Log in Waleed Junaidy 11 years agoPosted 11 years ago. Direct link to Waleed Junaidy's post “why is the radius, or x, ...” why is the radius, or x, 2 - x? • (38 votes) Ethan Dlugie 11 years agoPosted 11 years ago. Direct link to Ethan Dlugie's post “The radius of each cylind...” The radius of each cylindrical shell is the horizontal distance from the current x value to the axis of rotation. So if we rotate about the line x=2, the distance between our current x position and the axis of rotation is 2-x. (46 votes) Minh Anh Nguyen 11 years agoPosted 11 years ago. Direct link to Minh Anh Nguyen's post “At 2:22 why can't the rad...” At • (8 votes) Mack Andrews 11 years agoPosted 11 years ago. Direct link to Mack Andrews's post “For any given x-value, th...” For any given x-value, the radius of the shell will be the space between the x value and the axis of rotation, which is at x=2. If x=1, the radius is 1, if x=.1, the radius is 1.9. Therefore, the radius is always 2-x. The x^(1/2) and x^2 only come into play when determining the height of the cylinder. mad4soccer1 8 years agoPosted 8 years ago. Direct link to mad4soccer1's post “How do you decide whether...” How do you decide whether to use disks or whether to use shells? Is there something to look for that will indicate what problems should be solved using one vs the other? • (8 votes) keegan kimbrough 11 years agoPosted 11 years ago. Direct link to keegan kimbrough's post “What it we rotated about ...” What it we rotated about -2 instead of 2. Would it then be x+2? TY!! • (7 votes) Ethan Dlugie 11 years agoPosted 11 years ago. Direct link to Ethan Dlugie's post “Indeed it would be x+2 wh...” Indeed it would be x+2 which is really a simpler way of writing x–(-2), the distance from the current x coordinate to the line x=-2. (3 votes) Matthew Daly 11 years agoPosted 11 years ago. Direct link to Matthew Daly's post “How did Sal just throw in...” How did Sal just throw in the factor of dx at • (4 votes) Matthew Dai 11 years agoPosted 11 years ago. Direct link to Matthew Dai's post “The volume of the shell, ...” The volume of the shell, as stated in previous videos, is the circumference times the height of the shell times the width. Here, the width is dx. However, it does not really matter in the end, because you are just taking the definite integral to find the area. Hope that helps! (2 votes) Eddy Zavala 7 years agoPosted 7 years ago. Direct link to Eddy Zavala's post “If rotation occurs about ...” If rotation occurs about the Y-axis or X=0, the radius would simply be X correct? • (2 votes) POWA 7 years agoPosted 7 years ago. Direct link to POWA's post “correct!keep in mind th...” correct! keep in mind that if we rotate around y=c in D we change the integral: v(K)=π a(integral)b |(g(x)-c)^2-(f(x)-c)^2| dx, for the volume of body K where D={(x,y):a<=x<=b,f(x)<=y<=g(x)} this means that we subtract the value of c if we move the axis of rotation up and that we add the value of c if we move the axis of rotation down i hope this was a little helpful! (2 votes) Lezzlly Real 7 years agoPosted 7 years ago. Direct link to Lezzlly Real's post “At 4:04 how did you find ...” At • (2 votes) Alan Lam 7 years agoPosted 7 years ago. Direct link to Alan Lam's post “Using the intersection po...” Using the intersection points of two functions. (1 vote) matt.claude.lafrance 10 years agoPosted 10 years ago. Direct link to matt.claude.lafrance's post “How is the radius 2-x. He...” How is the radius 2-x. He always says that the radius is the function minus what it's being rotated around so wouldn't it be x-2? Because in every other video that is the case but in this one it's the value 2- x, why is that? • (1 vote) Creeksider 10 years agoPosted 10 years ago. Direct link to Creeksider's post “Many of these problems ca...” Many of these problems can be solved in multiple ways. For this one, you're rotating around the line x = 2, so you can find the radius by going from that point to the left, as Sal did, getting 2-x, or from that point to the right, getting x - 2, the version that seems more natural to you. Either one produces the same result, but Sal has chosen the easier way because when you measure on the left your integral is from 0 to 1 (the x-values on the left that are within range of the figure we're analyzing), whereas if you measure on the right your integral is from 3 to 4. It's almost always easier to evaluate expressions at 0 and 1 than at 3 and 4. It would be good practice for you to calculate the integral the way you suggest, using x - 2 as the radius and evaluating the integral from 3 to 4, and confirming that the result is the same. (3 votes) kma1660 8 years agoPosted 8 years ago. Direct link to kma1660's post “at 4:04, why are we integ...” at • (1 vote) wouter-muyres 8 years agoPosted 8 years ago. Direct link to wouter-muyres's post “In this example the x va...” In this example the x values only resemble the distance from the origin of the axis, you will have to synthesize the axis of revolution(x=2) and the x values to make the radius of the cilinder. (1 vote) nematwaseem 6 years agoPosted 6 years ago. Direct link to nematwaseem's post “These videos are inaccess...” These videos are inaccessible for me for some reason. I was only able to watch the videos previous to this one. What do I need to do to access it? • (1 vote) gundericusthemighty 6 years agoPosted 6 years ago. Direct link to gundericusthemighty's post “That sounds like a techni...” That sounds like a technical problem; I'd check the help center or contact the site rather than commenting. (1 vote)Want to join the conversation?
Likewise, if we rotate about the y axis (aka x=0) the radius is x-0=x.
2:22
Just make sure that your distance/radius factor is always positive when you take the shell method. By that I mean not to use -2–x. You will end up with a negative volume if you perform that integration. 3:45 4:04 4:04
Because the graph is twisted around the x=2 axis, the radius = the difference between the axis we twist the graphs around(x=2) and the x values between x=0 and x=1(to include all the possible x values we integrade from x=0 to x=1).
We are only interested in the solid of revolution of the shape on the left side. Graph the formulas between x=1 and x=2 for your visualization, the shape looks different.
Video transcript
This right here is asolid of revolution whose volume we were able tofigure out in previous videos, actually in a differenttutorial, using the disk method and integrating in terms of y. What we're goingto do right now is we're going to find the samevolume for the same solid of revolution, but we're goingto do it using the shell method and integratingwith respect to x. So what we do is we havethe region between these two curves, y is equalto square root of x and y is equal to x squared. And we're going to rotate itaround the vertical line x equals 2. And we're going to dothat at the interval that we're going to rotate thisspace between these two curves is the interval when square rootof x is greater than x squared. And so it's between 0 and 1. And so let's try to doit with the shell method. And so to do that, what we dois we want to construct a shell. Let me do this ina different color. Let's imagine a rectangleright over here. It has width dx. and its height is the differenceof these two functions. And so if I were todraw it right over here, it would looksomething like this. It'd be there, andthen it is a shell, it's kind of ahollowed-out cylinder. So it would looksomething like this. Just like that. And it has some depth,that's what the dx gives us. So we have the depth thatlooks something like that. And then let me shadeit in a little bit, just so we can see alittle bit of its depth. So when you rotatethis rectangle around the line x equals 2,you get a shell like this. So let's think abouthow we can figure out the volume of this shell. Well, we've alreadydone this several times. The first thing we mightwant to think about is the circumference ofthe top of the shell. We know circumferenceis 2 pi times radius. We just need to know whatthe radius of the shell is. What is thatdistance going to be? Well, it's thehorizontal distance between x equals 2 and whateverthe x value is right over here. So it's going to be2 minus our x value. So this radius, thisdistance right over here, is going to be 2 minus x. And so the circumference isgoing to be that times 2 pi. 2 pi r gives us thecircumference of that circle. So 2 pi times 2 minus x. And then if we wantthe surface area of the outside of ourshell, so the area is going to be the circumference2 pi times 2 minus x times the height of each shell. Now, what is theheight of each shell? It's going to be thevertical distance expressed as functions of y. So it's going tobe the top boundary is y is equal to square rootof x, the bottom boundary is y is equal to x squared. So it's going to be squareroot of x minus x squared. Let me do this in the yellow. So it's going to be squareroot of x minus x squared. And so if you want thevolume of a given shell-- I'll write all thisin white-- it's going to be 2 pitimes 2 minus x times square root of xminus x squared. So this whole expression,I just rewrote it, is the area, the outside surfacearea, of one of these shells. If we want the volume, we haveto get a little bit of depth, multiply by how deep theshell is, so times dx. And if we want the volumeof this whole thing, we just have tosolve all the shells for all of the x'sin this interval and take the limit as thedx's get smaller and smaller and we have moreand more shells. And so, what's our interval? Well our x's are goingto go between 0 and 1. So that right over there isthe volume of this figure.
