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The base f a solid is the region enclosed by y = Sin x and the x-axis on the interval [0,z]: If cross sections perpendicular to the x-axis are semicircles; write an...

Question

The base f a solid is the region enclosed by y = Sin x and the x-axis on the interval [0,z]: If cross sections perpendicular to the x-axis are semicircles; write an integral that represents the volume of the solid. (A) 8 [" (sinx): dx [S (sin.)dx 8 8 (C) E[, (sinx)' < dx (D) I ['(sin.):. dxIt is known that g(x)dx = 1 and that the four regions shown are all equal in area_ The area of theregion bounded y = f(x).y=g(x), and the x-axis in Quadrant ] is:f(x)(A) ; (C)(B) 3 3

The base f a solid is the region enclosed by y = Sin x and the x-axis on the interval [0,z]: If cross sections perpendicular to the x-axis are semicircles; write an integral that represents the volume of the solid. (A) 8 [" (sinx): dx [S (sin.)dx 8 8 (C) E[, (sinx)' < dx (D) I ['(sin.):. dx It is known that g(x)dx = 1 and that the four regions shown are all equal in area_ The area of the region bounded y = f(x).y=g(x), and the x-axis in Quadrant ] is: f(x) (A) ; (C) (B) 3 3



Answers

Find the volume of the solid generated by revolving the region
in the first quadrant bounded by $y=x^{3}$ and $y=4 x$ about
(a) the $x$ -axis$
(b) the line $y=8 .

They asked us to integrate this function or find the volume of the solid underneath this function over this region. So the function is four minus X minus why? And the region goes from um how did they give us the region? I guess they gave us two Ears and a bunch of lines. So we have uh this line is where we hear hawaii cause four minus X. And this is X equals zero. And this because why it goes zero. That's wait a minute. Uh Oh yeah yeah so that's you know basically this I wrote it in set notation. So why goes from 0 to 4 X. Or form hopes that it will be four minus X. Wild Coast. Um 0 to 4 minus X. And X. Co X is greater than zero. Um We don't really have to we could put X's great between zero and 4 if we wanted to. Um But again that pinches off here so um yeah no let's see here actually I think probably ought to write that explicitly out because theoretically um you can have kind of this region out here, right? Which would kind of be, you know, if X went if you had X between zero and five, I guess you would extend this out and you have this kind of double two triangles anyway. They said they said we don't have to plot with this um volume looks like, but I did that anyway because I don't know, I wanted to see what it looked like. I knew this was a plane and we have a plane and then over find over a triangle. So we have this kind of wedge here. Okay. Doing the integral. Hopefully we can, you know, I'm not going to go through all the details here because this is a pretty simple integral. And hopefully you can do this pretty easily by now and you get 35 or 32/5, which is interesting because if you say, okay, well what is the volume of this cube? And that's four by four by four, which is 64. So if you take 1/10 of that, look at this, which is kind of surprising, I think This is only the 10th the volume of this volumes. People have very poor intuitions about volume. Um You know say if you look at the sphere that has a volume of one cubic meter and then a square that has the same volume, they don't look the same at all. Basically people have very poor instincts and intuition on volume. So the fact that this is 1/10 of the volume of this of the total cube, it seems a little surprising. But again, I can see it, you know, I could see how that could possibly be.

Mhm. We were asked to indicate dysfunction for X. Times Y. Over the region defined bounded by Y equals square root of one minus X squared. So that's the semicircle here. Yeah. And why it was zero and X equals zero. So that's this kind of quarter circle here. Okay. And then if we actually caught the volume again they didn't ask you to do that, but we can see we get something that looks kind of like this, I don't know what you call it anyway. You know this looks like kind of um you don't parabolic kind of you know, kind of a saddle shape. So we did this saddle shape over this search in the region. So the region we need to integrate by sending this making this a set, you go from zero to square root of one minus X squared, the next goes from 0 to 1. And so um again we just set up our integral and this is a fairly simple integral. I'm not gonna uh huh go through all the details of it. I think you can should be able to do that by this time and what we get is one half. So that's uh again if you took the area of this um semicircle Chance a radius of one so pi r squared over four. So that's pi over four is the area of that in times a c by two. So pi over two would be the area of this solid cylinder. And that is definitely bigger than 1/2. So again that gives us some some confidence that we have um at least right order magnitude here cube here is just a unit cube. So we have this volume is one. And so this thing is taking up um the second is that no no its volume is um this is too so its volume is too so it's this is taking up a quarter of the volume of the the cube here

Mhm. We were asked to indicate dysfunction for X. Times Y. Over the region defined bounded by Y equals square root of one minus X squared. So that's the semicircle here. Yeah. And why it was zero and X equals zero. So that's this kind of quarter circle here. Okay. And then if we actually caught the volume again they didn't ask to do that, but we can see we get something that looks kind of like this, I don't know what you call it anyway. You know this looks like kind of um you don't parabolic kind of you know, kind of a saddle shape. So we did this saddle shape over this search in the region. So the region we need to integrate by sending this making this a set you go from zero to square root of one minus X squared. The next goes from 0 to 1. And so um again we just set up our integral and this is a fairly simple integral. I'm not gonna uh huh go through all the details of it. I think you can should be able to do that by this time and what we get is one half. So that's uh again if you took the area of this um semicircle Chance a radius of one so pi r squared over four. So that's pi over four is the area of that in times a c by two. So pi over two would be the area of this solid cylinder. And that is definitely bigger than 1/2. So again that gives us some some confidence that we have um at least right order magnitude here cube here is just a unit cube. So we have this volume is one. And so this thing is taking up um the second is that no no its volume is um this is too so its volume is too so it's this is taking up a quarter of the volume of the the cube here.

Okay, So for this particular problem were bounded by white equal. Seek it X, and it's bounded between Pi over four and pi over three. So what we're going to do is I'm a set up. The volume is being the integral from pi over. Ah, 42 pi over three of seeking X squared dx. And this is going to be, um, Tangent X from pi over 42 pi over three. And so, of course, Tangent of Pi over three is going to be the square root of three and then, of course, minus one.


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