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Use the shell mathod find the volume ol tne sclid generated by revo ving Ihe region bounded Dy the given curve and lines about bhe y-axisYe7VK,Y=0,*=1140 B.357...

Question

Use the shell mathod find the volume ol tne sclid generated by revo ving Ihe region bounded Dy the given curve and lines about bhe y-axisYe7VK,Y=0,*=1140 B.357

Use the shell mathod find the volume ol tne sclid generated by revo ving Ihe region bounded Dy the given curve and lines about bhe y-axis Ye7VK,Y=0,*=1 14 0 B. 357



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Find the volume of the solid generated when the region $R$ bounded by the given curves is revolved about the indicated axis. Do this by performing the following steps. (a) Sketch the region $R$. (b) Show a typical rectangular slice properly labeled. (c) Write a formula for the approximate volume of the shell generated by this slice. (d) Set up the corresponding integral. (e) Evaluate this integral. y=9-x^{2}(x \geq 0), x=0, y=0 ; \text { about the line } x=3

Okay. What we want to do is walk through the process of being able to find the volume of the region banded by Weichel the nine minus X squared. Um For X values greater than or equal to zero, X equals zero. Y equals zero about the Y axis. And any time we were talking about finding areas or volumes, it's always a good idea to draw the graph of the region. And so we're going to go ahead and draw that graph, it doesn't have to be perfect. Um We know it's going to be up here at nine. So if this is 369 and then it's going to cost the X axis that positive Three. So here is that region right here um without three and that is nine and we're revolving it about the Y axis. Um And so what we want to do is so it's going to be this region right here. X equals zero. Why called zero? And the graph um then the next thing we want to do is always draw that representative red rectangle. Right? So we have our little representative rectangle right here. Whoops. Uh huh. I can't draw a straight line for the life of me. Um So there we go. Um where this with is delta X. The height is of course nine minus X squared. And the distance from the axis of rotation is X. So um there is are a kind of a good quick sketch. So delta V is equal to two pi two pi X because that's the distance from or X is the distance from the axis of rotation. Um And then we are doing the height which is nine minus X squared. And then of course delta X. The width of that rectangle. So the volume is the integral from 0 to 3 of this. Let's go ahead and let me go ahead and distribute the experts. So this is two pi nine x minus execute. So this is going to be two pi times nine x minus x cubed and of course the delta X goes into dx OK, so now we can go ahead and integrate some a factor out that two pi and then this is going to be nine halfs X squared minus 1/4 X to the fourth and we're going to evaluate both of those at three and zero. Um So this is going to be two pi this will be 81/2 minus 81 over four. I'm in of course minus minus zero. Right, so everything is over zero. Um and so then this will be actually equal to, this is actually gonna be equal to um 81 pi over two and there is our volume.

Okay. What we want to do is to walk through the process of using the shell method to find the volume of a solid. Um That is generated by revolving the region bounded by X equal to the square root of two, I plus one. Why could the two X equals zero, Y equals zero. About the line Why quarter three? Um And the very first thing we need to do is kind of sketch a graph of this region. Um So if um and we're just we're just it doesn't have to be perfect. Um and so when I'm just going to kind of quickly step through a couple of points, if um yeah, why is zero then exits up here at one? So it's right here, if if y is too then excess three. So we know our region is going to be doing something like this possibly. Um And then why people to to is this line right here, X equal to zero and Y equals zero. So it's this little region right in here that we're interested. And then we're gonna revolve it about the line why called a three. So this is to. Okay. Um and we're gonna evolve it like this. So since we're using the shell method and we're evolving it about horizontal line then that means the rectangles have to be horizontal. Everything has been in terms of why? So then we're gonna draw in our representative rectangle where the width of that rectangle is delta Y. Um The heights of that rectangle then is actually the square root of two. Y. Plus one. And then this distance right here from the axis rotation to a rectangle is three minus why. Okay. Um And so now um delta V. Is equal to two pi the distance from the axis of rotation to that representative rectangle. Um The height at the right tangle. And then Delta. Why? Which is the worth of the rectangle. Okay, so let's go ahead and multiply all of those out. Um Not all of those but the two binomial. Um And so this is gonna be two pi this will be um let's see this will be three. Um and then this will be three square to two times wider the one half and then this will be minus Y. And then this will be actually minus the square root of two. Why? To that three house? And then we have this delta Y. Here. Ok so my volume then is the integral and my region is going from 0 to 2 in the Y. Direction. And then of course I have my two pi out here and this will be three plus three route to why do the one half minus Y minus screwed it to? Why did this three halves um and integrating with respect to y. Um And that's why it's always good to get that um sketch developed and put things in their label things because they are setting up your integration becomes no problem whatsoever. So this will be to pie, this will be three Y yeah plus. Um and then this will be to route to why do the three halves minus one half Y squared? And then this will be minus two route to over five wide of the five paths. And we're going to evaluate at two in it zero. And you notice if I evaluate everything at zero it's zero to begin with. So this is going to be two pi this will be six Y plus two. Um And then if I do this this will be four times to this will be eight. Mhm loops, that should be six, I should have a Y there, this would be plus an eight minus a two and then minus. And then if I do here, then a four comes out, so that would be eight, and then I have a square one or two times the square, too, which is another two. So this would be a 16 over five. Um and so common denominators, this is 88 pi over five. So there is my volume right there.

Okay. What we want to do is we want to find the volume of a solid generated um about the Y axis of the following region. And the first thing we need to do is to kind of get that sketch developed. Um and just to kind of help us out. And so the first thing we do is schedule region and so this is one, this is zero um and appear at one. And so we know, oops that went off at smart, didn't it? Um We know it looks something like this, that is why you go to exclude. And so we know it's why equal to zero. So it's going to be the X. Axis and X equal to one. So it's gonna be this region right in here that we're evolving about the Y axis. Okay. And of course we're gonna be doing um the shell method. So let's get our representative rectangle drawn. So let's just put a rectangle right here. Um We know the with that rectangle is delta X. The heights of that rectangle is why equal is X squared. And the height of the rectangle of course is is to this X squared. And then this distance to the center of the right tangle is X. Right? Um and so we know that the changing the volume is actually equal to two pi X. So it's gonna be this distance right here, two pi um X times the height time sir with that rectangle which is about two pi X cubed delta X. Right? So here is our representative volume and so now the volume is equal to the integral and we're going from 0 to 1 and this will be two pi X. Hope that should be a cubed dx and there we have them so now we can kind of integrate and find that volume. Um bring out the two pi and then of course this is going to be 1/4 x to the fourth. And we're gonna evaluate at zero and one. So this is gonna be pi over two times one minus zero. So this is just gonna be pi over two. So there is a volume right there.

So we have a region bounded by Y equals one over X. X equals one. X equals four and Y equals zero. What we're going to do is we're going to rotate this region bounded by this around the y axis. And we're gonna find out what the volume is by using integral. So first let's sketch this. So I'll sketch one. This is one over X. So that's why it was one over X. And then we have X equals one, So x equals one and then we have x equals four. So it's the we also have it pounded by this Y access down here. I won't right over that because it's already there or the X axis. Excuse me. Why? Uncle? Zero? So the X axis. So we're looking at this region right here. And what we're gonna do is we're going to rotate this region around the Y. Access using shells. So we're going to take as for example, this is just one shell and we're going to take an integral to add up all the shells and then we're going to rotate it around. Yeah. That why access to create a shell. So what does it look like is we can actually lay this shell out and it'll look kind of like a rectangle and let's just say that this is at X equals two. So the radius of that shell would be too, and the height of it would be well between this red line and the X axis. So that would be one over two. Following this equation of the red line zero from the X axis. So if we want an equation for that rectangle, the equation would look kind of like this. It would be so the shell is going to equal two pi times the radius. So two pi r. So that's the circumference equation, which is what we're doing by rotating around. So the radius in this case would be to, Because this is that X equals two. So the radius would be too, and the height of this Rectangle would be 1/2. So laying out that shell, we can make a new rectangle and this would be an equation for that show. So what we want to do with the integral, we want to add up all of these rectangles, so what the integral would look like? Yeah, Well we're going to add up all of the shells between four and 1 And we use this equation so two pi or what is the radius going to be? It's going to be X. In this case whatever exes and then the height is going to be one over X. So now adding up all these shells, we have the integral, which is kind of the hard part. Now all we need to do is take the integral so we can pull this to pie out so two pi for one and then these X's cancel so that that will just be DX. or one DX. So this is a pretty easy integral. So what we end up getting is two pi uh times X Between four and 1. All right. Which will equal to pi four minus one. And that will equal six pi. Thank you. So the volume of of this region rotated about the y axis is six pie or about 18 point 85


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