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### Course: Integral Calculus > Unit 3

Lesson 11: Volume: washer method (revolving around x- and y-axes)# Generalizing the washer method

Looking at the example from the last video in a more generalized way. Created by Sal Khan.

## Want to join the conversation?

- How would the volume be calculated if the shape was only partly rotated?(50 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.(79 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(25 votes)
- ^^Exactly, you would have to find their point of intersection and then split it there :)(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.(11 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)
- 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)

- 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.(4 votes)

- 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)

- 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)

- if you had something like f(x^2) = x^3 would this be equivalent to f(x) = x^8 or f(x) = x(2 votes)
- Why there's no lessons for shell method?(3 votes)

## Video transcript

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.