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Let v=?_ Consider the rcgion endosed by the graph ol this Iunction Jnd the I-axis where 0 < 1 < [let S; be the solid obt Jined by rotating the region about th...

Question

Let v=?_ Consider the rcgion endosed by the graph ol this Iunction Jnd the I-axis where 0 < 1 < [let S; be the solid obt Jined by rotating the region about the line 1 = 1.(a) Use the method of cylndrlcal shells t0 glve an Integral for the volume of the solld S: 2 [ = (2) &(b) Evaluate vour integral from (a) t0 find the volume 0f 51 :# 1 6 { 1Let Sz be the solid obtalned by rotatlng the reglon In Problem about the axis_(a) Use the disk method to give an integral for the volume 0f the so

Let v=?_ Consider the rcgion endosed by the graph ol this Iunction Jnd the I-axis where 0 < 1 < [let S; be the solid obt Jined by rotating the region about the line 1 = 1. (a) Use the method of cylndrlcal shells t0 glve an Integral for the volume of the solld S: 2 [ = (2) & (b) Evaluate vour integral from (a) t0 find the volume 0f 51 : # 1 6 { 1 Let Sz be the solid obtalned by rotatlng the reglon In Problem about the axis_ (a) Use the disk method to give an integral for the volume 0f the solid $ (b) Evaluate rour Integral from (a) t0 find the volume 0f 57: 5 1 { 5



Answers

Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Then use your calculator to evaluate the integral correct to five decimal places.

$ y =x^2 $ , $ x^2 + y^2 = 1 $ , $ y \ge 0 $

(a) About the x-axis
(b) About the y-axis

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.

Okay. What we want to do is walk through the process of being able to determine the volume of the region banded by X. Equal to y squared. Why go to one X. Equals zero about the X. Axis. And we're using the shell method. Um And so the first thing we should always do is to draw a graph of our region and then label things and so forth. So if this is X. And this is why Xing otherwise squared is if why is one X. Is one. So it really actually looks something like uh huh. Something like this. Um And if I if it's easier for you to kind of think of what I call the square root of X. That's what it would look like. Um And we're going up here at one as well, whoops forgot I was in eraser Road. And then this is one, right? So this is a little region that we're interested in and we're evolving it about the ex taxes. Um And so first of all we want to do then draw representative rectangle. And so here would be our representative rectangle where this is delta Y. Um the height or the length of the rectangle of course is why squared and the distance from the axis of rotation is why. Um And so there we there we have him. Um And so now um delta V is equal to two pi the distance from the axis of rotation. Which is why times the heights of the right tangle, times the width of the rectangle. And so this is going to be equal to two pi. Why cubed delta Y. So the or the volume is equal to the integral. And in the y direction we're going from 0 to 1 of two pi. I'm gonna bring that to pie out of Y cubed and delta Y. Becomes a. T. Y. Okay so now we know how to integrate white cubes. So this is going to be actually 1/4. So this would be um pi over two times y to the fourth and we're going to evaluate it at one and at zero. So this is just going to be equal to pi over two is our volume.

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

Okay. What we want to do is we want to go ahead and find the volume. Using the shell method. Uh The solid that is generated by the region bounded by X. Equal to screw to want Y plus one. Why I called the four X equals zero Y. Zero about the X. Axis. Um So the very first thing we should do is is draw that graph, right? And so we're in quickly drawl um the region that we're interested in, so if y is zero, X. Is one, so we're right here. Um If why is one, why is one then X is over here at two. And so it's kind of kind of looks something like of course I miss my mark, something like this. Um So that is what that's gonna look like. I just quickly did a couple points in my head um to 34 This is why I go to four right here and we're doing X equals zero, which is the Y axis and why go zero which is the X. Axis. So it's this region bright in here that we're interested in and we're evolving it about the X. Axis. So don't forget our rectangle for doing the shell method and we're evolving about the horizontal, the rectangles horizontal. So then the second thing we need to do is to draw that representative rectangle. So let's just draw right tangle in here. And so um the width of the rectangle is delta Y. And the distance from the axis rotation is why? And the height of the rectangle or the length of the rectangle is this square root of why? Plus one. Right? Um And so the change in volume is equal to two pi the distance from the axis of rotation times the height that that rectangle times the width of the rectangle. So our volume is the integral and ry is going from 0 to 4. Um And I probably need to put the two pie in here to let's let's change that over. So the volume is equal to two pi where I bring that out 024 And this actually is going to be why to the three halves plus. Why do you why? And there we haven't so now I know how to integrate this. Right? So this is two pi times 2/5. Why did the fifth five halves? Sorry about that? Um Plus one half Y squared. And we're going to evaluate at four and then again at zero. So this will be equal to two pi. This will actually be two to the fifth which is 32 times another two which is 64/5 plus. And then this will be mhm. 16/2 which is eight. Um So if you put it over a common denominator it's going to be two oh eight pi over five and there is our volume of that solid


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