In the last video we took essentially the length of y
equals x square, not the length, but we went from zero
to 1 on the x-axis and you can view it as this area.
And we rotated around the y-axis to get this figure here
and we figured out the volume, I think our answer if I
remember was pi over 2.
And we use what I called and what everyone calls
the Shell method.
I want to show you that you could actually used
the disk method here.
But then we'll just have to switch the ys and xs.
Instead of writing this the function as a function of x,
we'll just take the inverse of it and write it
as a function of y.
so this curve y equals x squared, what it can also be
written as, we can just take the squared root of both
sides as x is equal to the square root of y.
Just take the square root of both sides.
And now we can use this information to do the disk
method, but everywhere where we had an x in the past
we'll now have a y.
So let's think about how we would do it.
Once again it's this area-- really the hardest thing about
all of these problems is just the visualization.
I think that's why they do it.
Just to make sure that you know how to visualize things and
maybe understand the calculus.
It's more visualization then calculus really.
So we're still dealing with-- if we take a cross section, if
we were to cut this figure this area would be the
cross section.
And we're still rotating around the y-axis just like we
did in the last video.
We're still rotation around the y-axis, we get this figure.
So how would we deal with disks.
Well a disk would look like this at some point-- let
me pick another color.
At some point we'd have a disk like that.
That would be the top of the disk, and it would have some
depth like we did before.
And its height, or the radius of that disk,
would be equal to x.
It's equal to x.
I know you're thinking it looks like the shell method.
But what is x equal to?
It's equal to the square root of y.
It equals the square root of y.
And what would be the width of that disk?
Now we are making everything as a function of y, so the width
would be just a very small distance d y, the
differential y.
That's essentially all we need to know.
So the volume of that disk would be the radius squared
times pi times d y.
Hopefully that makes a little bit sense, but there's another
hitch on this problem.
This is actually similar to what we did two videos ago.
Because when you view it in the y-axis, from this y frame of
reference, what we're going to do is we're going to take the
volume, we take the volume of x is equal to 1.
So what would be the volume of x is equal to 1 rotated
around the x-axis?
It would just be the entire cylinder.
It's really important that you visualize this
right, what we're doing.
We're going to figure out this volume, volume
of x is equal to 1.
Let me draw the axis just so you know what we're doing.
So this would be the y-axis, that would be the x-axis
as best as I can draw.
Originally, we can figure out the volume of
the entire cylinder.
This is x is equal to 1, x is equal to 1 from what y points?
From the point, well what is this?
This is y equals 1.
y equal 1 to zero.
So we would figure out this volume from y equals 1 to zero.
And how would we do that?
What would be the integral?
Remember everything is in y now, so it might seem
a little confusing.
For each of these disks, this is going to be made up of
a bunch of disks, let me draw one of them.
Let me draw the top one.
The top disk, and it has a width, it's not going to be
this entire cylinder, its width is d y, its width
is just going to be this.
d y, a very small sliver.
d y, and its radius is x or you could say f of y.
What would be the volume of that disk?
What would be the surface area of the top?
It would be f of y squared.
Remember we're dealing everything with y. f of y
squared times pi, that gives the area of the top.
Put the pi outside of the integral times d y.
d y, and we said y is going from zero to 1.
y is going from zero to 1.
So that's that entire cylinder.
And what we'll want to do is subtract out, essentially cut
out the volume of the inner bowl.
So minus.
How would we figure out the inner bowl?
That's where we will deal with the x is equal to the
square root of y function.
Because here, what's each disk?
Once again it's f of y.
I wrote this generally, this is f of y of the outside.
I'm going to now do it for the inside.
I'm going to cut out the inside volume and because I kept
it general I could do it.
We're still going from zero to 1, from y equals
zero to y equals 1.
Remember we're doing with y now.
I will take f of the inside f of y squared d of y.
You can imagine in this case the inside function, we're
going to take the volume.
That original disk I drew is actually one of the disks
for the inside function.
Where the height of that disk is d y, that should be d y.
The radius of the disk is f of y.
And of course the area of the top of the desk is pi
times the radius squared.
How do we apply it to this particular situation?
What is the outside function?
f of y, x is equal to f of y.
We are just switching the variables.
We say that x is equal to 1.
This is just a big cylinder, right?
So f of i is equal go to 1 in this function, so we get pi--
let me switch colors-- times from zero to 1.
1 squared d y minus pi, and we're still going
from zero to 1.
Remember, that's the y boundaries.
What's f of y?
Here f of y is square root of y squared d y.
Let's take the pi out.
So pi times, well 1 squared is 1.
What's the anti-derivative of 1?
What function's derivative is 1?
It's x, right?
x, that's the anti-derivative there.
We can merge these.
We took the pi out.
This square root of y squared, this is just y.
Sorry, we have that derivative 1.
See, even I get quite confused here.
We're taking this with respect to y, right?
The anti-derivative of 1 now is y.
We are setting up a bunch of ys.
I'm sorry.
I find this even a little perplexing when I start
switching x and y.
The second function.
Remember, we can merge these because it's the same
boundaries and they're both in respect to y.
Square root of y squared is y.
What's the anti-derivative of y, the function y or y d y.
Was y squared over 2, minus y squared over 2.
We are going to evaluate that at 1 and zero.
So what does that get us?
We get pi-- we get 1 minus 1/2 minus zero minus zero
minus zero plus 0/2.
Whatever these are zero.
1 minus 1/2 that's 1/2 and times pi is
equal to pie over 2.
Which is the exact same result.
And I was worried, because you never know when you might make
a careless mistake as I often do.
We got the exact same result as I got in the previous video
when we used the shell method.
The hardest thing here is just remembering that you're doing
everything in terms of y.
When we did the disk method traditionally, the disks
were vertical disks.
Now they're horizontal disks but it's the exact same thing.
You just have to get your brain around the idea that we're
dealing with the ys, that the boundaries on the integrals
are now y values.
We're taking the width of the disks, or the height of the
disks are d y and the radius is now the function of y.
Hopefully I didn't confuse you too much.
My own brain malfunctioned a bit when I took the
anti-derivative of 1 d y, it should have been y.
I will see you in the next video where we will do even
more complicated problems, where I'm sure I'll make
even more mistakes.
See you soon.