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# Volume with cross sections perpendicular to y-axis

AP.CALC:
CHA‑5 (EU)
,
CHA‑5.B (LO)
,
CHA‑5.B.1 (EK)

## Video transcript

let R be the region enclosed by y is equal to four times the square root of nine minus x and the axes in the first quadrant and we can see that region R in gray right over here region R is the base of a solid for each Y value the cross-section of the solid taken perpendicular to the y axis is a rectangle whose base lies in R and whose height is y express the volume of the solid with a definite integral so pause this video and see if you can do that all right now let's do this together and first let's just try to visualize the solid and I'll try to do it by drawing drawing this with a little bit of perspective so if that's our y-axis and then this is our x-axis right over here and I can redraw a region R it looks something like this and now let's just imagine a cross-section of our solid so it says the cross-section solid taken perpendicular to the y-axis so let's pick a y value right over here we're gonna go perpendicular to the y-axis it says whose base lies in R so the base would look like that it would actually be the x value that corresponds to that particular Y value so I'll just write X right over here and then the height is y so the height is going to be whatever our Y value is and then if we wanted to calculate the volume of just a little bit a slice that has an infinitesimal depth we could think about that infinitesimal depth in terms of Y so we could say its depth right over here is dy dy and we could draw other cross-sections for example right over here our Y is much lower it might look some sort aiight will be like that but then our base is the corresponding x value that sits on the curve right over that XY pair that would sit on that curve and so this cross-section would look like this and once again if we wanted to put if we wanted to calculate its volume we could say there's an infinitesimal volume and it would have depth d why and so as we've learned many times in integration what we want to do is think about the volume of one of these I guess you could say slices and then integrate across all of them now there's a couple of ways to approach it you could try to integrate with respect to X or you could integrate with respect to Y I'm going to argue it's much easier to integrate with respect to Y here because we already have things in terms of dy the volume of this little slice is going to be Y times x times dy now if we want to integrate with respect to Y we want everything in terms of Y and so what we need to do is express X in terms of Y so here we just have to solve for X so one way to do this is let's see we can square both sides of actually let's divide both sides by 4 so you get Y over 4 is equal to the square root of 9 minus X now we can square both sides Y squared over 16 is equal to 9 minus X and then let's see we could multiply both sides by negative 1 so negative Y squared over 16 is equal to X minus 9 and now we could add 9 to both sides and we get 9 minus y squared over 16 is equal to X and so we could substitute that right over there so another way to express the volume of this little slice right over here of infinitesimal depth dy depth is going to be Y times 9 minus y squared over 16 dy and if we want to find the volume of the whole figure that's going to look something like something like that we're just going to integrate from y equals 0 to Y is equal to 12 so integrate from Y is equal to 0 2y is equal to 12 and that's all they asked us to do to express the volume as a definite integral but this is actually a definite integral that you could solve without a calculator if you if you multiply both of these terms by Y well then you're just going to have a polynomial in terms of Y and we know how to take the antiderivative of that and then evaluate a definite integral
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