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Current time:0:00Total duration:9:07

AP.CALC:

CHA‑5 (EU)

, CHA‑5.C (LO)

, CHA‑5.C.3 (EK)

let's do some more volumes of solids of revolutions so let's say that I have the graph y is equal to square root of x so let's do it so it looks something like this so that right over there is y is equal to the square root of x and let's say I also have the graph of y equals x so let's say y equals x looks something like this it looks just like that y equals x and what I care about now is the the solid I get if I were to rotate if I were to rotate the area between these two things around the x axis so let's try to visualize it so the outside is going to be kind of a a truffle shape it's going to look like a truffle shape and then we hollow out a cone inside of it so let me my best attempt to draw this shape so it's going to look something like it's going to look something like okay so the outside is going to look something like this it's going to look something like that and we care about the interval we care about the interval between the points that they intersect so between this point and this point here so the outside is going to look something like this so this is the base of the truffle that is the base of the truffle it's going to have this kind of truffle shape on the outside but I guess maybe we're on some type of a diet we don't want to eat the entire truffle so we carve out a cone on the inside so the inside of it is essentially hollow except for this kind of shell part so we carve we carve out we carve out a cone we carve out a cone in the center so we rotate it around the x-axis truffle on the outside carved out a cone on the inside so what's going to be the volume of that thing so it's essentially it's essentially if we take a slice of our figure it's going to be this is going to be the wall and we're essentially going to take the volume of this entire wall that we're rotating that we're rotating around the X axis so how do we do that well it might dawn on you that if we found the volume of the truffle if it was not carved out and then subtract from that the volume of the cone we would essentially find out the volume of the space in between the the outside of the truffle and the the cone part of the truffle so how would we do that well so to find the volume of the outer shape so let me draw it over here so actually let me draw it over here the if we think about the volume of the outer shape once again we can use the disk method so at any given point in time our radius for one of our disks is going to be equal to the function let's rotate that disk around actually let me do it in a different color it's hard to see that disks in the same magenta so this is a radius and let's rotate it around so I'm rotating the disk around this is our face of the disk that's our face of the disk it's going to have a depth of DX we've seen this multiple times it's going to have a depth of DX so the volume of this disk is going to be our depth DX times the area of the face the area of the face is going to be pi times the radius squared the radius is going to be equal to the value of the outer function in this case it's square root of x so it's going to be pi times our radius squared which is pi times square root of x squared and so if we want to find the volume of the entire outer thimble or truffle or whatever we want to call it before we were before we even carve out the center we just take a sum of a bunch of these a bunch of these disks that we've created so that's one disk we would have another disk over here another disk over here for each X we have another disk and as we go as X's get larger and larger the disks have a larger and larger radius so we're going to sum up all of those disks all of those disks and we take the limit as each of those disks gets infinitely thin and we have an infinite number of them but we have to figure out our boundaries of integration so what are our boundaries of integration what are the two points right over here where they intersect well we could just set these two things to equal to each other if you just said X is equal to square root of x when does X equal square root of x I mean you could square both sides of this you could say when does x squared equal X you could well we could we could keep it there you could kind of solve this there's multiple ways you could do it but you could solve this kind of just thinking about it if X is equal to 0 x squared is equal to X and you see that on the graph right over here X is equal to 0 and also 1 squared is equal to 1 1 is equal to the square root of 1 you could have done other things you could say ok x squared minus X is equal to 0 you could factor out an X you get x times X minus 1 is equal to 0 and so either one of these could be equal to 0 so X is equal to 0 or X minus 1 is equal to 0 and then you get x equals 0 or X is equal to 1 X is equal to 0 or x equals 1 which gives us our boundaries of integration so this goes from x equals 0 to x equals 1 and so for the outside of our shape we can now figure out the volume but we're not done we also need to figure out the volume of the inside of our shape that we're going to take out so we're going to subtract out that volume so we're going to subtract out a volume our X values once again are going between 0 & 1 and so let's think about those disks so let's think about it let's construct the disk on the inside right over here so if I construct a disk on the inside so now I'm carving out the cone part of it what is what is the area of the face of one of those disks well it's going to be pi times the radius squared in this case the radius is going to be equal to the value of this inner function which is just X so pi times and so this is just Y is equal to X and then we're going to multiply it times the depth times the depth of each of these disks and each of these disks are going to have a depth of DX if you imagine 1/4 that has a infinitely thin and infinitely thin depth right over here so it's going to be DX and so the volume of our kind of our our truffle with the cone carved out is going to be this integral minus this integral right over here and we could evaluate it just like that or we could even say okay we could factor out a pie out of both of them it actually there's multiple ways that we could write it but let's just evaluate it like this and then I'll generalize it in the next video so this is going to be equal to the definite integral from 0 to 1 you take the pie outside the square root of x squared is going to be X DX minus the integral we can factor the pie out from 0 to 1 of x squared x squared DX and we could say this is going to be equal to actually let me just do this is equal to pi times the antiderivative of X which is just x squared over 2 evaluated from 0 to 1 minus pi times the antiderivative of x squared which is X to the third over 3 evaluated from 0 to 1 this expression is equal to and I'm going to arbitrarily switch colors just because the greens getting monotonous pi times 1 squared over 2 minus 0 squared over 2 I could write it squared 1 squared over 2 minus 0 squared over 2 minus pi times 1 to the 3rd over 3 minus 0 to the third over 3 and so we get this is equal to this is equal to let me do that same blue color it's equal to so this is just simplify this is just 0 right over here this is 1 squared over 2 which is just 1/2 so it's just PI over 2 1/2 times pi minus well this is just this is just 0 this is 1/3 minus PI over 3 minus PI over 3 and then to simplify this it's just really subtracting fractions so we can find a common denominator common denominator is 6 this is going to be 3 PI over 6 this is 3 PI over 6 minus 2 PI over 6 minus 2 PI over 6 PI over 3 is 2 PI over 6 over two is three PI over six and we end up with we end up with three of something - two of something you end up with one of something we end up with one PI over six and we are done we were able to find the volume of that wacky kind of gutted out truffle