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By cutting: a cone of slant height $s$ and base radius $r$ along a lateral edge and laying the surface flat. we can form a sector of a circle of radius. $x$. (Sec t...

Question

By cutting: a cone of slant height $s$ and base radius $r$ along a lateral edge and laying the surface flat. we can form a sector of a circle of radius. $x$. (Sec the figure.) Use this idea to verify $(10.8 .1)$

By cutting: a cone of slant height $s$ and base radius $r$ along a lateral edge and laying the surface flat. we can form a sector of a circle of radius. $x$. (Sec the figure.) Use this idea to verify $(10.8 .1)$



Answers

Use the Theorem of Pappus to find the volume of the given solid.
A cone with height $ h $ and base radius $ r $.

All right. The way of online y equals age. Um and we have this line down here. Why host H over r times X. So it's been is going to be revolved around. Why access open here and uniform a surface off revolution. And it's a corns, basically. So we want to find the surface area of that cone. So we'll need a few things here. Well, you know, radius from the excess of revolution Morgan years their formula that the surface area is gonna be two pi ready to be are square root of one pause. I'm gonna go with with in grand with respect to axe. Initially, I'll set this up. If I decide it's not going to work for me, then I'll I'll switch to d y. But right now I'm going to try this. So one of the two pi and course are, uh, were this these to intercept over here because that X equals are So this is our h right in here. So are our x will. We will integrate from zero. So are and our is actually acts in this case and we have one pause. Why? Prime? Oh, squared. Sorry. That's gonna be squared up there. If that's why, then why Prime must be h over are so we'll have h over are quietly squared, t x. Okay, so it's what? Let's work on this. This is a constant inside here. This is This is a constant. So much was simplified and a little bit And bring it out to the front. I have a two pi, uh, have squarer off r squared plus h squared over r squared. And then you have zero to our x t x, some liberal, more work on this guy's two pi Well, dio the, um Square Rudd's of the R squared plus h squared and then over our Now let's integrate this thing here. This is X squared or two from zero R. So that becomes to pile or are square run on r squared plus h squared. And then over here we'll just have an r squared over two minus zero so it comes off. And then that's going to be that will be able to, uh, let's say two pi are actually that's Who's that. Who's gone so well to will be Go on so we'll just have the pie are and where the scoreboard of R Squared plus h squared. And, uh, that's it, right. It's her final formula for that surface area. Um, no. Notice that this is, uh, this is in, uh, it's consistent with our formula for for this the surface area of the corn, which is pi r l notice said this to this number that we have in here. This number here, if you'll hit our diagram, it really is just that length in here. In other words, it's l. So it is consistent. Staggering serum will show you that are square puts h squared is gives you that l squared in there. So l is the square root of r squared plus a square. Okay, so it is consistent with our well known geometric form to us.

Okay, so let's consider the following curve. Why is equal to X over two plus one half? Trying to find the surface area of the cone that's generated by revolving This line segment where we have one less than or equal t X and X is less than or equal to three about the X axis. So the area of the surface generated by revolving about the X axis is given by where our surface s is equal to the integral from a to B off to pie. Why times the square root off one plus Ah, the derivative square. So the derivative here is D Y D X square. Okay, so here we have that wise equal to x over two plus one half. Um so therefore are derivative f prime off X is just going to be equal to one half derivative off X over two. That's just one half times x. So the derivative off X over two is just the coefficient one half and the derivative of one half is zero delivered for constant zero. So the derivative here is just equal toe one half okay. And we have that a well is equal to one and big is equal to three going from 1 to 3. So therefore, um, our surface is gonna be equal to the integral, going from 1 to 3 off two pi times X over to plus one half times thes square root of 5/2 D x. Okay, so we're all the pie, The pie squared or five over to that whole thing could come out front of our integral. So we have pi times the square to 5/2. That's just a constant. And then, um, times the integral going from 1 to 3 off just X plus one. Yeah, the X. Okay, that's not too bad. That so, eh? So what we have here? Well, again, we have a constant pi times the square or five over to Okay. The integral here is just gonna be well, um driven integral of access X squared over two. So we get X squared over two, and the derivative of one is just Oh, that. The integral of one is X. So we have our constant here. Outfront has multiplied by this which is our integral evaluated. And then we're evaluating from 1 to 3. So we plug in three. First it's cracked off. What we get, we plug in one. So therefore our surface is gonna be equal to again. We have pie Times Square or 5/2. And then what do we have? Well, plugging in three first, um, we get nine halves plus three and then minus one half minus one started out. If we have nine halves plus three and then we have a minus. Plug in one to get minus one half and then one gives us a minus one. Well, plus one. But then it Zoe, get a minus one here. Okay. Eso therefore, our surface gonna be equal to well, pi times the square to five over to times while times are four plus two or times six So therefore, while six over to that's just three. So therefore our surface is equal to our surface s is equal to three pi times the square root of five. All right, Thank you

We have an equation. Why equals X over two plus 1/2? And it says that this is a, uh has some sort of geometric meaning, but we don't really care so much. All we really need to know is that this equip this curve is defined for one less than are equal to X less, they're able to three and we're revolving around the X axis. This is all the information that we need really to do this problem. So because we're revolving around the X axis here that tells us that are integral is gonna be a d x integral. All right. And then these numbers here tell us the limits of integration so are integral is going to be from 1 to 3. And we're gonna be integrating, um, with respect to X now because it's an area of revolution. Your surface area. Excuse me. It's a surface area for a surface of revolution. There's a formula for that. There's always a two pi involved. And then there's two things that go inside the integral. So for one thing, since we're revolving around the x axis, we would put the why co ordinate, uh, for any value of X. We want the corresponding y coordinate, which is just the y value here. And then we need to multiply by the arc length element. Now, what is that? That's this guy. One plus why? Prime of X squared, We need to include this factor. Well, what is what is this? In our present case, this is the square root of one. Plus why? Prime is just 1/2. So this becomes one plus 1/4. So that's going to be 5/4 square roots. So we get five over too. Okay. Uh, well, this is just a constant. So, actually, we could pull this out. We can pull this out of the integral and have route five times pi integral from 1 to 3. X over two plus one over to D X. And now we're supposed to integrate this. So that's gonna be route five times pi. And then this will be If we integrate this, it'll be X squared over four plus X over too, Uh, from 3 to 1 by the fundamental theorem of calculus. And then that's pretty easy to evaluate. Uh, just my plugging stuff in. So what do we get if we plug everything in we're going to get, um I guess five I ah, route 55 times, Three squares. So 9/4 plus three halves minus what we get when plugging one. So 1/4 plus 1/2 and then this comes out to equal. Ah, Route five pie. So this is too, uh, plus two. Or I guess it's too plus one. So I guess it's three. So I guess this is our final answer are I guess there should be a four, actually, let's be a little more careful here. So, uh, when I subtract the one, I get a two year and then when I subtract No. Okay. Well, I I'm getting three here, but it should be four, according to the answer. So, um, here you get. I'm computing three, but you guys can double check your work and see what you get

So we're told to take the line segment. Why? Equals are over Age X and we're told to rotate that around the X axis. So if we do that, we indeed get a cone. Which, uh, we're supposed to be using this to find the area, Uh, a cone. So we're going to go from zero to H and we're gonna integrate around the X axis. Okay, So despite the fact that they're not giving us numbers, they're just giving us These letters aren't age. Those represent constant, so we can actually go about this in the usual way. So, for example, we note that Y prime of X is equal to our over h. So that's gonna be useful. So when we get the surface area, it's gonna be two pi times again. A girl from zero to h, and then we need the y coordinate, which is are over Age X, and we also need to multiply by the arc length element, which is this Well, that's gonna be square root of one plus r squared over h squared. Okay, well, now that since we're integrating with respect to X, all of this other stuff is just a constant so we could pull that outside of the integral. And then we just integrate from zero to age X d x, which is going to be equal to two h are over age one plus r squared over h squared. And this is gonna be X squared over two evaluated at H. So it's gonna be this now we can simplify this a little bit. Um, so we have. So the two's gonna cancel this two cancels with this, too. And, um, this age cancels with this h here. So we end up with pie. Ah, yeah, times. And then we have an H. But we can also think of this as, um, we can bring the H inside the square root as an h squared. And then we get this, um, as your final answer, which agrees.


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