If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Disc method around x-axis

AP.CALC:
CHA‑5 (EU)
,
CHA‑5.C (LO)
,
CHA‑5.C.1 (EK)

## Video transcript

over here I've drawn part of the graph of y is equal to x squared and what we're going to do is use our powers of definite integrals to find volumes instead of just areas so let's review what we're doing when we take just a regular definite integral so if we take the definite integral between say zero and two of x squared DX what does that represent let's look at our endpoints so this is X is equal to zero let's say that this right over here is X is equal to two what we're doing is at eight for each X we're finding a little DX around it so this right over here is a little DX and we're multiplying that DX times our function times x squared so what we're doing is we're multiplying this with times this height times this height right over here the height right over here is x squared and we're getting the area of this little rectangle and the integral sign is literally the sum of all of these rectangles all of these rectangles for all of the X's between X is equal to zero and X is equal to two but the limit of that as these DX is get smaller and smaller and smaller get infinitely small would not being equal to zero and we have an infinite number of them that's the whole power of the definite integral and so you can imagine as these DX is get smaller and smaller and smaller these rectangles get narrower and narrower and narrow and we have more of them we are getting a better and better approximation of the area under the curve until at the limit we are getting the area under the curve now we're going to play it apply that same idea not to find the area under this curve but to find the volume if we were to rotate this curve around the x axis so this is going to stretch our powers of visualization here so let's think about what happens when we rotate this thing around the x axis so let me so let's figure to rotate it and I'll look at it say we're looking at a little bit from the right so we get kind of a base that looks something like this so it's my best attempt to draw it so you have a base that looks something like that and then the rest of the function it looks kind of if we just think about it between zero and two it looks like one of those pieces from and if you ever played the game sorry or it looks like a little bit of a kind of a weird hat so it looks like this and let me shade it in a little bit shade it in a little bit so it looks something like that and just so that we're making sure we can visualize this thing that's being rotated we care about the entire volume of the thing let me draw it from a few different angles so if I drew it from the top if I do it from the top it would look something like this become a little more obvious it looks something like a hat point up like this and goes down like that it would look something like that so we're not seeing in this angle we're not seeing the we're not seeing the bottom of it and if you were to just to orient yourself if you were to orient yourself the axes in this case look like this so this is the y axis that is the y axis and then the x axis goes right inside of this thing and then pops out pops out the other side and if this thing was transparent then you could see you could see the backside it would look something like that the x axis the x axis you could see through it would pop the base right over there would go right through the base right over there and then come out on the other side so this is one orientation for the same thing you could visualize it from different angles so let's think about how we can take the volume of it well instead of thinking about the area of each of these rectangles what happens if we rotate each of these rectangles around the x axis so let's do it so let's take each of these let's say you have this DX right over here you rotate it around the x axis so if you were to rotate this thing around the x axis so I'm trying my best to around the x axis you rotate it what do you end up with what do you end up with well you get something that looks kind of like a coin like a like a disc like a like a quarter of some kind let me draw it out here so if you were that same disc out here would look something like this it would look something it would look something like this and it has it has a depth of of DX so how can we find the volume of that disk let me redraw it out here too it's really important to visualize this stuff properly so this is my x-axis my disk looks something like my disk looks something like this my best attempt the x-axis hits it right over there comes out of the center and then this is the surface of my disk and then this right over here is my depth DX so that looks pretty good and then let me just show it shade it in a little bit to give you a little bit of the depth so how can we find the volume of this well like any disk or cylinder you just have to think about what the area of this face is and then multiply it times the depth so what's the area of this face well we know that the area of a circle is equal to PI R squared so if we know the radius of this face if we know the radius of this face we can figure out the area of the face well what's the radius well the radius is just the height of that original rectangle and for any X the height over here is going to be equal to f of X and in this case f of X is x squared so over here our radius is equal to x squared radius is equal to x squared so the area of the face for a particular X is going to be equal to area is going to be equal to pi times f of X squared in this case f of X is x squared f of X is x squared now what's our volume going to be well our volume is going to be our area times the depth here it's going to be that times the depth times DX times DX so the volume of this thing right over here so my the volume just of my of this this this coin I guess you could call it is going to be equal to so my volume is going to be equal to my area times DX times DX which is equal to pi times x squared squared so it's equal to PI x squared squared is X to the fourth PI X to the fourth DX X to the fourth DX now this expression right over here this gave us the volume just of one of those disks but what we want is the volume of this entire hat or this entire kind of bugle or cone looking or gets together the front of a trumpet looking thing so how could we do that well the exact same technique what happens if we were to take the sum of all of these things so let's do that take the sum of all of these things and I'll switch to one color pi times X to the fourth DX we're going to take the sum of all of these things from X is equal to zero to two those are the boundaries that we started off with I just defined em arbitrarily we can do this really for any two x values between X is equal to zero and x equals two and we're gonna take the sum of all of these all the volumes of all of these coins with the limit as these as the depths get smaller and smaller and smaller and we have more and more and more coins at the limit we're actually going to get the volume of our cone or our bugle or whatever we want to call it so if we just evaluate this definite integral we have our volume so let's see if we can do that so this is a now this is just taking a standard definite integral so this is going to be equal to and I encourage you to try it out before I do it so we could take the PI out so it's going to be equal to PI times the integral from 0 to 2 of X to the fourth DX I don't like that color now the antiderivative of X to the fourth is X to the fifth over five so this is going to be equal to PI times X to the fifth over five now we're going to go from zero to two so this is going to be equal to PI times this thing evaluated at two let's see 2 to the third is 8 2 to the fourth is 16 2 to the fifth is let make sure limits write it down 2 to the fifth over 5 minus 0 to the fifth over 5 and this is going to be equal to 2 to the fifth is 32 so it's going to be equal to PI times 32 over 5 minus well this is just 0 minus 0 which is equal to thirty two PI over five and we're done we were able to figure out this the volume of this kind of wacky shape
AP® is a registered trademark of the College Board, which has not reviewed this resource.