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# Disc method rotation around horizontal line

AP.CALC:
CHA‑5 (EU)
,
CHA‑5.C (LO)
,
CHA‑5.C.2 (EK)

## Video transcript

here we've graphed the function y is equal to the square root of x and we're going to create a solid of revolution but we're not going to do it by rotating this around the X or the y axis instead we're going to rotate it around another somewhat arbitrary line and in this case I will rotate it around the line y is equal to one so let's say that this right over here is the line y equals one so this is what we're going to rotate it around y equals one so the first thing we want to do is visualize what we're doing and we actually care about the interval so let's say that the interval is between this point right over here where the two points intersect and let's say between that and X is equal to four so let's say that this right over here so this is x equals four so it's going to be just right over here so this is the interval that we're rotating we're going to rotate it around y equals one not around the x axis so what would our figure look like well we're going to rotate it around we're going to rotate it around like this I'm going to rotate it around something like this and so your figure is going to look I guess you could call it it'll almost look like a conehead if you view it on the side or something like a bullet but not quite looking like a bullet but it would look it would be a shape that looks something something like that so hopefully we can visualize this we've done this several times already trying attempting to visualize these shapes but let's think about how we could actually figure out the volume of this solid of revolution well let's just think about it disk by disk so let's construct a disk right over here let's construct a disk right over here and we've done this many many many times already where we essentially want the volume of each of these disks and then we're going to take this for each of these X's in our interval and then we're going to take the sum of the volumes of all of the disks to find the volume we're gonna stack them all up or I guess put them next to each other in this case and find the volume of our entire figure and so to figure out the volume of each disk we just have to figure out the area of its face the area of its face and I'll do that in purple we just have to figure out the area of the face and multiply it by the depth so what's the area of this face well it's going to be I times the radius of the face squared times the radius of the face squared now what's the radius of the face here well it's not just square root of x square root of x would tell us the different the distance between the x-axis and our function it's now square root of x minus 1 this length this length right over here this length right over your square root of x minus 1 for any given X in our interval so it's going to be equal to square root of let me do that same orange color just to make it clear where I'm getting it from it's going to be equal to square root of x minus 1 minus 1 it's essentially our function minus what we are rotating around that gives us the radius and so this will give us the area of each of our faces and then we just multiply that times the depth the depth of course is DX we've done that multiple times so times DX this is the thing that we want to sum up this is the thing that we want to sum up over our interval and our interval let's see this point right over here where the square root of x is equal to 1 that's just going to be x equals 1 this is just x equals 1 over here and we said that we would do this all the way until x equals 4 this was kind of our end of our interval so this is until X is equal to I'm going to be careful this is the x axis right over here this is the x axis right over here all the way until x equals 4 so you get confused this is the X X so we're going from x equals 1 to x equals 4 and we're essentially taking this area one way to think about it between the our square root of x + 1 and we're rotating it around 1 so we're using these little disks right over here so the intervals between 1 & 4 and so this is going to give us the volume of our solid of revolution and so now we just have to evaluate this definite integral so let's give a shot at it so this is going to be equal to the integral from from 1 to 4 1 to 4 and we can factor out where we could put the pie outside the integral sign and then we can expand this out so square root of x-squared now all we're going to do is expand to this binomial so square root of x minus 1 times square root of x minus 1 square root of x times square root of x is X square root of x times negative 1 is negative square root of x negative 1 times square root of x is another is negative square root of x and the negative 1 times negative 1 is equal to positive 1 so this part right over here will simplify to it'll simplify to X minus 2 square roots of X this is 2 minus 2 square roots of x minus 2 square roots of X and then you have plus 1 and then all of that times DX so this is going to be equal to let's put our PI out there the antiderivative of this business so the antiderivative of X is x squared over 2 the antiderivative so minus 2 times the antiderivative of square root of x well square root of x is just X to the one-half power so we increment the power by 1 so it's going to be X to the 3 halves power times 2/3 so times 2/3 or we could say it's going to be it's going to be times 2/3 X to the 3 halves let me make it clear this negative 2 this negative 2 right here is this negative 2 and this expression right over here is the antiderivative of square root of X and you can verify this if you take the derivative of this 3 halves times 2/3 is 1 decrement the 3 halves you get X to the 1/2 now let's take the antiderivative of 1 well that's just going to be equal to X and we're going to evaluate this from 1 to 4 so we're in the homestretch so this is going to be equal to it's going to be equal to PI times well first let's do it at 4 see if 4 squared I'll write it all out 4 squared over 2 minus what's 4 to the 3 well let me do this part first so minus 4/3 4/3 and that was for to the three-halves so 4 to the 1/2 is 2 and then you raise that to the third power you get 8 so times 8 times 8 plus 4 and then you subtract out all of this stuff evaluated at 1 so 1 squared over 2 so that's just going to be minus 1/2 and then we're subtracting this now actually I want to skip too many steps so let's do let's do it's going to be it's going to be this this is when we evaluated at 4 and then we're going to subtract when we evaluate this whole thing this whole thing at 1 so when we evaluated it at 1 we do this in this green color or that's well that's let me do it that's not the green color I thought I was going to do so when you evaluate it at 1 you get 1 squared over 2 which is 1/2 and then you get 1 to the three-halves which is just 1 so this becomes minus 4/3 minus 4/3 and then you have plus 1 plus 1 and so let's simplify this and I'll do it all in the same color now this is going to be equal to PI times well 4 squared over 2 that is 16 over 2 which is equal to 8 and then you have 4 times 8 which is 32 over 3 so minus 32 over 3 plus 4 and then you have minus 1/2 we're just distributing the negative plus 4/3 plus 4/3 and then you have minus 1 and now we just have to add up a bunch of fractions to simplify this thing so what do we get this is going to be equal to PI times see our least common multiple looks like 6 we're going to put everything over a denominator of 6 everything over a denominator of 6 and so 8 is the same thing as 48 over 6 32 over 3 is the same thing as 64 over 6 so minus 64 over 6 4 is the same thing as 24 over 6 1/2 is the same thing as 3 over 6 so this is my negative 3 over 6 the same thing as negative 1/2 4/3 is the same thing as 8 over 6 8 over 6 and the negative one is the same thing as negative six over six a little bit of arithmetic here and let's see what we get so if we take if we take 48 and we subtract 64 from 48 we get negative 16 is that right yes we get negative 16 and then if we add 24 to negative 16 you get positive 8 positive 8 minus 3 is 5 5 plus 8 is 13 13 minus 6 is 7 so this whole numerator simplifies to 7 so we get 7 PI over 6 as our volume and let me just verify that right so this is going to be negative 16 we get to positive 8 positive 8 plus positive 8 is 16 16 minus 9 is yes as it is indeed 7 so we get 7 PI over 6 and we're done we figured out our volume of this little sideways cone looking thing
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