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### Course: Calculus, all content (2017 edition) > Unit 6

Lesson 10: Washer method- Solid of revolution between two functions (leading up to the washer method)
- Generalizing the washer method
- Washer method rotating around horizontal line (not x-axis), part 1
- Washer method rotating around horizontal line (not x-axis), part 2
- Washer method rotating around vertical line (not y-axis), part 1
- Washer method rotating around vertical line (not y-axis), part 2
- Washer method: revolving around x- or y-axis
- Washer method worksheet
- Disc & washer methods challenge

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# Washer method rotating around vertical line (not y-axis), part 1

Setting up the definite integral for the volume of a solid of revolution around a vertical line using the "washer" or "ring" method. Created by Sal Khan.

## Want to join the conversation?

- How does one determine whether it's best to integrate with respect to x vs. with respect to y? It isn't always obvious, at least not to me...(10 votes)
- If rotating around x-axis or line parallel to x-axis: Integrate with respect to x.

If rotating around y-axis or line parallel to y-axis: Integrate with respect to y.(45 votes)

- what if the rotating axis is not parallel to either the x axis or y axis? For example, x+y=2 or something.(12 votes)
- I dont think there is a quick solution to that. You would have to transform your function such that the axis of rotation becomes the x-Axis.(11 votes)

- At5:36, the outer radius is defines as 2-y^2.

Can anybody explain why ones takes 2 - the function. My first guess was the function + 2 so that i would reach the line where x =2. This is wrong, but i can´t say i really understand why.

And, taking 2-y^2 would that be the gap between the undefined point on the x-axes and x=2? Where is this on our shape?

Thanks.(12 votes)- At this stage in the process we're trying to find the radius of the outer disk. The radius is the distance from the center of the disk to the perimeter. At the beginning of the video we're told that the figure is created by rotation around the line x=2, so we know the center of the disk is at x=2. The outer edge of the larger disk is at the point x=y^2, so the distance between the two points is 2-y^2.(9 votes)

- how did you come up with the values of the limits of integration right towards the end of the video?(6 votes)
- those are the y values where both functions intersect.(15 votes)

- at7:36how do you get your limits of integration? I understand that you did by inspection, but how would you do it by making the two equations equal to each other ?(3 votes)
- In that example, Sal is integrating with respect to y. To solve for zero and one analytically, we could make the equations equivalent to each other and solve. Here's the example:

sqrt(x) = x^2

x = x^4

This last equation is true only when x = 0 or x = 1.(4 votes)

- What would the solid look like of the sine function around the y-axis? A series of co-joined balls with a radius of pi/2 perhaps?(3 votes)
- Be careful: what you were visualizing is a rotation about the
**x**-axis (not y-axis).

In order to find out how it would look like about the y-axis, you'd have to specify how to construct the area whose rotation will produce the solid. That's because "sin(x)" by itself doesn't produce any area, it's just a continuous line. If, for example, you were to specify that it's "the area between the curve sin(x) and the x-axis", then the rotation about the y-axis would yield a "ripples on a pond" sort of figure.(1 vote)

- I just want to say thank you for doing this. It is helping me a ton, but I still have one question. How do you know when you don't have to integrate the outer volume but have to integrate the inner volume (and then subtract it) or vice versa? I hope that makes sense...it just seems that sometimes there's only one volume you need to integrate while the other volume is staying constant. I'm having a hard time figuring out when that is. Thank you so much!(2 votes)
- I recommend to watch the exercises and think them through on your own.

As a quick guide,

1. Look at the rotational axis, is it parallel to the x or y-axis.

2.Check the offset ( distance of your axis of rotation)

3.Determine the boundaries.

Integrate and calculate the result.

(Practice makes perfect )(3 votes)

- Do i ever HAVE to use the shell method, or can i always use disc/washer if i want?(2 votes)
- I think you can but if its a shell method problem its going to be really hard to do disk or washer method on it.(1 vote)

- So if we are rotating around the y-axis or x= a number then we make the bounds in terms of y?(1 vote)
- Nearly all of these problems can be attacked at least two different ways. When rotating around the y-axis or other vertical line we may solve by the shell method, in which case we integrate with respect to x, or by the disk or washer method, in which case we integrate with respect to y. The reverse would be true if rotating around the x-axis or other horizontal line. Rather than try to memorize these relationships, learn to think in terms of which direction will have a small increment of thickness for the method you've chosen. If it's a horizontal thickness, it's dx, so you integrate with respect to x; if vertical, you're dealing with y.(3 votes)

- Find the volume (in units3) generated when the region between the curves is rotated around the given axis.

y=1/6−x, x=1, and x=2 rotated around the line x=6.(2 votes)

## Video transcript

What we're going to
do in this video is take the region
between the two curves, y is equal to square
root of x on top and y is equal to x
squared on the bottom and rotate it around a vertical
line that is not the y-axis. So we're going to rotate it
around the vertical line x is equal to 2. We're going to rotate it
right around like that. And if we rotate
it like that, we will get this shape, so
this strange-looking shape. It's hollowed out in the middle. You can kind of use the y equals
x squared part kind of hollows out the middle. And it's really the
stuff in between that forms the wall
of this rotated shape. So let's think about how we
can figure out the volume. And we're going to do it using
the disk method, sometimes called the ring
method-- actually, it's going to be more
of the washer method-- to do this one right over here. So we're rotating
around a vertical line. We want to use our disk
or ring or washer method. And so it'll be
really helpful to have a bunch of rings stacked
up in the y direction. So we're probably going to want
to integrate with respect to y. Let me make it clear
what I'm talking about. Each of our rings are going
to look something like this. So let me do my best
attempt to draw. So that's the inner
radius of the ring defined by y is equal to x squared. And then the outside
radius of the ring might look something like this. My best attempt to
draw it reasonably. The outside radius of the ring
might look something like that. And the depth of the ring would
be a little dy just like that. So let me draw the depth. So let me see, right over here. You could kind of view it
as the height of the ring, because we are thinking
in a vertical direction. And that would be
a ring for a given y for, say, this
y right over here. It's essentially
the ring generated if you were to take this
rectangle of height dy and rotate it around
the line x equals 2. And we're going to
construct a ring like that for each of
the y's in our interval. So you could imagine stacking
up a whole set of rings just like this and then taking
the sum of the volumes of all of those rings and
the limit of that side as you have an infinite number
of rings with-- or close to infinite-- or
really, infinite number of rings with
infinitesmally small height or depth or dy. So let's figure out
how we would do this. Well, we know that
all we have to do is figure out what the
volume of one of these rings are for given y
as a function of y and then integrate along all
of them, sum them all up. So let's figure out what the
volume of one of these rings are. And to do it, we're
going to express these functions
as functions of y. So our purple function, y is
equal to square root of x, if we square both sides,
we would get y squared is equal to x. And let me swap the sides so
it makes it clear that now x is a function of y. x is equal to y squared. That's our top
function the way we've drawn a kind of outer
shell for our figure. And then y is
equal to x squared. If you take the principal root
of both sides of that and it all works out because we're
operating in the first quadrant here. That's the part of it
that we care about. So you're going to get x is
equal to the square root of y. That is our yellow
function right over there. Now, how do we
figure out the area of the surface of one of these
rings or one of these washers? Well, the area-- let
me do this in orange because I drew that
ring in orange. So the area of the
surface right over here in orange as a
function of y is going to be equal to the
area of the circle if I just consider
the outer radius, and then I subtract out the
area of the circle constructed by the inner radius, just
kind of subtract it out. So the outer-circle
radius, so it's going to be pi times outer
radius squared minus pi-- let me write it this
way-- outer-- well, I'll just write
it-- outer radius-- this is going to be as a
function of y-- outer radius squared minus pi times
inner radius squared. And we want this all
to be a function of y. So the outer radius
as a function of y is going to be what? Well, it might be easier
to visualize-- actually, I'll try it both places. So it's this entire
distance right over here, essentially, the distance
between the vertical line, the horizontal distance
between our vertical line, and our outer function. So if you think about
it in terms of x, It's going to be
2 minus whatever the x value is right over here. So the x value right over
here is going to be y squared. Remember, we want this
is a function of y. So our outer radius,
this whole distance, is going to be 2 minus the x
value here as a function of y. That x value is y squared. So the outer radius
is 2 minus x-- sorry. 2 minus y squared. We want it as a function of y. And then our inner radius is
going to be equal to what? Well, that's going
to be the difference, the horizontal distance,
between this vertical line and our inner function,
our inner boundary. So it's going to be the
horizontal distance between two and whatever x value this is. But this x value as a function
of y is just square root of y. So it's going to be 2
minus square root of y. And so now, we can come up
with an expression for area. It's going to be-- and I'll
just factor out-- actually, I'll leave the pi there. So it's going to be
pi-- right over here-- it's going to be pi times
outer radius squared. Well, the outer
radius is 2 minus y squared-- and let me just--
well, I'll just write it. 2 minus y squared--
and we're going to square that-- squared,
minus pi times the inner radius squared. Well, we already
figured that out. The inner radius is 2
minus square root of y. And we're going to
square that one, too. So this gives us the
area of one of our rings as a function of y, the
top of the ring, where I shaded in orange. And now, if we want the
volume of one of those rings, we have to multiply it by its
depth or its height the way we've drawn it right over here. And its height-- we've done this
multiple times already right over here-- is an
infinitesimal change in y. So we're going to multiply
all that business times dy. This is the volume
of one of our rings. And then we want to sum up all
of the rings over our interval. So we're going to sum up all
of the rings over the interval. And when you take
the integral sign, it's a sum where you're
taking the limit as you have an infinite number of rings
that become infinitesimally small in height or depth,
depending on how you view it. And what's our interval? So we've looked at
this multiple times. These two graphs-- you
could do it by inspection. You could try to
solve it in some way, but it's pretty obvious
that they intersect at-- remember we care
about our y interval. They intersect at y is equal
to 0 and y is equal to 1. And there you have it. We've set up our
integral for the volume of this shape right over here. I'll leave you there,
and in this next video, we will just evaluate
this integral.