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Main content
Current time:0:00Total duration:8:34
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Video transcript

let's see if we can imagine a three-dimensional shape whose base could be viewed as this shaded in region between the graphs of y is equal to f of X and Y is equal to G of X so that's the base of our this purple this I guess move or purple color is the base of it but then it's kind of popping out of our screen and what I've drawn here in blue you could view this kind of the top Ridge of the figure and if you were to take cross sections of the figure and that's what this yellow line is if you were to take cross sections of this figure that are vertical that are that are I should say perpendicular to the x axis those cross sections are going to be isosceles right triangles so this cross section is going to look like this if you were to look at if you were to flatten it out so over here it's sitting it's popping out of your page or out of your screen but if you were to actually flatten it out the cross section would look like this it's going to be an isosceles right triangle where the hypotenuse of the right I saw sleaze right triangle sits along the base sits along the base so it's isosceles so that's equal to that it's a right triangle and then this distance this distance between that point and this point is the same as the distance between f of X and G of X f of X and G of X for this for this x value right over there and obviously that changes as we as we change our x value and to help us visualize the shape here I've kind of drawn a picture of our coordinate plane if we're kind of if we view it as an angle if we're kind of above it and you can kind of start to see how this figure would look I've once again I've drawn the base I've drawn the base of it I've drawn the base of it right over there and maybe I should to make it clear and we can like this let me shade it in kind of parallel to these cross sections so I've drawn the base right over there there's two other sides there's the side that's on at least the way I've drawn it here I guess you could view it as its top side or the left side right over there and over on this picture that would be this when we're looking at it from above and then you have this other side I guess on this view with this would you could call the kind of the right side and over here this is kind of the when you view it over here this is the bottom side so the whole reason why I set this up and we're attempting to visualize this figure is I want to see if you can come up with a definite integral that describes the volume of this figure that kind of kind of almost looks like a football if you cut it in half or a rugby ball but it's cued a little bit as well but what's a what's an expression a definite integral that expresses the volume of this and you I encourage you to use the fact that it intersects at the point these functions intersect at the point 0 0 and C comma D so can you come up with some expression a definite integral in terms of zeros and c's and d's and f's and G's that describe the volume of this figure so assuming you've paused the video and have had a go at it so let's let's think about it so if we want to find the volume one way to think about it is we could take we could take the volume of we could approximate the volume as the volume of these individual triangles so that would be the area of each of these triangles times some very small depth some very small depth so let me just shade it in to show the depth some very small depth which we could call we could call DX so once again we could find the volume of each of these by finding the area the cross-sectional area there then multiplying that times a little bit a little DX a little DX which would give us 3 so this is a little DX which would give us three dimensional that's our DX I could write that a little bit neater DX to give us a little bit of three-dimensional depth so how could we what is good value of one of these figures is going to look like well if we say let me just call this height H and we know that H is going to be F of X minus G of X that's this distance right over here so let's call that H we know that H is going to be and maybe I should say H of X because it is going to be a function of X H of X is going to be f of X f of X minus G of X minus G of X but given an H what is going to be the area of this triangle well this is a 45-45-90 triangle if this is 90 then this is going to have to be 45 degrees that's going to have to be 45 degrees and we know that the sides of a 45-45-90 triangle or square root of 2 times the hypotenuse so this is going to be square root of 2 I'm sorry square root of 2 over 2 times the hypotenuse square root of 2 over 2 times the hypotenuse and you can get that straight from the Pythagorean theorem if the side let's say it's length a then this side has length a so you're going to have a squared plus a squared is equal to the hypotenuse squared or 2a squared is equal to the hypotenuse squared or that a squared is equal to the hypotenuse squared over 2 or that a is equal to H over the square root of 2 which is the same thing as the square root of 2 H over 2 I just rationalize the denominator multiplied both the numerator and the denominator by square root of 2 so that's where I got this from and so what is the area going to be well the area is just going to be your base times your height times 1/2 so let me write that down so the area the area there the area is just going to be the base which is square root of 2 over 2 times our hypotenuse times the height which is square root of 2 over 2 times our hypotenuse times 1/2 times 1/2 if we didn't do that what half we'd be figuring out the area of this entire this entire square but this is obviously we're concerned with with the triangle so what's this going to be this is going to be square root of 2 over 2 times square root of 2 over 2 is going to be 1/2 and then you're going to have another 1/2 so it's 1/4 1/4 H weird did I do that right this is yes it's going to be 2 over 4 which is 1/2 and then times another 1/2 is 1/4 H squared is the area now what's going to be the volume of each of these triangles well the volume of each of these triangles right over here the volume is just going to be our area times DX or it's just going to be 1/4 h squared times our depth times our depth DX and so if we just integrated a bunch of these from our x equals 0 all the way to x equals C we essentially have our volume of the entire figure so how could we write that so we want to write we want to essentially define the volume of the figure and it was kind of in the homestretch homestretch here actually let me write it right value this is volume of a section volume of a cross section so what's the volume of the entire thing going to be well the volume volume of the figure of the figure is going to be the definite integral the definite integral from x equals 0 to x equals c x equals 0 to C of 1/4 H squared 1/4 but we know that H is equal to f of X minus G of X so instead of H I'm going to write f of X minus G of x squared squared DX D X and we're done we just found an expression for a definite integral expression I guess you could say for the volume of this strange figure that we have defined