Finding the solid of revolution (constructed by revolving around the x-axis) using the disc method. Created by Sal Khan.
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- How could he take pi out of the integral?(23 votes)
- Why is the depth denoted by dx? Shouldn't it be delta x, like previous integral problems?(16 votes)
- delta x is a change in x.
dx means derivative x which is basically a super small change in x.(2 votes)
- I do not quite get the intuition of the shape when he rotates it, is there some other visualization ?(4 votes)
- Yes, there are many helpful computer programs by Peter Collingridge, that are quite helpful. Here is one of them:https://www.khanacademy.org/cs/3d-surface-viewer-2/1092842126 ,this one helped me understand. If this is not that good, see his other solids of revolution programs, as there are many. Yet, this is probably one of the most helpful as it was SPECIFICALLY made for this video.(17 votes)
- what are the units? were there any?(4 votes)
- Would y=x^2 the same as y=x^-2? on the interval from x=1 to x=3.(2 votes)
- But isn't the edge of the disk curved?(3 votes)
- Yes, but we are finding the volume of an infinite number of infinitely thin cylinders. So thin that the curvature of the edge is negligible. Just like in Riemann sums where the square edge of the rectangle is so narrow that the error goes away.(9 votes)
- Is there a specific or technical name for these kind of problems? Like Volume Calculation of... I don't even know how to give an example...(0 votes)
- Where's the +C? Is it unnecessary for solids? I would expect it at8:56.(3 votes)
- +c is unnecessary for definite integrals because when you evaluate it at it's limits you will have one +c and one -c which will cancel out.(2 votes)
- I understand this video well. But, what happens if the function that is being rotated about the x-axis has a negative portion? The function x^2 in this video did not go below the x-axis at any point, but a function such as f(x)=-x^4+2x^3-1 has a positive and a negative portion. Why do I have to make the limits of the integral the x-intercepts where f(x) is positive (i.e. x=1 to x=1.84)?(2 votes)
- If I'm understanding your question correctly, it's because we're looking for the solid of revolution formed by the area between the curve f(x) and the x-axis; the other area (say from -infinty to 1) would be infinite.(3 votes)
- What if the limits were not given? How can you solve for a and b in order to solve the definite integral?(2 votes)
- I am assuming you are asking about area and/or volume problems involving graphs of regions. Generally it is a good idea to sketch a graph of the region. Then algebraically determine the x-coordinates of the relevant intersection points of the curves forming the region, to find the integration limits a and b. Remember that the x-coordinates of the intersection points of the curves y=f(x) and y=g(x) are the solutions of the equation f(x)=g(x).
Have a blessed, wonderful day!(1 vote)
Over here I've drawn part of the graph of y is equal to x squared. And what we're going to do is use our powers of definite integrals to find volumes instead of just areas. So let's review what we're doing when we take just a regular definite integral. So if we take the definite integral between, say, 0 and 2 of x squared dx, what does that represent? Well, let's look at our endpoints. So this is x is equal to 0. Let's say that this right over here is x is equal to 2. What we're doing is for each x, we're finding a little dx around it-- so this right over here is a little dx. And we're multiplying that dx times our function, times x squared. So what we're doing is we're multiplying this width times this height right over here. The height right over here is x squared. And we're getting the area of this little rectangle. And the integral sign is literally the sum of all of these rectangles for all of the x's between x is equal to 0 and x is equal to 2. But the limit of that as these dx's get smaller and smaller and smaller, get infinitely small, but not being equal to 0. And we have an infinite number of them. That's the whole power of the definite integral. And so you can imagine, as these dx's get smaller and smaller and smaller, these rectangles get narrower and narrower and narrower, and we have more of them, we are getting a better and better approximation of the area under the curve until, at the limit, we are getting the area under the curve. Now we're going to apply that same idea, not to find the area under this curve, but to find the volume if we were to rotate this curve around the x-axis. So this is going to stretch our powers of visualization here. So let's think about what happens when we rotate this thing around the x-axis. So if we were to rotate it, and I'll look at it and say that we're looking it a little bit from the right. So we get kind of a base that looks something like this. So this is my best attempt to draw it. So you have a base that looks something like that. And then the rest of the function, if we just think about between 0 and 2, it looks like one of those pieces from-- I don't know if you ever played the game Sorry-- or it looks like a little bit of a weird hat. So it looks like this, and let me shade it in a little bit so it looks something like that. And just so that we're making sure we can visualize this thing that's being rotated. We care about the entire volume of the thing. Let me draw it from a few different angles. So if I drew from the top, it would look something like this. It'll become a little more obvious that it looks something like a hat. It would point up like this, and it goes down like that. It would look something like that. So in this angle, we're not seeing the bottom of it. And if you were to just orient yourself, the axes, in this case, look like this. So this is the y-axis. And the x-axis goes right inside of this thing and then pops out the other side. And if this thing was transparent, then you could see the back side. It would look something like that. The x-axis, if you could see through it, would pop the base right over there, would go right through the base right over there. And it'd come out on the other side. So this is one orientation for the same thing. You could visualize it from different angles. So let's think about how we can take the volume of it. Well, instead of thinking about the area of each of these rectangles, what happens if we rotate each of these rectangles around the x-axis? So let's do it. So let's take each of these. Let's say you have this dx right over here, and you rotate it around the x-axis. So if you were rotate this thing around the x-axis-- so I'm trying my best to-- around the x-axis, you rotate it. What do you end up with? Well, you get something that looks kind of like a coin, like a disk, like a quarter of some kind. And let me draw it out here. So that same disk out here would look something like this. And it has a depth of dx. So how can we find the volume of that disk? Let me redraw it out here, too. It's really important to visualize this stuff properly. So this is my x-axis. My disk looks something like this. My best attempt at the x-axis sits it right over there. It comes out of the center. And then this is the surface of my disk. And then this right over here is my depth dx. So that looks pretty good. And then let me just shade it in a little bit to give you a little bit of the depth. So how can we find the volume of this? Well, like any disk or cylinder, you just have to think about what the area of this face is and then multiply it times the depth. So what's the area of this base? Well, we know that the area of a circle is equal to pi r squared. So if we know the radius of this face, we can figure out the area of the face. Well, what's the radius? Well, the radius is just the height of that original rectangle. And for any x, the height over here is going to be equal to f of x. And in this case, f of x is x squared. So over here, our radius is equal to x squared. So the area of the face for a particular x is going to be equal to pi times f of x squared. In this case, f of x is x squared. Now, what's our volume going to be? Well, our volume is going to be our area times the depth here. It's going to be that times the depth, times dx. So the volume of this thing right over here-- so the volume just of this coin, I guess you could call it, is going to be equal to. So my volume is going to be equal to my area times dx, which is equal to pi times x squared squared. So it's equal to pi-- x squared squared is x to the fourth-- pi x to the fourth dx. Now, this expression right over here, this gave us the volume just of one of those disks. But what we want is the volume of this entire hat, or this entire bugle or cone-looking, or I guess you could say the front-of-a-trumpet-looking thing. So how could we do that? Well, the exact same technique. What happens if we were to take the sum of all of these things? So let's do that, take the sum of all of these things. And I'll switch to one color-- pi times x to the fourth dx. We're going take the sum of all of these things from x is equal to 0 to 2. Those are the boundaries that we started off with. I just defined them arbitrarily. We could do this, really, for any two x values-- between x is equal to 0 and x equals 2. And we're going to take the sum of the volumes of all of these coins. But the limit-- as the depths get smaller and smaller and smaller and we have more and more and more coins, at the limit, we're actually going to get the volume of our cone or our bugle or whatever we want to call it. So if we just evaluate this definite integral, we have our volume. So let's see if we can do that. And now this is just taking a standard definite integral. So this is going to be equal to-- and I encourage you to try it out before I do it. So we can take the pi out. So it's going to be equal to pi times the integral from 0 to 2 of x to the fourth dx. I don't like that color. Now, the antiderivative of x to the fourth is x to the fifth over 5. So this is going to be equal to pi times x to the fifth over 5. And we're going to go from 0 to 2. So this is going to be equal to pi times this thing evaluated at 2. Let's see. 2 to the third is 8. 2 to the fourth is 16. 2 to the fifth is-- let me just write it down. 2 to the fifth over 5 minus 0 to the fifth over 5. And this is going to be equal to-- 2 to the fifth is 32, so it's going to be equal to pi times 32/5 minus-- well, this is just 0-- minus 0, which is equal to 32 pi over 5. And we're done. We were able to figure out the volume of this kind of wacky shape.