Math 2B. Calculus. Lecture 08. Calculating The Volume of Solids.

good afternoon let's begin section 6.2 calculating the volume volume of solids so i'm going to ask you about cylinders and to see whether you know what the cylinder is so is this a cylinder yes or no yes okay how about this is this a cylinder yes perfect how about this yes or no well in fact it is a cylinder if you talk about generalized cylinders okay and so today we are going to talk about generalized cylinders what is a generalized cylinder uh let's suppose we have a region on a plane anything at all like this okay now i'm going to put a lot of vertical lines through this and replicate the same thing on top okay okay so this is a cylinder so it doesn't look very good but uh the the top cover should be identical to the bottom cover okay and everything in between uh are perpendiculars of the same height okay so this is a cylinder whose base is a rectangle okay this is a cylinder whose base base is a circle and so is this okay uh and the cylinders are great because we know how to calculate their volume okay and today's lecture is devoted to volumes so for instance uh this one if we have the height r uh the height h and radius r then the volume of this is pi r squared times h right and the same formula holds here okay it's a very thin cylinder so pi r squared h now for this one i don't have a general formula like this but if i know the area of the base and if i know the height then the volume of the cylinder is given by the area times height okay so for example here a b h the volume is given by the area of the base which is a times b times the height and obviously that's just the multiplication of the three dimensions of the of this rectangular body but this is the general formula we're going to use and uh the reason to use it is uh that it gives us a way to calculate uh volumes of anything almost so let's suppose we have uh in three dimensional space okay x y and z going out of the blackboard some somehow here i don't want to draw it and we have a solid situated somewhere between a and b okay so something it's it's hard to draw in 3d but just imagine a loaf of bread or something here okay uh i want to calculate uh its volume so this is solid s okay so i'm going to divide the solid into n slabs of equal with delta x so basically remember about partitions i'm going to have a partition here with a step delta x and i'm going to take a knife and slice this body into slices vertical slices like this okay um so each slice will be like a slice of bread okay so each slice is in fact a generalized cylinder okay uh so the way i do it i don't take a knife i take uh planes okay which are perpendicular to the x-axis so divide the sole in the n slabs of equal width h by planes intersecting the x axis at points x naught equals a x one x two dot dot dot x n equals b okay perpendicular planes like this a plane is perpendicular to the x-axis okay and so at each of those slices i need to determine their area okay so they're going to look like this what is h what is the height here that's delta x right each slice has a height of delta x okay each slice has the height delta x and the area is not really constant okay but the area can be calculated at the sample point the area the areas of the slices is calculated at sample points x i star okay therefore the volume of each slice is v of slice i is the area evaluated at the sample point times the height area times the height generalize the cylinder and now i can approximate the volume which is the sum from 1 to n the area of each slice times its height volume of slice i and i take the limit as n goes to infinity to make these slices infinitely thin and to make my approximation better and better and better okay and of course this limit of the riemann sum can be denoted by an integral an integral from a to b a of x dx okay so a of x is a function of x which tells me the area of each cross section and i will give you an example to show how to calculate it so this is area of cross sections okay so uh before uh i explain this further i'll just give you an example uh and in this in this first example again we will know the answer such that we can check that we are doing the right thing so calculate example calculate the volume of a sphere of radius r okay so let me place the sphere place the center of the sphere at the origin and when i draw it like this it looks like a circle right but it's really a 3d sphere this is x okay so i can say that this is radius r okay um and let's write down the answer before we forget it so the answer what is the volume of that sphere yes do you know four thirds pi r cubed okay we have it down so whatever we'll get should coincide with this so what's the algorithm we have our sphere we have to slice it um we are going to slice it in this direction okay so my planes are perpendicular to the x-axis let's suppose so let me remove my attempts to draw 3d so this smudges were supposed to represent a three-dimensional sphere you just look at the projection here right so if i take a knife and slice it and slice it like this what does the cross section look like it's a circle right it's a circle that goes like this okay so uh so slice it look at it it's a circle uh so the volume is going to be given by an integral of a of x dx which these are the areas of my cross sections what are the limits of integration this is uh the extent of my 3d solid okay so x has to go from minus r to r right um so from minus r to r all i need to figure out is the area of each circle so let's practice um you can first of all you can see that each slice has a different area the biggest area corresponds to this slicing and when i slice here i just have a little circle that has a small area and here so the very last point corresponds to the area of zero okay so i have to figure out the area of those circles let's suppose that i'm here this is my current coordinate x calculate a of x what is the radius at point x what is the radius of the circle what is the radius of the cross section well that's given by this right this is the the radius of my cross section this is the center of my cross section it goes like this and this is the radius it's given by y it is y okay and we can calculate y from the pythagorean theorem what is y y squared is r squared minus x squared correct so this is y this is r this is x okay therefore y the square root of r squared minus x squared r squared minus x squared for example when x is 0 what is y substitute here r minus 0 square root so that's just r and that's correct when i slice my sphere in the middle the radius of the cross section is equal to r when x is equal to r when i'm here i have r squared minus r squared so my radius of the cross section is zero okay so that gives me correct formula so therefore i can calculate a of x that's pi y squared that's the area of a circle of radius y okay area of circle of radius y so it's pi times this squared square of square root r squared minus x squared so this is always the hardest step to calculate the area of each cross section okay and i've done it okay from now on it's going to be easy let me know if you have questions so i can substitute this area into my integral so i have pi r squared minus x squared dx okay and before i go on i want to apply uh a theorem that we proved last time remember symmetric functions well it's the same it's like if i can go from left to right or i can start from here do it once and multiply by a factor of two so it's twice from 0 to r pi r squared minus x squared the rest is really easy so 2 pi and i have to take the antiderivative of what r squared minus x squared and remember r is just a constant so it this is very easy integral to take i have r squared is a constant times x minus x cubed over three integrated from zero to r two pi when i substitute here i get r cubed and this is r cubed over three and when i substitute zero i get zero that's why it's nice to replace it by twice the integral from 0 to r so what's the answer 2 pi times 2 3 r cubed does this look right okay so we did get the same answer okay questions okay so we only have to do uh the sphere once uh it's the rest of the problems we don't know the answer okay so uh let's try some of those so here's a typical problem calculate the volume of the solid obtained by rotating the curve y equals x cubed for between 0 and 1 around the x axis okay so we don't say a sphere or any known thing like this this is something that uh we have to imagine what it is we have to figure out what the solid is okay so to do that we start by drawing that curve they're talking about very easy curve so y equals x cubed and i'm only interested in the in the interval between 0 and 1. it looks like this now what does it do we take this curve and rotate it okay rotating we rotate this uh think of it as a piece of wire okay and we rotate it around the x-axis in this direction what do you see it's kind of like a cone martini cup or material cup or rehearsal's kiss okay this this shape hershey's kiss martini cup yes um like this and so this is a 2d projection of it of course it's circular it's a cone it's kind of a cone okay so so i'll draw it once for so it looks like this right i'm going to erase my three-dimensional drawing because we will need this plot to figure out the things for my integral okay so this is x y um so what do we do we have to for each point x we have to figure out the area of the intersection i'm going to slice this martini cup at uh each x what is the shape of the intersection circle right because it's obtained by rotation uh of this thing the intersections are circles so intersections are circles what is the radius what is the radius of this circle one intercept here so this is x y is x cubed so the radius is just given by the function at this point right the radius is x cubed the area is a of x equals pi x cubed squared because it's a circle okay so it's 5 x to the sixth and i'm done with the hardest part okay so this is the part with which you need to understand uh so let me just tell you that with problems like this we want to obtain a circle by rotating something it's called this the resulting object is called a solid of revolution a solid a solid of revolution okay when something is revolving around something else now uh so is a revolution are great because the intersections are circles by construction intersections are circles makes it totally easy to calculate their area if we know the radius and we do know the radius here because we know the function we know the radius and we know the area uh now we can just go through the motions the the volume is given by an integrating of a of x dx from where to where well that's given right from zero to one so go from zero to one of this thing pi x to the sixth pi over 7 is the answer question um i was wondering about the radius how when you're going through each slice the radius is changing right absolutely but we can assume that it's a constant no it changes so when i slice it here the radius is uh x cubed okay so when i when i slice near zero it's very very small when i slice near one it's one cubed which is one so in this integral it's changing with x it's not a constant okay i thought you said like when you're integrating it like that you said it was a constant the radius no so for here uh here the radius for each intersection is x cubed and area is pi x to the sixth so for each x the radius changes uh it wouldn't change if i had a solid of this kind let's suppose my solid is given by y equals 1. okay so it's this function from zero to one when i rotate this around the x-axis what do i get a real cylinder right a real cylinder so in this case my radius is equal to 1 and area is equal to pi 1 squared so the volume of this would be from 0 to 1 pi times 1 dx so that's pi so in this particular case you can see that the radius is the same for each slice but not here most of the time it's not a constant now let's rotate the same thing around the y-axis and see what we get so rotate the same curve around the y axis and i have to draw again now okay so so here y is x cubed okay and now i have to rotate it in this direction so this is one this is one what does it look like it does not look like a hershey's kiss anymore right it's a different shape it's a ball yes uh and now it's a solid of revolution obtained by rotation rotating around the y axis so if i slice it in this direction my intersections are going to be things like this and i really don't know how to calculate their area okay but if i slice it in this direction what are my intersections circles because it's a ball in this direction so it's to my advantage to integrate over the y axis okay so we can slice in the direction perpendicular to the y axis so the volume will be given by the same formula only in terms of y and we need to figure out the radius and the area of each intersection for each point y so what are the limits of integration x goes from zero to one how about y because y equals x cubed when x equals zero y is zero when x equals one y is also one so by coincidence we have the same uh limits of integration they didn't have to be if this was 2 x cubed then my limit of integration would be from 0 to 2 right so fine so now we have to figure out for a given point y what's the radius of of the intersection here okay so the this is given by x right so given why uh this is why the radius is given by x i have to figure out x given y so if y is x cubed then x is y one third right so this is the radius and area is pi times this squared so pi y two-thirds questions so my operation here is slightly different i have to express everything in terms of y given y i'm going to change my y from zero to one given y what's this length this length is given by x y and x are related like this so i have to solve for x in terms of y so i reverse my function and you always do this when you uh integrate in the y direction okay you kind of have to resolve everything in terms of y so now uh the hardest part is over and i just write down the formula pi y two thirds okay so pi y five thirds five thirds from zero to one three pi over five so which one is bigger when i rotate around the x-axis or when i rotate around the y-axis this one is bigger you can see 3 pi over 5 is greater than pi over 7. because this one is skinny and this one is fat right you can you can tell okay fine uh it all these problems they get increasingly worse so the next step is the following uh calculate the volume of the solid obtained by rotating region r around the x-axis what is region r region r is given i'm sorry is is enclosed by curves y equals x to the fourth y equals 8x and uh x between zero and four okay so when they start talking about region r this is what we draw first okay uh so one curve is a straight line 8x like this and the other one is a parabola and this is region r so this is y equals 8 x this is y equals x to the fourth okay let's find the intersection okay find the intersections we learned it last time we have to equate them right x to the fourth is equal to eight x so i have x x cubed minus eight equals zero what are the answers x equals zero and x equals two right so this point here is two i don't see the point of this condition actually let's erase it um okay now how do we visualize uh the solid we have to take this piece like this wing type thing and we rotate it around the x-axis so this is the x-axis we're going to spin it like this what do we see so if we look from the outside we actually see a straight cone right the outer surface is straight now it's not just a cone because i can put my hand here and there's nothing there so it's a kind of a hollow cone right it has a certain width to its walls but it's hollow inside so what happens if i make a cross section in this direction what is the shape of that cross section is this a circle no it's something else what does it look like draw it in your in your notes figure it out does it look like this when it what do you call this a disc a disc is just like this right but it's not it's hollow inside a washer or a donut okay this is a washer when i started teaching this class i didn't know what the washer was except for a dishwasher this is not it it's it's it's a thing that looks like this okay you can call it a donut but it's flat okay so this is a cross-section when i slice this cone in this direction uh because of this structure and it's totally easy to calculate the area of this object because it's the outer circle minus the inner circle right so if i know the outer radius and i know the inner radius i can calculate the area of this cross section so let's do that so what is the area of a washer of radii so a washer is characterized by two radii right an inner and one and an outer one so let's draw it like this this is y out and this is y in this is bad sorry so the area is given by of course pi y out squared minus pi y n squared i take the whole area and i take out the area of the hole in the middle right whole of the doughnut gets subtracted okay so to what i'm after here is the areas of the cross sections and i have to figure out the inner and the outer radii of the washers given that solid perhaps i'll go back to here and mark it here so let's suppose i'm making my cross section here this is x so i have to go from the center what is the radius of the outer uh circle y out that's given by the outer function the one that's furthest away from the axis eight which one is it 8x the straight line okay and y in is therefore x to the fourth okay so this is x to the fourth and this is 8 x now i go back and i can calculate a of x is pi and i have 8 x squared minus 5 x to the fourth squared at this point i can simplify a little bit and then i'm almost done so i have 16 x squared minus x 2 oh yes 64. thank you like this and i'm done with the hard part so now the volume of the whole uh solid of revolution is an integral from 0 to 2 y2 because my solid runs from 0 to 2. we'll calculate at the point of intersection a of x dx so from 0 to 2 phi can go outside 64 x squared minus x to the 8th and the rest is easy can i just give you the answer here or they want me to go through the steps skip the steps okay so i all i need to do times pi all i need to do is calculate the anti-derivative of these simple powers okay okay it gets worse okay now i'm going to take the same region and rotate it around something else rotate the same region r around the axis y equals 20. so that's going to be quite different so let me leave this picture and figure out what this is my region r is the same x so let's find out what this point is in terms of x we have x equals 2 what is y here so because this is both lines x to the fourth or eight x this is sixteen and my axis corresponds to y equals twenty so it's here i have to draw this axis then take this petal or this wing and start rotating it around this axis like this okay so what do we see what is it this is a bowl without a bottom can you see this so let me draw it here just kind of it's something that looks like this so the the projection looks like this so if you put a uh a disc here that would be a ball i could eat from here right but it doesn't have a bottom do you see this is it clear what we have here okay so um what are the cross sections in uh this direction washers again okay i want to drill better so imagine a bowl like this now the walls are a little bit thick like this and the bottom is missing okay so if i slice it like this i'm going to get a washer so cross sections are washers the hard bit here is to figure out the radii okay so let us try to do this so the radius is the distance between the center and the uh edge right so we are going to measure our distances from the center not from here we are not rotating around the x axis we are rotating around this axis so what is the inner radius corresponding to x so it's the distance between this and the second denoted y inner okay so why you know how do i obtain this distance what is this distance this is 8x right so of corresponding to x the top function is 8x so this is 20 minus 8x okay and the outer radius is given by the distance from the center all the way down to this function so it's 20 minus x to the fourth x to the fourth okay so now the area of the cross section is pi i out squared minus pi y in squared okay so pi and i go 20 minus 8x squared minus 20 minus x to the fourth squared and now i have to i know i'm wrong because everybody says it but it doesn't is this correct yes oh yes yes yes thank you so much thank you uh x to the fourth yeah so you see if you get them confused then you get a negative volume like this right thank you now tedious part begins because so before i use this in my integral i have to simplify it this is the stage at which i simplify okay so i have iai um look at this 400 goals uh and that's all the simplification i get so let's go here so the volume is an integral from 0 to 2 a of x dx so it's pi zero to two of this whole thing and this is totally easy but tedious so i skip one step to give you the answer okay and you can check it because i know this is correct i did it at home okay so question am i mad at you why this is gross you don't like it well it gets worse are we expected to simplify to the final stage or can we just like plug it in so for my midterms i'll try so these are uh problems that you can expect for the final and midterms okay uh maybe not so bad for this midterm but for the final for sure and we try to give you algebra that is better than this okay such that you can do it without the calculator okay so uh let me just talk to you a little bit about the plan okay so i'm going to uh have a whole uh almost the whole lecture on wednesday devoted to the midterm the midterm is on friday okay and we still have one very small section to cover because we have one fewer lectures than than uh sections that we need to cover so uh in the last five minutes i'm going to explain to you uh one of the simplest things okay compared to this it's very easy and then they'll let you go and wednesday we do review okay so the the section that i need to cover now is section 6.5 average value of a function okay so if i were to calculate the average height of students in this room the rule is simple i take everybody's height i add them up and i divide by the number of students so this is clear now if i want to calculate the average temperature it's harder okay because temperature as a function of time is a continuous function what am i going to add up and divide by how many right so uh an average value of a function can be calculated approximately by pretending that i have a number of discrete values so what i do is i'm going to split this function into number of points okay uh x naught x 1 x 2 and so on okay i'm going to add these and divide by n as many as i have uh values so let me give you an example uh temperature so time in hours and temperature in fahrenheit so let's suppose that take this graph and i use values one two we're going to only do six hours okay so let's suppose that the temperature was 70 72 80 72 75 76 and 60. okay so i i split the the six hours into uh one hour period uh recorded the temperature at the top of each hour and now i can calculate the average temperature obviously it's given by the sum of the of the all the measurements divided by actually i have 7 because i also counted 0. so it's 72.86 now of course i could have measured the temperature every second and that would make this uh calculation of average more precise in fact i can take the limit so for a continuous function we are going to say that f average is going to be equal to f of x 1 plus f of x 2 and so on f of x n divided by n okay these are sampling points i have a partition i measured the temperature for each of the intervals i add them up and i divide by how many i have okay so this is sorry this is approximate okay and i can take the limit to make it exact so it almost looks like our integrals but not quite what the difference we divide by n and we don't multiply by delta x but this is a very easily fixed so let's recall that delta x is nothing but b minus a over n okay therefore one over n is the same as delta x over b minus a i'm going to rewrite this like this the limit as n goes to infinity of the same thing and instead of 1 over n i put delta x divided by b minus a so now i have everything that is in the integral and i divided by b minus a so by definition i rewrite it as an integral from a to b f of x dx divided by the length of the interval question 77 oh i don't know 80 72 right i'm sorry i was hoping somebody would ask okay so to summarize how do we calculate the average value of a function on an interval between a and d we take the inter integral f of x dx and we divide it by the length of the interval one example and i'll let you go so um find average value of the function sine x for the interval between 0 and 2 pi let's do this so f average by definition i have to integrate from 0 to 2 pi sine x dx and i have to divide by 2 pi minus zero right so that's all we do we just have to evaluate the interval uh the integral over the whole uh interval and interval is always specified and divide by the length what's the answer here so i get minus cosine x of zero and two pi i have minus one minus negative one okay is zero let me illustrate on i have to give you this example um let's draw a sign okay what's the average value of this function between zero and two pi think of it as snow or sand for californian people with sand for russians it's snow okay i want to make the average level so i take a shovel and i take it from here and fill it up fill it here and so when i make the actual average level it is zero okay and that that is the procedure that is explained by this interval integral thank you very much

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Keywords: UCI, UCIrvine, OCW, OpenCourseWare, Komarova, Volume, Solids, Math, 2b, calculus, single-variable, intersection, circle, radius, area, washer, average value, continuous function

Id: B1qXIkr05tk

Length: 55min 12sec (3312 seconds)

Published: Wed Oct 16 2013