Main content

## AP®︎ Calculus AB (2017 edition)

### Course: AP®︎ Calculus AB (2017 edition) > Unit 11

Lesson 8: Volume: shell method (optional)- Shell method for rotating around vertical line
- Evaluating integral for shell method example
- Shell method for rotating around horizontal line
- Shell method with two functions of x
- Calculating integral with shell method
- Shell method with two functions of y
- Part 2 of shell method with 2 functions of y
- Shell method worksheet
- Shell method

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Shell method with two functions of x

Using the shell method to rotate around a vertical line. Created by Sal Khan.

## Want to join the conversation?

- why is the radius, or x, 2 - x?(37 votes)
- The radius of each cylindrical shell is the horizontal distance from the current x value to the axis of rotation. So if we rotate about the line x=2, the distance between our current x position and the axis of rotation is
**2-x**.

Likewise, if we rotate about the y axis (aka x=0) the radius is x-0=**x**.(45 votes)

- At2:22why can't the radius be 2-x^(1/2). Isn't the distance from the axis of rotation to the furthest point on the shell (the point on the function x^(1/2)), the value of the radius?(8 votes)
- For any given x-value, the radius of the shell will be the space between the x value and the axis of rotation, which is at x=2.

If x=1, the radius is 1, if x=.1, the radius is 1.9.

Therefore, the radius is always 2-x.

The x^(1/2) and x^2 only come into play when determining the height of the cylinder.(3 votes)

- How do you decide whether to use disks or whether to use shells? Is there something to look for that will indicate what problems should be solved using one vs the other?(8 votes)
- What it we rotated about -2 instead of 2. Would it then be x+2? TY!!(7 votes)
- Indeed it would be x+2 which is really a simpler way of writing x–(-2), the distance from the current x coordinate to the line x=-2.

Just make sure that your distance/radius factor is always*positive*when you take the shell method. By that I mean not to use -2–x. You will end up with a negative volume if you perform that integration.(3 votes)

- How did Sal just throw in the factor of dx at3:45? The volume of a real shell is not the area of its outer surface times the depth of the shell, so why should that be the volume when the shell is infinitesimally thin?(4 votes)
- The volume of the shell, as stated in previous videos, is the circumference times the height of the shell times the width. Here, the width is dx. However, it does not really matter in the end, because you are just taking the definite integral to find the area.

Hope that helps!(2 votes)

- If rotation occurs about the Y-axis or X=0, the radius would simply be X correct?(2 votes)
- correct!

keep in mind that if we rotate around y=c in D we change the integral:

v(K)=π a(integral)b |(g(x)-c)^2-(f(x)-c)^2| dx,

for the volume of body K where D={(x,y):a<=x<=b,f(x)<=y<=g(x)}

this means that we subtract the value of c if we move the axis of rotation up

and that we add the value of c if we move the axis of rotation down

i hope this was a little helpful!(2 votes)

- At4:04how did you find the interval from 0 to 1 when looking at volume?(2 votes)
- Using the intersection points of two functions.(1 vote)

- How is the radius 2-x. He always says that the radius is the function minus what it's being rotated around so wouldn't it be x-2? Because in every other video that is the case but in this one it's the value 2- x, why is that?(1 vote)
- Many of these problems can be solved in multiple ways. For this one, you're rotating around the line x = 2, so you can find the radius by going from that point to the left, as Sal did, getting 2-x, or from that point to the right, getting x - 2, the version that seems more natural to you. Either one produces the same result, but Sal has chosen the easier way because when you measure on the left your integral is from 0 to 1 (the x-values on the left that are within range of the figure we're analyzing), whereas if you measure on the right your integral is from 3 to 4. It's almost always easier to evaluate expressions at 0 and 1 than at 3 and 4.

It would be good practice for you to calculate the integral the way you suggest, using x - 2 as the radius and evaluating the integral from 3 to 4, and confirming that the result is the same.(3 votes)

- at4:04, why are we integrating from 0 to 1? That would mean our radius (2-x) goes from 2 to 1, but don't we really want the radius from x=0 to x=1? Sorry if I'm not explaining myself well, but I thought we should integrate from 1 to 2 so the radius goes from 1 to 0.(1 vote)
- In this example the x values only resemble the distance from the origin of the axis, you will have to synthesize the axis of revolution(x=2) and the x values to make the radius of the cilinder.

Because the graph is twisted around the x=2 axis, the radius = the difference between the axis we twist the graphs around(x=2) and the x values between x=0 and x=1(to include all the possible x values we integrade from x=0 to x=1).

We are only interested in the solid of revolution of the shape on the left side. Graph the formulas between x=1 and x=2 for your visualization, the shape looks different.(1 vote)

- These videos are inaccessible for me for some reason. I was only able to watch the videos previous to this one. What do I need to do to access it?(1 vote)
- That sounds like a technical problem; I'd check the help center or contact the site rather than commenting.(1 vote)

## Video transcript

This right here is a
solid of revolution whose volume we were able to
figure out in previous videos, actually in a different
tutorial, using the disk method and integrating in terms of y. What we're going
to do right now is we're going to find the same
volume for the same solid of revolution, but we're going
to do it using the shell method and integrating
with respect to x. So what we do is we have
the region between these two curves, y is equal
to square root of x and y is equal to x squared. And we're going to rotate it
around the vertical line x equals 2. And we're going to do
that at the interval that we're going to rotate this
space between these two curves is the interval when square root
of x is greater than x squared. And so it's between 0 and 1. And so let's try to do
it with the shell method. And so to do that, what we do
is we want to construct a shell. Let me do this in
a different color. Let's imagine a rectangle
right over here. It has width dx. and its height is the difference
of these two functions. And so if I were to
draw it right over here, it would look
something like this. It'd be there, and
then it is a shell, it's kind of a
hollowed-out cylinder. So it would look
something like this. Just like that. And it has some depth,
that's what the dx gives us. So we have the depth that
looks something like that. And then let me shade
it in a little bit, just so we can see a
little bit of its depth. So when you rotate
this rectangle around the line x equals 2,
you get a shell like this. So let's think about
how we can figure out the volume of this shell. Well, we've already
done this several times. The first thing we might
want to think about is the circumference of
the top of the shell. We know circumference
is 2 pi times radius. We just need to know what
the radius of the shell is. What is that
distance going to be? Well, it's the
horizontal distance between x equals 2 and whatever
the x value is right over here. So it's going to be
2 minus our x value. So this radius, this
distance right over here, is going to be 2 minus x. And so the circumference is
going to be that times 2 pi. 2 pi r gives us the
circumference of that circle. So 2 pi times 2 minus x. And then if we want
the surface area of the outside of our
shell, so the area is going to be the circumference
2 pi times 2 minus x times the height of each shell. Now, what is the
height of each shell? It's going to be the
vertical distance expressed as functions of y. So it's going to
be the top boundary is y is equal to square root
of x, the bottom boundary is y is equal to x squared. So it's going to be square
root of x minus x squared. Let me do this in the yellow. So it's going to be square
root of x minus x squared. And so if you want the
volume of a given shell-- I'll write all this
in white-- it's going to be 2 pi
times 2 minus x times square root of x
minus x squared. So this whole expression,
I just rewrote it, is the area, the outside surface
area, of one of these shells. If we want the volume, we have
to get a little bit of depth, multiply by how deep the
shell is, so times dx. And if we want the volume
of this whole thing, we just have to
solve all the shells for all of the x's
in this interval and take the limit as the
dx's get smaller and smaller and we have more
and more shells. And so, what's our interval? Well our x's are going
to go between 0 and 1. So that right over there is
the volume of this figure.