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## Integral Calculus

### Unit 3: Lesson 10

Volume: disc method (revolving around other axes)# Disc method rotating around vertical line

AP.CALC:

CHA‑5 (EU)

, CHA‑5.C (LO)

, CHA‑5.C.2 (EK)

Volume of solid created by rotating around vertical line that is not the y-axis using the disc method. Created by Sal Khan.

## Want to join the conversation?

- What is a 'principle root?' Is that just another name for a square root?(18 votes)
- Principle root is the
**positive**square root. For example, √25 asks the question "what number when multiplied together gives 25?" This question has two answers: 5 and -5. The**principle**root of 25 however is unambiguously 5. Unless you see ± before the radical symbol, it usually means the principle root.(63 votes)

- The diagram is confusing to me because the parabola at -2 would be at y = +3 and revolve from the left piece to a cup or from the right piece to an inverted rim. This diagram looks like the parabola was transformed so that it minimum occurs at -2,-1.(15 votes)
- You aren't the first person to find this example confusing, and I suspect Sal would spend a little more time explaining the setup, or maybe change his diagram a bit, if he were to redo it. The main thing you need to understand for proper visualization is that we/re using only the part of the parabola from x = 0 to x = 2 (which translates into the right branch of the parabola from y = -1 to y = 3).

If that clue isn't enough to give you a clear visualization, step away from this example for a minute and imagine a different one, where we have the line segment traced by x = 2 from y = -1 to y = 3, and we rotate it around the vertical line at x = -2. This would simply be a cylinder centered at x = -2. Now imagine that we change the vertical line segment that we rotated by bending in the bottom of it, not all the way to the center of the cylinder (at x = -2) but only to x = 0, and we rotate**that**curve around the same vertical at x = -2. The bottom of the figure would still have a flat region, just as the cylinder did, but now the flat region is smaller because its radius extends only from x = -2 to x = 0. That's the "gumdrop-shaped" figure Sal drew in his diagram.

To give it 3 dimensions he drew it at an angle instead of a straight-on cross section. Drawing it this way called for the bottom of the figure to be shown as curved instead of straight as it would be in a cross section, and that makes it look as if the parabola continues downward to the left from x = 0, when in reality the parabola ends at x = 0 and the curved line to the left of that point is a perspective view of the round, but flat, bottom of this figure.(10 votes)

- Why is this evaluated using the disk method and not the washer method? If you're evaluating it from -1 to 3 on the y-axis, wouldn't that leave a hollowed out section in the middle around the axis of rotation?(11 votes)
- Yeah I kinda agree with you in this, it wasn't really clear for me too till I watched it couple times, but Sal did mention that he is integrating about the "rotational axis" which is in this case x=-2, so it's a solid face shape and there is no hole so he used the disk method not the washer method! hope I didn't confused you more! :/(3 votes)

- considering y=x^2 - 1 is a parabola with a local minimum at x = -1, wouldn't the solid created by rotating around x=-2 require the shell method?(3 votes)
- You could do this with the shell method (integrating in x)--or you could solve for x = sqrt(1+y) and then use the disk/washer method (integrating in y).

The shell method is only required when it is not possible to solve for x in terms of y (though sometimes it is easier anyway).(5 votes)

- How do we get -1 and 3?(4 votes)
- Hi!

We rotated`y = x^2-1`

around the line`x = -2`

.`x = -2`

is parallel to our y-axis. Therefore, Sal picked 2 points on the y-axis as our interval (`y = -1`

,`y = 3`

).

Think about it, the sum of all the discs that are being created when we let`y = x^2-1`

rotate around the line`x = -2`

is the total volume of the figure created when`y = x^2-1`

rotated around`x = -2`

. This is basically the same thing as letting`y = x^2-1`

rotate around the y-axis but instead, we rotated it around a line that is*parallel*to it, in this case`x = -2`

. The difference is that the radius becomes different but the same sort of discs are created inside of the figure.**To answer your question**: By picking`-1`

and`3`

we will calculate the volumes of all the discs between`y = -1`

and`y = 3`

, which gives us the volume of the figure between`y = -1`

and`y = 3`

.

Hope I could be of any help!

// Kris(1 vote)

- Why use a principle root only and not include the negative root?(4 votes)
- The equation y=x^2-1 is that of a shifted parabola with point of contact with the line x= (-2) at (-2,3). Then on rotating the figure around the axis x= -2 shouldn't we get a cup like figure ( as shown in the video) along with a conical figure in it ( not shown in the video ) ?(3 votes)
- Earlier, you subtracted a function by a constant (x=2 or something) but now, you are subtracting a constant (in this case -2) from your function. How do you know if the given function should be subtracted by the given constant or vice versa?(1 vote)
- Due to the squaring in the integral, it does not matter if i do ( 2-f(x) )^2 or ( f(x)-2 )^2 because they are mathematically equivalent.

The difference is like saying the distance from the function to the line y=2 and the distance from the line y=2 to the function. They are both correct.(4 votes)

- Based on how the area is between y = x^2 - 1 and the y axis, and the rotation is around x =-2, shouldn't there be a hole in the middle of the gumdrop shape?(2 votes)
- The y-axis is not a boundary for the problem. If it was, then yes, there would be a hole in the object.(1 vote)

- I have a problem where I have to revolve y=x^3/2, x=4, and the x-axis about the line x=4. I'm confused on how to do it although I imagine it would be similar to the solution in this video.(2 votes)

## Video transcript

Let's do another example,
and this time we're going to rotate our function
around a vertical line that is not the y-axis. And if we do that--
so we're going to rotate y is equal
to x squared minus 1-- or at least this
part of it-- we're going to rotate it
around the vertical line x is equal to negative 2. And if we do that, we
get this gumball shape that looks something like this. So what I want to
do is I want to find the volume of this
using the disc method. So what I want to do is
construct some discs. So that's one of the
discs right over here. It's going to have some
depth, and that depth is going to be dy
right over there. And it's going to have
some area on top of it that is a function of
any given y that I have. So the volume of a
given disc is going to be the area as
a function of y times the depth of
the disc times dy. And then we just have to
integrate it over the interval that we care about, and we're
doing it all in terms of y. And in this case, we're
going to integrate from y is equal
to-- well, this is going to hit-- this
y-intercept right over here is y is equal to negative 1. And let's go all the
way to y is equal to, let's say y is equal
to 3 right over here. So from y equal negative
1 to y equals 3. And that's going to
give us the volume of our upside-down
gumdrop-type-looking thing. So the key here, so
that we can start evaluating the double
integral, is to just figure out what the area of each of these
discs are as a function of y. And we know that area
as a function of y is just going to be pi times
radius as a function of y squared. So the real key is, what is
the radius as a function of y for any one of these y's? So what is the radius
as a function of y? So let's think about
that a little bit. What is this curve? Well, let's write it
as a function of y. If you add 1 to both sides--
and I'm going to swap sides, so you'll get x squared
is equal to y plus 1. I just added 1 to both sides
and then swapped sides. And you get x is equal
to the principal root of the square root of y plus 1. So this we can write as x-- or
we can even write it as f of y if we want-- f of y is equal
to the square root of y plus 1. Or we could say x is equal
to a function of y, which is the square root of y plus 1. So what's the distance
here at any point? Well, this distance-- let
me make it very clear. So it's going to be
our total distance in the horizontal direction. So this first part as we're--
and I'm going to do it in a different
color so we can see. So this part right
over here is just going to be the value
of the function. It's going to give
you an x value. But then you have to add another
2 to go all the way over here. So your entire radius
as a function of y is going to be equal to the
square root of y plus 1. This essentially will give
you one of these x values when you're sitting on this curve,
this x as a function of y, it'll give you one
of these x values. And then from that, you
add another 2, so plus 2. Another way of thinking about
it, you get an x value here, and from that x value
you subtract out x is equal to negative 2. And when you subtract x
is equal to negative 2, you're adding 2 here. But hopefully, this
makes intuitive sense. This is the x value-- let
me do this in a better color-- this right over here,
this distance right over here, is the x value you
get when you just evaluate the function of y. But then if you wanted
the full radius, you have to go another
2 to go to the center of our axis of rotation. Once again, if you just take
a given y right over there, you evaluate the y,
you get an x value. That x value will just
give you this distance. If you want the full distance,
you have to subtract negative 2 from that x value,
which is essentially the same thing as adding
2 to get our full radius. So our radius as a function of
y is this thing right over here. So substituting
back into this, we can now write our definite
integral for our volume. The volume is going to be
equal to the definite integral from negative 1 to 3 of pi
times our radius squared dy. So I can write the
pi out here-- we've done this multiple times--
times radius squared-- so it's going to be square root
of y plus 1 plus 2 squared-- that's our radius-- times dy. So we've set up the
definite integral. And now we just have
to evaluate this thing. And I'll save that
for the next video. And I encourage you to
try this out on your own.