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.

Main content
Current time:0:00Total duration:6:39

Video transcript

Part C right but do not evaluate an integral expression for the volume of the solid generated when our so that's this region right over here when R is rotated about the horizontal line y is equal to one so y equals one is right over there so the way I'd like to think about it let's think about what the volume if I were to just take the bottom function if I were to just take f of X if I were to just take f of X and if I were to rotate that function around y equals x if I were to rotate this thing around sorry around Y is equal to one what would the volume of that be and then I'm going to subtract from that the volume if I were to take the top function if I were to take G of X and rotate it around so let's first of all think about what volume I would get and this is really the disk method and then go into it in much more detail earlier in the calculus playlist but let's think about the volume if f of X is rotated around that axis and to do that let's imagine each of each sliver of that volume so this is let me draw a little thing right over here and you could imagine once this little sliver is rotated it forms the bot it forms that or that you should you could imagine you could imagine this length is the radius of a disc and so and just to imagine that let me draw the entire disk so if this is rotated around if this is rotated around it will become a disc it will become a disc it will become a disc that looks something like that and I'll just call the depth of the disc so the disc if you imagine it a coin this is kind of the side of the coin the depth of the coin the depth of the coin right over there and I can draw that better than that let me so the depth of the coin is just like that it's a fixed depth and I'm going to call that DX so it's just this distance right over here it is DX and what is going to be the area of that coin well the area of the surface of this coin so let me let me do it like this in a different color so I want to do it in blue the area of this coin is just pi times the radius of that coin squared and what is the radius of the coin well the radius of the coin is this height is this height right over here and what is that height well it is 1 minus f of X so that is equal to the radius so the area the surface the area of kind of the face of this coin is going to be pi the area of the face of this coin is going to be pi times the radius squared which is equal to pi times 1 minus f of X 1 minus f of x squared that's this blue area right over here and then if I want to find the volume of this coin I would multiply it by the depth of the coin so x so x DX and if I wanted to find the volume of this entire solid this entire rotate' solid I would want to find the sum of all of these volumes so this is just the disc right over here but I could have another disc a similar disc that I do right over here I could have another disc right over here and I want to take the sum of all of those discs so I want to take I want to take so the volume is going to be the sum over all of those discs so X goes from 0 which is this bounding point to X is equal to 1/2 times pi times 1 minus f of X squared this is the area of eat the the area of the face of each of those discs and then I multiply it times the depth of each of those discs now this is the volume of each of those discs and I'm taking the sum of all of them so this is the volume this is the volume if I were to just rotate f of X around y is equal to 1 and actually I should just write D of DX here and so this right over here this expression I just did that - they really are equal this is obviously just the volume of each of those discs so this is a volume if I take the f of X around y equals 1 let's figure out so let me call this volume of f of X and by the same logic the same exact logic we can figure out the volume if we take G of X if we rotate G of X around if we rotate G of X if we construct disks like this and rotate them around Y is equal to 1 and so the volume if I take G of X around y equals 1 would be 0 to 1/2 times pi times 1 minus G of X because 1 minus G of X is each of these radiuses right over here each of these radiuses that squared DX and so the volume of what they're asking us the volume of the solid generated when R is rotated well R is kind of the space in between f of X and G of X so it's going to be it's the volume is going to be the difference between these volumes it's going to be this volume it's this is kind of the outer volume and we're gonna take out its hollow core we're gonna hollow it out by subtracting out this volume so the volume of that region is going to be the integral I'll do this in a new color integral from 0 to 1/2 of pi times 1 minus f of x squared DX minus the integral from 0 to 1/2 of pi times 1 minus G X G of x squared DX and this is a completely valid answer but you might want to simplify it we have the same bounds of integration we have the same variable of integration and actually we have this PI over here so we could factor that out and so this is the same thing as pi times the integral from 0 to 1/2 of 1 minus f of X 1 minus f of x squared minus 1 minus G of X 1 minus G of x squared and then all of that DX and then an actually you probably would want to do this you probably would want to do this if you're taking the AP exam not just leave it in terms of f of X and G of X you would actually want to write the expressions for what f of X f of X and G of X are so really the best answer would probably be pi times the integral from 0 to 1/2 times 1 minus well f of X is 8x to the 3rd power 8x to the 3rd power squared minus 1 minus G of X G of X is sine of PI X that squared that squared times DX and that would be our answer and you could see why they didn't want us to go through the trouble of evaluating it they just wanted us to set up this integral
AP® is a registered trademark of the College Board, which has not reviewed this resource.