Looking at the example from the last video in a more generalized way. Created by Sal Khan.
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- How would the volume be calculated if the shape was only partly rotated?(49 votes)
- This is a cool question!
If you know how far you want to rotate the shape (in radians) , you're area would be
A = ([angle of rotation]/2pi) * pi * ((f(x))^2-(g(x))^2)
You are essentially finding the area of a sector of a washer this way. Then you can proceed with your integral as usual.(75 votes)
- what happens if we evaluate the integral over interval [a,b] where on part of the interval f(x)>g(x) and on the other part g(x)>f(x)? i can visualize what the solid would look like but am not sure how the math would work out(23 votes)
- How would I find the volume of a shape like a bundt cake formed by a parabola with zeros at 1 and 5 being rotated around the y-axis(3 votes)
- When would you use the washer method as opposed to the disc method? Why is it hollow in the middle?(6 votes)
- The washer method should be used if there is "air" between the shaded region and the axis of rotation. When you rotate the shaded region, this air becomes a void in the shape.(10 votes)
- How do we know when to use the washer method or the disc method?(4 votes)
- You can always use either, the difference is that the washer method takes the cross-section of your final shape, then rotates it, while the disk method subtracts the entire volume of the shape enclosed by g(x) from the shape enclosed by f(x). If you think about it, both are the same thing, except in a slightly different order (using f(x)-g(x) at the end or the beginning).(1 vote)
- So how exactly would you find the interval? Is it given to you or is it where the two functions intersect or something completely different? I wasn't too sure what he meant when discussing that portion.(2 votes)
- If two functions bound a region, then their intersection are usually the bounds. Finding the bounds can be tricky sometimes, especially in multivariable calculus - often my students have more difficulty with finding the bounds than doing the integration once the bounds have been identified. Why is that? It is usually due to being weak in algebra and not putting enough importance on graphing when it was being covered.(4 votes)
- how do you know when it's dy or dx and in which direction the washers are cut?(2 votes)
- When it's a dy integral, it's when you are rotating about a line that is vertical - for example, about x=2. (Sounds counterintuitive, dy for x=something, XD)
When it's a dx integral, it's rotated about a horizontal line, such as y = 5.(3 votes)
- I got to wondering why we are using for the squared radius f(x)^2 - g(x)^2 instead of [f(x) - g(x)]^2, and I played with it a little. Does using [f(x) - g(x)]^2 for the squared radius give you the volume if the axis of rotation is y=g(x)?(3 votes)
- The expression for the volume of a solid of revolution of the area trapped between f and g about the x-axis is ∫ π f(x)^2 dx - ∫ π g(x)^2 dx.
∫ π f(x)^2 dx calculates the volume of the convex hull of the solid, and ∫ π g(x)^2 dx calculates the negative space inside to be removed. Because the integral is linear, the expression can be simplified to π ∫ (f(x)^2 - g(x)^2) dx.(1 vote)
- What would happen if we switch the functions in the integral? We know that sqrt(x) is above x in the coordinate plane but what if, instead of integral of sqrt(x) - x, we integrate x - sqrt(x) and revolve that around the x-axis?(2 votes)
Let's now generalize what we did in the last video. So if this is my y-axis and this is-- that's not that straight. This right over here is my x-axis. And let's say I have two functions. So I'm just going to say it in general terms. So let's say I have a function right over here. So let's say it looks something like this. So that's one function. So this is y is equal to f of x. And then I have another function that y is equal to g of x. So let's say it looks something like this. So the blue one right over here is y is equal to g of x. And like we did in the last video, I want to think about the volume of the solid of revolution we will get if we essentially rotate the area between these two, and if we were to rotate it around the x-axis. So we're saying in very general terms. This could be anything, But. In the way we've draw it now, it would literally be that same truffle shape. It would be a very similar truffle shape, where on the outside it looks like a truffle. On the outside, it looks something like a truffle, and on the inside, we have carved out a cone. Obviously, this visualization is very specific to the way I've drawn these functions, but what we want to do is generalize at least the mathematics of it. So how do we find a volume? Well, we could think of disks. But instead of thinking of disks, we're going to think about washers now, which is essentially the exact same thing we did in the last video mathematically, but it's a slightly different way of conceptualizing it. So imagine taking a little chunk between these two functions, just like that. What is going to be the width of this chunk? Well, it's going to be equal to dx. And let's rotate that whole thing around the x-axis. So if we rotate this thing around the x-axis, we end up with a washer. That's why we're going to call this the washer method. And it's really just kind of the disk method, where you're gutting out the inside of a disk. So that's the inside of our washer. And then this is the outside of our washer. Outside of our washer looks something like that. Hopefully, that makes sense. And so the surface of our washer looks something like that. I know I could have drawn this a little bit better, but hopefully it serves the purpose so that we understand it. So the surface of our washer looks something like that. And it has depth of dx. So let me see how well I can draw this. So depth dx. That's the side of this washer. So a washer, you can imagine, is kind of a gutted out coin. So how do we find the volume? Well once again, if we know the surface, if we know the area of the face of this washer, we can just multiply that times the depth. So it's going to be the area of the face of the washer. So the area of the face of the washer-- well, it could be the area if it wasn't gutted out. And what would that area be? Well, it would be pi times the overall, the outside radius squared. It would be pi times the outside radius squared. Well, what's the outside radius? The radius that goes to the outside of the washer? Well, that's f of x. So it's going to be f of x is the radius. And we're just going to square that. So this expression right over here would give us the area of the entire face if it wasn't a washer, if it was a coin. But now we have to subtract out the inside. So what's the area of the inside? This part right over here? Well, we're going to subtract it out. It's going to be pi times the radius of the inside squared. Well, what's the radius of the inside squared? Well, the inside, in this case, is g of x. It's going to be pi times g of x squared. That's the inner function, at least over the interval that we care about. So the area of this washer, we could just leave it like this, or we could factor out a pi. We could say the area is equal to-- if we factor out a pi-- pi times f of x squared minus g of x squared. I don't really have to write a parentheses there. So we could write f of x squared minus g of x squared. And then if we want the volume-- put that in the same yellow. If we want the volume of this thing, we just multiply it times the depth of each of those washers. So the volume of each of these washers are going to be pi times f of x squared minus g of x squared. The outer function squared over our interval, minus our inner function squared over the interval, and then times our depth. That'd be the volume of each of these washers. And that's going to be defined at a given x in our interval, but for each x at these interval, we can define a new washer. So there could be a washer out here, and a washer out here. And so we're going to take the sum of all those washers, and take the limit as we have smaller and smaller depths. And we have an infinite number of infinitely thin washers. So we're going take the integral over our interval from where these two things intersect, the interval that we care about. It doesn't have to be where they intersect, but in this case, that's what we'll do. So let's say x equals a to x equals b. Although this could have been a, that could have been b. But this is our interval we're saying in general terms, from a to b. And this will give our volume. This right over here is the volume of each washer. And then we're summing up all of the washers and taking the limit, as we have an infinite number of them. So let's see if we apply this to the example in the last video whether we get the exact same answer. Well, in the last video, y equals g of x was equal to x, and y is equal to f of x was equal to the square root of x. So let's evaluate that given what we just were able to derive. So our volume-- do it up here. The volume is going to be the integral-- what are the two intersection points? Well, over here, once again, we could have defined the interval someplace else, like between there and there. And we would have gotten a different shape. But the points that we care about the way we visualize it is between x is equal to 0 and x is equal to 1. That's where these two things intersect. We saw that in the last video. Of pi times-- what's f of x squared? f of x squared, square root of x squared is just x, minus g of x squared. g of x is x, that squared is x squared. And then we multiply times dx. So this is going to be equal to-- we can factor out the pi. 0 to 1, x minus x squared dx, which is equal to pi times-- let's see, the antiderivative of x is x squared over 2. The antiderivative of x squared is x squared over 3-- sorry, x to the third over 3. And we're going to evaluate this from 0 to 1. So this is going to be equal to-- I'm running out of my real estate a little bit. Let me scroll over to the right a little bit. So this is going to be equal to pi times-- well, when you evaluate this whole thing at 1, you get-- let's see, you get 1/2 minus 1/3. And then you subtract it, evaluate it at 0, but that's just going to be 0. 0 squared over 2 minus 0, 0 to the third power. That's just all going to be 0. So when you subtract out 0, you're just left with this expression right over here. What's 1/2 minus 1/3? Well, that's 1/6. This is 1/6. And so we're left with this is equal to pi over 6, which is the exact same thing that we got in the last video. And that's because we did the exact same thing that we did in the last video. We just conceptualized it a little bit differently. We generalized it in terms of f of x and g of x. And we essentially conceptualized it as a washer, as opposed to doing the disk method for an outer shape and an inner shape like we did in the last video.