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## Calculus 2

### Unit 4: Lesson 9

Volume: disc method (revolving around x- and y-axes)# Disc method around y-axis

AP.CALC:

CHA‑5 (EU)

, CHA‑5.C (LO)

, CHA‑5.C.1 (EK)

Finding the volume of a figure that is rotated around the y-axis using the disc method. Created by Sal Khan.

## Want to join the conversation?

- How will I know if I'll use the disk, ring or shell method? And when do I use vertical or horizontal?(24 votes)
- It really depends on the situation you have. If you have a function y=f(x) and you rotate it about the x axis, you should use disk (or ring, same thing in my mind). If you rotate y=f(x) about the y axis, you should use shell.

Of course, you can always use both methodsyou can find the inverse of the function. If I wanted to rotate y=x^2 about the y axis, that would be equivalent to rotating x=√(y) about the x axis. I prefer to not bother with finding the inverse of the function. I would just identify the situation and use the appropriate method.**if**

By the way, if you have a function that is reverse, that is to say x=f(y), everything I said would be reversed. You would use disk about the y axis and shell about the x axis.(44 votes)

- Anyone that might explain why Sal takes the principle root of y at3:50? For some reason it is completely lost on me..(11 votes)
- He did that because he is rotating around the y-axis and he is trying to find the volume. As Sal showed, you need to find the radius of each disk so as to apply it into A = (pi)r^2 and then V = A(dy). Notice that it is in terms of dy, not dx. Therefore, the equation y=x^2 needed to be changed into terms of x, otherwise you would be finding a radius and thus an area and thus a volume of a solid that is irrelevant to this problem.(7 votes)

- At5:55, why did we use the interval from 1 to 4 instead of from 0 to 4?(7 votes)
- See0:12- it is 1 to 4 because that is how he decided to define the shape of the object whose volume is being measured.(10 votes)

- What are the practical uses of the disk method (or finding the volume of solids of revolution in general)? What are some physical applications of these problems?(2 votes)
- Edison's volume or displacement techniques won't work to estimate the volume of material in coronal loops (http://news.discovery.com/space/sun-throws-huge-and-beatiful-plasma-arc-into-space-140925.htm), the volume of brick in underwater tels showing a massive ancient flood in the Black Sea, or the volume of seamounts (http://en.wikipedia.org/wiki/Hawaiian%E2%80%93Emperor_seamount_chain).

Newton wrote calculus and physics in the same book as one idea. The book's title "Il Principia" means "The Idea." Whenever you wonder an application, for calculus, think "physics." Whenever the math in a physics question seems inefficient, try attempting the same question, after completing first-year calculus. Sometimes second-year calculus is required. When you have the math, physics is pretty simple.(10 votes)

- The integral used in the problem is in terms of y. However, since y=x^2 and thus dy=2x dx, can you redefine the integral so that it is from 1 to 2 and its expression is 2pix^3 dx?(6 votes)
- That's a very creative approach. It seems to work, but involves substituting things around and updating the bounds of integration. You still have to solve for x in terms of y (the square root bit) to transform the bounds from y to x and there is a little more simplification required at the end. This adds points at which stupid algebra errors can creep in.

If it helps you get the right answer and**it does**give you the right answer every time, go for it. Though you may have to argue with your teacher for a couple of "show your work" points because you do it differently. I leave it to someone more experienced to show proof.(7 votes)

- how do we calculate the volume if the question said roteted about the line y=3, x=5 on the same equation of y=x^2(2 votes)
- This isn't right as (1/2)^2 is 1/4 not 1/2. So the answer would be 31/4.(2 votes)

- how do you decide the limits of the integral?(1 vote)
- It will be stated if there is one curve and/or they will be the intercepts of two curves(4 votes)

- If we are integrating with respect to y, why don't we have to adjust the boundaries?(2 votes)
- Why do we? We integrate from y=1 to y=4 to find the volume, those are the boundaries regarding to y-axis. And Sal already manipulated the equation of the function into y: x=radical y or y^1/2 before carry out the integration.(2 votes)

- is the volume always positive?(1 vote)
- Volume is defined as the amount of space inside a solid. This is always a positive number because an object cannot take up a negative amount of space. Hope this helps!(3 votes)

- could you have done this by integrating in terms of x? instead of taking horizontal slices from top to bottom in terms of the y value with the depth being dy couldn't you take vertical slices in terms of x with the width being dx?

I'm confused because I hear if you're rotating about the y axis you should always try to put your equations in the form of x=...something in terms of y, and do the differentiation in terms of y but it seems unnecessary.(1 vote)- Depending on the function, it is just as viable. Sometimes it's easier to go one way than another, however.(3 votes)

## Video transcript

Here we have the graph,
or part of the graph, of y is equal to x squared again. And I want to find the volume
of another solid of revolution. But instead of rotating
around the x-axis this time, I want to rotate
around the y-axis. And instead of going
between 0 and some point, I'm going to go between y is
equal to 1 and y is equal to 4. So what I'm going
to do is I'm going to take this graph
right over here. I'm going to take this curve. And instead of rotating it
around the x-axis, like we did in the last few
videos, I'm going to rotate it around the y-axis. So I'm going to rotate
it around just like that. So what's the shape
that we would get? So let me see if we
can visualize this. So the base is going
to look something like that if we
could see through it. And then this up
here, the top of it, would look something like that. And we care about
the stuff in between. So we care about this
part right over here, not the very bottom of it. And let me shade
it in a little bit. So it would look
something like that. So let me draw it separately,
just so we can visualize it. So I'll draw it at
different angles. So if I were to draw it with
the y-axis kind of coming out the back, it would look
something like this. It would look something like--
it gets a little bit smaller like that. And then it gets cut off right
over here, right over here like that. So it looks-- I don't know
what shape you could call it. But I think hopefully
you're conceptualizing this. Let me do it in that
same yellow color. The visual-- that's not yellow. The visualization here is
probably the hardest part. But as we can see
it's not too bad. So it looks something like this. It looks like maybe a truffle,
an upside-down truffle. So this right here, let
me draw the y-axis just to show how we're oriented. So the y-axis is popping out
in this example like that. Then it goes down over here. And then the x-axis
is going like this. So I just tilted this over. I tilted it over
a little bit to be able to view it at
a different angle. This top right over here is
this top right over there. So that gives you an idea
of what it looks like. But we still haven't thought
about how do we actually find the volume of this thing? Well, what we can do,
instead of creating discs where the depth is in
little dx's, what if we created discs where the depth is in dy? So let's think about
that a little bit. So let's create-- let's
think about constructing a disc at a certain y-value. So let's think about
a certain y-value, and we're going to construct
a disc right over there that has the same radius of
the shape at that point. So that's our disc. That's our disc right over here. And then it has
a depth-- instead of saying it has a depth of dx,
let's say it has a depth of dy. So this depth right
over here is dy. So what is the volume of
this disc in terms of y? And as you could
imagine, we're going to do this definite integral,
and it is a definite integral, with respect to y. So what's the volume
of this thing? Well, like we did
in the last video, we have to figure out the
area of the top of each of these discs. Or I guess you could say
the face of this coin. Well, to find that
area it's pi r squared. If we can figure out this
radius right over here, we know the area. So what's that radius? So to think about that
radius in terms of y, we just have to solve
this equation explicitly in terms of y. So instead of saying it's
y is equal to x squared, we can take the principal
root of both sides, and we could say that the
square root of y is equal to x. And this right over here is only
defined for non-negative y's, but that's OK, because we are
in the positive x-axis right over here. So we can also call of this
function right over here x is equal to the
square root of y. And we're essentially
looking at this side of it. We're not looking at this
stuff right over here. So we're only looking at
this side right over here. We've now expressed
this graph, this curve, as x as a function of y. So if we do it that way, what's
our radius right over here? Well, our radius right over
here is going to be f of y. It's going to be the
square root of y. It's going to be the square
root of y is our radius. So it's going to
be a function of y. I don't want to confuse you if
you thought this was f of x, and actually this is f of y. No, it would be a function of y. We could call it g of y. It's going to be the
square root of y. So area is equal to
pi r squared, which means that the
area of this thing is going to be pi times
our radius, radius squared. Our radius is square root of y. So this thing is going
to be equal to pi-- the square root of y
squared is just pi times y. Now, if we want
the volume, we just have to multiply the area of
this surface times the depth, times dy. So the volume of
each of those discs is going to be pi y times dy. This gives you the
volume of a disc. Now, if we want the volume
of this entire thing, we just have to sum
all of these discs for all of the y-values
between y is equal to 1 and y is equal to 4. So let's do that. So we just take the definite
integral from y is equal to 1 and y equals 4. Just as a reminder,
definite integral is a very special type of sum. We're summing up
all of these things. But we're taking the
limit of that sum as these dy's get
shorter or get, I guess, squatter and squatter
or smaller and smaller, and we have a larger and
larger number of these discs. Really, as these
become infinitely small and we have an infinite
number of discs, so that our sum doesn't
just approximate the volume, it actually is the
volume at the limit. So to figure out the volume
of this entire thing, we just have to evaluate
this definite integral in terms of y. And so how do we do that? What's it going to be equal to? Well, we could take
the pi outside. It's going to be pi times the
antiderivative of y, which is just y squared over 2,
y squared over 2 evaluated from 1 to 4, which is
equal to pi times-- well, if you evaluate it
at 4, you get 16 over 2. Let me just write
it out like this. 4 squared over 2 minus
1 squared over 2, which is equal to pi times
16 over 2 is 8, minus 1/2. And so we could view
this as 16/2 minus 1/2, which is equal to 15/2. So this is equal
to 15/2 times pi. Or another way of thinking
of it is 7 and 1/2 times pi. But this is a
little bit clearer. So we're done. We found our volume not
rotating around the x-axis, but rotating around the y-axis,
which is kind of exciting.