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Main content
Current time:0:00Total duration:8:17
AP.CALC:
CHA‑5 (EU)
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CHA‑5.B (LO)
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CHA‑5.B.3 (EK)

Video transcript

this right over here is the graph of X plus y is equal to one and let's say that the region that's below that that's below this graph but still in the first quadrant that this is the base of a three dimensional figure so this region right over here is the base of the of a three dimensional figure and what we know about this three dimensional figure is if we were to take cross sections that are perpendicular to the x axis so cross-sections I guess we could say that are parallel to the y-axis so let's say we were take a cross section like that we know that this will be a semicircle so if we take a slightly different view of the same three-dimensional figure we would see something like this I've kind of laid the coordinate plane down flat and I'm looking at it from above so this cross section if we're looking at it at an angle and if this if the figure were transparent it would be this cross-section right over here it would be that cross-section right over here which is which is a which is a semicircle if we were to take this cross-section right over here along the y-axis that would be this cross-section this larger this larger semicircle so given that given what I've just told you I encourage you to pause the video and see if you can figure out the volume of this of this thing that I've I could shade it in a little bit the volume of this thing that I've just started this three-dimensional figure that we are that we are attempting to visualize so I'm assuming you've had a go at it and one way to think about it is well let's just think about each of these let's let's split up the figure into a bunch of disks into a bunch of disks if you figure out the volume of each of those disks if you figure out the volume of each of those disks and sum them up then that would be a pretty good approximation for the volume of the whole thing and then if you took the limit as you get an infinite number of disks that are infinitely thin then you're going to get the exact volume so let's just say let's just take the approximating case first so let's say that right over here at X what is going to be what is going to be the diameter what is going to be the diameter of the disk at X well to think about that we just have to re Express X plus y is equal to 1 that's the same thing as saying that the function that f of X or Y is equal to f of X is equal to 1 minus X and so the diameter of this circle right over here let me make it clear the diameter of this circle is going to be this height it's going to be the difference between 1 minus X and the x axis or between 1 minus X and x equals 0 and so this is just going to be as a function of X the diameter is going to be 1 minus X now if you want to find the surface area for a circle or if we want to find the area I guess you could say of a circle we know that area is PI R squared now for our semicircle you're going to divide by 2 so what's the radius going to be so let me zoom in a little bit so what is the radius of one of these semicircles going to be so the radius I could draw it like this my best attempt to draw one of these things so it's going to look something like this I'm trying to draw it at an angle so the radius what's going to be half the diameter the diameter is 1 minus X the radius the radius is going to be 1 minus x over 2 right that distance is 1 minus x over 2 that distance is 1 minus x over 2 that distance is 1 minus x over 2 so that is 1 minus x over 2 is equal to the radius and so what would be the what would be the area of what would be the area of this side right over here well if it was a full circle would be PI R squared but it's a it's a semicircle so it's PI R squared over 2 so let me write this so the area is equal to PI R squared R squared over 2 because it's a semicircle so in terms of X I guess is it going to say our area as a function of X right over there that way our areas of function of X is going to be PI over 2 let me just write that it's going to be PI over 2 times 1 minus x over 2 squared so it's 1 minus x over 2 over 2 squared and so if I wanted a volume of just this just this this disk right over here I then multiply that area times the depth I just multiplied the area times this depth here and we could call that DX or we could call that Delta X so we could call the depth there Delta X Delta X now that's that's the actually me be clear that's the area the volume the volume of one of those shells is going to be equal to and I'll just write this in one color it's going to be PI over 2 times 1 minus x over 2 squared times the depth the area times the depth so this is volume of one of these half disks half disks so the volume of the whole thing we can approximate is the sum of these or we could take the limit is our Delta X's get much much much much smaller and we have an infinite number of these things which is essentially if we're taking that limit we're going to take the definite integral so we could take the definite integral so the volume that we care about the volume of this figure is going to be equal to the definite integral from x equals 0 to x equals 1 that's where we intersect the x-axis its equals 0 to x equals 1 of we're integrating an infinite number of these things that are that are infinitely I guess thin so it's going to be let me write it it's going to be PI over 2 times I'll just write it what's 1 minus x squared let me just expand it out for fun right over here so that's going to be that's the same thing as X minus 1 squared so it's going to be x squared minus 2x plus 1 and then 2 squared is 4 / / 4 and instead of Delta X I'm going to write DX now I'm going to write DX because I'm taking the limit as these become infinitely small and I have an infinite none I'm summing an infinite number of them so the volume is just going to be we just have to evaluate this definite integral and if you feel so inspired but if you feel inspired at the moment feel free to pause and try to solve and try to evaluate this definite integral so let's just take some of these constants out so let's take PI over 8 out so our volume our volume is going to be equal to PI over 8 times the definite integral from 0 to 1 of x squared minus 2x plus 1 DX DX which is equal to PI over 8 PI over 8 and so the antiderivative of this is X to the third over 3 - x squared plus X and we're going to evaluate that at 0 and 1 0 and 1 and so this is going to be equal to PI over 8 when you evaluate it at 1 you get 1/3 minus 1 plus 1 so it's going to be one-third and when you evaluate at 0 you get at 0 minus 0 plus 0 so it's going to be minus 0 so it's just PI over 8 times 1/3 which is equal to PI over 24 and we are done