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# 2011 Calculus AB free response #3 (c)

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Part C, write but do not evaluate an integral expression for the volume of the solid generated when R - - so that's this region right over here - when R is rotated about the horizontal line y=1. So y=1 is right over there. So, the way I like to think about it - let's think about the volume if I were to just take the bottom function - if I were to just take f(x) - if I were to just take f(x) - and if I were to rotate that function around y=x - if I were to rotate this thing around - sorry, around y=1 - 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(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 I go into it in much more detail earlier in the calculus playlist, but let's think about the volume if f(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 silver is rotated it forms the bo- it forms the or you should - you could imagine - you could imagine - this length is the radius of a disk. 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 disk - it will become a disk - it will become a disk that looks something like that, and I'll just call the depth of the disk - so the disk 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. I know I can draw that better than that. 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 the dx - and what is going to be the area of that coin? Well the area of the surface of this is 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-f(x). So that is equal to the radius. So the area - the surf- the area - the 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 one minus f of x - one minus f of x squared - that's this blue area right over here, and if I want to find the volume of this coin I would multiply it by - by the depth of the coin. So times - so times dx. And if I wanted to find the volume of this entire solid - this entire rotated, i would want to find the sum of all of these volumes. So this is just the disk right over here but I can have another disk, a similar disk that I do right over here, I could have another disk right over here, and I want to take the sum of ALL of those disks. So I want to take - I want to take - so the volume is going to be the sum over all of those disks - so x goes from 0 - which is this bounding point - to x is equal to 1/2, times pi(1-f(x))^2 - this is the area of e- the fa- the area of the face of each of those disks and then I multiply time the depth of each of those disks - no this gives the volume of each of those disks, 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(x) around y=1 - actually I should just write dx here - and so this right over here - this expression - I just did that so they really are equal - this is obviously just the volume of each of those disks. So this is the volume if I were to take f(x) around y=1. Let's figure out - so let me call this volume of f(x) - and by the same logic - the same exact logic - we can figure out the volume if we take g(x) - if we rotate g(x) around - if we rotate g(x) - if we construct disks like this and rotate them around y=1. And so the volume - if I take g(x) around y=1 would be 0 to 1/2 times pi(1-g(x) - cause 1-g(x) is each of these radius' right over here - each of these radius' - 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(x) and g(x) - so so what's going to be - is g- the volume is going to be the difference between these volumes - it's going to be this volume - this is kind of the outer volume, and we're going to take out its hollow core - we're going to hollow it out be subtracting 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(1-f(x))^2 dx MINUS the integral from 0 to 1/2 of pi(1-g(x))^2 dx and this is 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 just factor that out - and so this is the same thing as pi times the integral from 0 to 1/2 of (1-f(x))^2 MINUS (1-g(x))^2 - and then all of that dx. And then - and actually you probably would want to do the - you probably would want to do this while taking the AP Exam, not just leaving it in terms of f(x) and g(x) - you would actually want to write the expression for what f(x)... f(x) and g(x) are. So really the best answer would probably be: pi times the integral from 0 to 1/2, times [1(1-... well f(x) is 8x to the third power - 8x to the third power, squared. Minus, (1-g(x)... and g(x) is sin 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.