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## Calculus 2

### Unit 4: Lesson 12

Volume: washer method (revolving around other axes)- 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 other axes

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# Washer method rotating around horizontal line (not x-axis), part 1

AP.CALC:

CHA‑5 (EU)

, CHA‑5.C (LO)

, CHA‑5.C.4 (EK)

Washer method when rotating around a horizontal line that is not the x-axis. Created by Sal Khan.

## Want to join the conversation?

- why in the last video is the function minus the horizontal line and in this video is the horizontal line minus the function?(26 votes)
- Because in this video the original function is below the horizontal line.(59 votes)

- In my perspective all of these rotations around perfectly vertical or horizontal axes are simply translating all those functions up/down so essentially that axis you are rotating along is the new "x/y axis", in a relative sense. Is this concept correct?

Also (this is probably another separate question) so what if we were trying to rotate along, say, y=2x? And does rotation along an axis which is not a straight line exist (or is it just that I can't imagine it at the moment)?(12 votes)- First question: Yes, that is a way of thinking about it! :)

Second questions: No, you can do what you described but it would not be a rotation. A rotation must not alter the shape of the "object" that we are rotating. Imagine rotating something in real life, you never alter the shape of the object, just its position and orientation.

"rotating" something around a curved line in math would be like bending something.(9 votes)

- At3:44, why is the outer radius 4- x^2 +2x instead of x^2 - 2x -4? Or is it the same thing because even though it's the negative version of the terms, it's going to get squared anyway?(8 votes)
- No, it's because the horizontal line is above the function, instead of being below it like in the last video, so we subtract the function from the horizontal line. It's also because 4 is the larger value in the interval from 0 to 3, so it keeps the result positive.

Your intuition is correct that it doesn't matter since squaring the result will make it positive anyway, but it keeps the math easier without dealing with confusing sign changes.

(x^2 - 2x - 4)^2 = (- x^2 + 2x + 4)^2 = (4 - x^2 + 2x)^2

x^4 - 4x^3 - 4x^2 +16x + 16(6 votes)

- at around3:40, why does the height equal to 4 - (x^2-2x)? Why do we subtract by the function?(6 votes)
- It just so happens that there is another site with the exact same function with nice 3D graphical breakdown of the model.

See example 3:

http://tutorial.math.lamar.edu/classes/calcI/volumewithrings.aspx(5 votes)

- How does "rotating a function around a line and obtaining a solid of revolution" help us in our life? Where is this concept actually applied?(1 vote)
- Realistically it's unlikely you'll ever have to actually apply this technique. Many of the problems we tackle in calculus are designed to improve (or test) our overall understanding of the concepts and ability to apply them in a variety of contexts. There happens to be a story about this type of calculation, though. Thomas Edison hired scientists to work in his laboratory and one day he saw one of them doing a complicated calculation to figure out the volume of a glass bulb of some kind. Chances are the scientist was using this kind of calculus technique. Edison walked over, took the glass bulb, filled it with water, and poured the water into a graduated beaker to find the volume in about five seconds.(15 votes)

- I get the part about the inner radius. But I am still confused about how the outer radius is not 4+ x^2 -2x? so from x-axis to horizontal line the distance is 4 but then don't we have to add x^2+4 to get the radius??(2 votes)
- The outer radius is defined in a later video as the distance from the axis of rotation to the outer function. To get this, you would take the axis of rotation (in this case: 4) and subtract it by the outer function (x²-2x). Ultimately, as in before Sal simplifies it, the outer radius would be: 4-(x²-2x). Sal just simplified it right away to 4-x²+2x by distributing the negative. I hope that makes more sense...(5 votes)

- can you find the volume if you rotate the function around a slanted axis like y=x for example?(2 votes)
- Yes, but it would have to be done with triple integrals, which is something you learn in multivariable calculus.(4 votes)

- So even if the region was touching the horizontal axis, you would take the 4-f(x) in this particular problem?(3 votes)
- Yes, because you are rotating around y=4 instead of the horizontal axis. All distances have to be relative to the axis that the figure is being rotated around.(1 vote)

- if the rotating line is negative or below the x-axis and the function is positive or over the x-axis, how would you calculate the height between the rotating horizontal line and the function ?(3 votes)
- Imagine just a simple line, like
`y = 3`

. It'll lie parallel to the x axis. If you want to shift it up by one, you'd add one, right?:`y = 3 + 1`

. If you want to shift it down by one, you'd subtract one:`y = 3 - 1`

.

The shifting of weird curves works in the same way:

1) Take the rotating line to be your new x axis.

2) If your function`f(x)`

is below that line, then it was shifted by an amount`C`

downwards from that line. These are examples:`y = 3 - 1`

← A concrete, simple one.`y = f(x) - C`

← One more general.

3) If your function`f(x)`

is above that line, then it was shifted by an amount`C`

upwards. These are examples:`y = 3 + 1`

← A concrete, simple one.`y = f(x) + C`

← One more general.

Don't get confused by the weirdness of your function or whether the new line is above or below the original x axis. You can use this equation:`y = f(x) ± C`

, to figure out the value of`C`

in the first place and, at the end, replace`f(x)`

by the function you're working with.(1 vote)

- At4:05, when Sal is finding the outer radius, he writes 4 - x^2 + 2x. Shouldn't it be 4 - x^2 - 2x as in the original function?(2 votes)
- The original function is x² - 2x. The outer radius is 4 minus this function: that is, 4 - (x² - 2x), which is, after removing the parentheses, 4 - x² + 2x.(2 votes)

## Video transcript

Now let's do a really
interesting problem. So I have y equals x,
and y is equal to x squared minus 2x
right over here. And we're going to
rotate the region in between these two functions. So that's this region
right over here. And we're not going to rotate
it just around the x-axis, we're going to rotate it around
the horizontal line y equals 4. So we're going to
rotate it around this. And if we do that, we'll get
a shape that looks like this. I drew it ahead of time, just
so I could draw it nicely. And as you can see, it looks
like some type of a vase with a hole at the bottom. And so what we're going to do
is attempt to do this using, I guess you'd call it
the washer method which is a variant of the disk method. So let's construct a washer. So let's look at a given x. So let's say an x
right over here. So let's say that we're
at an x right over there. And what we're
going to do is we're going to rotate this region. We're going to give
it some depth, dx. So that is dx. We're going to rotate
this around the line y is equal to 4. So if you were to visualize it
over here, you have some depth. And when you rotate it
around, the inner radius is going to look like the
inner radius of our washer. It's going to look
something like that. And then the outer
radius of our washer is going to contour
around x squared minus 2x. So it's going to
look something-- my best attempt
to draw it-- it's going to look
something like that. And of course, our washer
is going to have some depth. So let me draw the depth. So it's going to
have some depth, dx. So this is my best attempt at
drawing some of that the depth. So this is the
depth of our washer. And then just to make the face
of the washer a little bit clearer, let me do it
in this green color. So the face of the
washer is going to be all of this business. All of this business is going
to be the face of our washer. So if we can figure
out the volume of one of these washers for a
given x, then we just have to sum up all of
the washers for all of the x's in our interval. So let's see if we can
set up the integral, and maybe in the
next video we'll just forge ahead and actually
evaluate the integral. So let's think about the
volume of the washer. To think about the
volume of the washer, we really just have to
think about the area of the face of the washer. So area of "face"--
put face in quotes-- is going to be equal to what? Well, it would be the
area of the washer-- if it wasn't a washer,
if it was just a coin-- and then subtract out
the area of the part that you're cutting out. So the area of the
washer if it didn't have a hole in the
middle would just be pi times the
outer radius squared. It would be pi times
this radius squared, that we could call
the outer radius. And since it's a washer,
we need to subtract out the area of this inner circle. So minus pi times
inner radius squared. So we really just
have to figure out what the outer and inner radius,
or radii I should say, are. So let's think about it. So our outer radius is
going to be equal to what? Well, we can visualize
it over here. This is our outer
radius, which is also going to be equal to
that right over there. So that's the distance
between y equals 4 and the function that's
defining our outside. So this is essentially,
this height right over here, is going to be equal to 4
minus x squared minus 2x. I'm just finding the distance
or the height between these two functions. So the outer radius is
going to be 4 minus this, minus x squared minus
2x, which is just 4 minus x squared plus 2x. Now, what is the inner radius? What is that going to be? Well, that's just going to
be this distance between y equals 4 and y equals x. So it's just going
to be 4 minus x. So if we wanted to find
the area of the face of one of these washers for a
given x, it's going to be-- and we can factor
out this pi-- it's going to be pi times the
outer radius squared, which is all of this
business squared. So it's going to be 4 minus
x squared plus 2x squared minus pi times
the inner radius-- although we factored
out the pi-- so minus the inner radius squared. So minus 4 minus x squared. So this will give us
the area of the surface or the face of one
of these washers. If we want the volume of
one of those watchers, we then just have to
multiply times the depth, dx. And then if we want to actually
find the volume of this entire figure, then we just have to
sum up all of these washers for each of our x's. So let's do that. So we're going to sum
up the washers for each of our x's and take the
limit as they approach zero, but we have to make sure
we got our interval right. So what are these-- we care
about the entire region between the points
where they intersect. So let's make sure
we get our interval. So to figure out our
interval, we just say when does y equal
x intersect y equal x squared minus 2x? Let me do this in
a different color. We just have to
think about when does x equal x squared minus 2x. When are our two functions
equal to each other? Which is equivalent
to-- if we just subtract x from
both sides, we get when does x squared
minus 3x equal 0. We can factor out an x
on the right hand side. So this is going to be when does
x times x minus 3 equal zero. Well, if the product is equal
to 0, at least one of these need to be equal to 0. So x could be equal to 0,
or x minus 3 is equal to 0. So x is equal to 0
or x is equal to 3. So this is x is 0, and
this right over here is x is equal to 3. So that gives us our interval. We're going to go
from x equals 0 to x equals 3 to get our volume. In the next video,
we'll actually evaluate this integral.