the figure we're working with is here the the radius of this circle is five negative five right all of our rectangles are running this way so we're going to fill it up from here all the way up to there so we can get the volume of this whole thing so from negative five it's a positive five does that make sense those are those are the limits of integration here so they could make things a little bit easier using symmetry and go from zero to positive five if we multiply the whole thing by two so it's totally up to you it's just if you're evaluating these things by hand especially it's always easier to plug zero in than it would be to plug negative five right yeah that's what I would do the root 3 is a constant so I would just pull that out front so I mean it doesn't look terrible after we're finally finished with it but it was quite involved maybe I don't know yes it's relative but I think it's it was pretty involved to find the area function itself once we've done that it's pretty straightforward [Music] alright yeah that should be 500 times root 3 over 3 is an exact answer that's our volume and you'll have any questions on that we're going to look at one more of those okay I guess deriving the area of a sphere so we're gonna derive the formula for the volume of a sphere with the radius of capital R and we want to run our slices to be parallel to [Music] so since we're you know since with this method we're only summing up an area times the thickness you never you never want to think about necessarily the full volume or working with the figure that is represented by a volume we always want to translate that one level one dimension down and work with that so if I slice parallel to the y-axis what are those cross-sections going to be they're going to be spear all surface so let's just draw this over here I'm not really dealing with a spear the I mean the the boundary of the sphere is this thing right here and the cross-sections that we generate are against circles themselves here so if I slice through a sphere you know that's giving me this circle who's gonna look at us in this two-dimensional setting because we're not concerned with the volume in order to find the volume we just sum up the area times the thickness so we're concerned with finding the area of these things of the cross-section specifically across sections let's label this the greater boundary here is coming from the sphere itself so this is capital R or negative R so we've identified not only how we're slicing this we're slicing this vertically which means we're integrating with respect to what 2x so let's write that down everything needs to be in terms of X and we've established that the area's rather the cross sectional areas are all circles PI R squared so what do we need from PI R squared what do we need to do we can't just leave it as PI R square you can't find a way to calculate our in terms of things or lowercase R in terms of X right so just like in that last one we said hey the area's 1/2 base times height being H had to be in terms of Y here our needs to be in terms of X because we're integrating with respect to X so let's just write that down we need and notice I'm using a lowercase R here because these things vary right if I'm slicing right down the middle of that thing then the that circle then this circle right here does have a radius of capital R but only at that particular location anywhere else I slice it it's gonna have a different radius so like this this slice here just gives me this radius this is the radius of that circle so which you just well label it as a different R because it's not the same as the radius for the boundary of the sphere okay and we need our in terms of both X because of this DX because of our slices so just keep in mind once again that our is simply the distance from here to here right for any one of those circles that we slice if it's this circle the radius of that circle is just this distance from here to here right so how do we measure that distance from there to there do what using the two points yet so this in point would be wild it's here on the x-axis so just in general how do you measure that length what top - bottom right so we can say that the radius of this circle the radius of this circle is what we really need to do top - bottom who really need to do the top - the bottom so the top minus the bottom isn't the actual radius of the circle what's the diameter of the circle right that's the circumference go to PI R squared I'm not sure what that is - PI R is the circumference yeah the the radius is only the distance from the top of the bounding function which in our case is just this circle - the the x-axis which is zero or we could say I'm going to take this whole link here this whole leaf here which would be top function minus bottom function and then I have to divide that in half because I want the radius not the diameter so however you want to look at that but you'd say it's R is half of the top minus the bottom or it's the same thing as the top - zero however you want to look at that we just need to measure half of it okay so here again when we say Tod we mean from this mistake you're here we know that this figure is represented as a sortable centered at zero zero whose radius is capital R and so before we take the top minus the bottom we're going to remember that our slices were this way and we want everything in terms of X so we want to solve this thing for y so everything's in terms of X plus minus again sorry this is x squared plus y squared equals 2r squared solving for y plus minus the square root of R squared minus x squared everybody okay with that that's the equation of that event circle which represents the boundary of the sphere which ever however you're looking at it any questions okay so we've got a wide that's positive and 2y that's negative so here's our here's our Y that's positive here's the Y that's negative but also our axis runs here right we're just measuring the radius of each one of these circles we just need to do the top function minus zero so the radius R little R is just a square root of R squared minus x squared [Music] it's necessary this right here that's all right so that you know if this is if this is confusing please email me please ask you know the way we're reasoning through this is how you're going to reason through every one of these prototypes of problems so we've got the radius in terms of X then that means the area function we can write out now the area function of X is pi times the radius small radius squared which of course just gives us PI times our spirit - so then finding the volume of a sphere radius capital R what is that going to look like volume is equal to the integral from negative capital or lowercase negative capital R to capital R the area DX right I know some people are sloppy with their differentials because like in your head you know what you're doing but please write them there it's it's going to be a lot more evident when we start doing in like trig substitutions and a lot be a lot more pertinent when you're in count 3 so if you're not in the habit of writing those differentials there you're gonna need to get into that habit ok negative R to R capital R of negative R negative capital R to positive R capital R does that make sense where we're going we're filling this region up with these rectangles so we have our slices that are running this way right so the the cross-sections all run from here to here negative R to R we're integrating with respect to X just because we were told to slice this way right so if you need if you need to label these those are both X values get the pious constant we'll pull that out what's what's the antiderivative of R squared with respect to X sorry someone over here x times R squared L agree what is R squared with respect to X is it a function of X clearly not R squared with respect to X is a constant right yeah you can pull this you can pull that out but you need to have two integrals here you'd have the integral of PI R squared DX minus PI times the integral negative R to R sorry expert yeah so R squared is just a constants we just tap on the X we have the antiderivative of x squared but we're evaluating that from an x value of negative R to an x value of positive R and hi times all of that so everywhere we've got an X we're gonna plug in a capital R and then subtract off all of that with a lowercase R so first one you've got an R squared times this are here for the upper limit of integration minus R cubed over 3 so that's the upper limit capital R in for X or antiderivative - now honestly we could have made this a little bit easier to you by using symmetry right instead of going from negative R to R we could have just doubled this and gone from zero to R but here we are so we'll just keep going negative R now in for X we have R squared plus negative R minus negative R cubed over 3 five times all of this so you know just just for time sake just work on simplifying combining like terms you do get 4/3 PI capital R cubed is that the volume of a sphere or thir it's PI R cubed okay so you know look over these if you have any questions email me or we can start a next class by answering some of these things but you know make sure you understand this as we move forward especially alright then see y'all really good you