Let's do a couple more rotational volume problems, and
I'm going to make these a little bit more difficult.
And hopefully after these if you've understood everything
we've done up to now and the ones I'm about to do, I think
you're pretty set for most of what you should face
in most math classes.
And definitely I think you'll be set for the AP exam,
and either ab or bc on this concept.
So let's do another example.
OK.
So let's say that I want to-- actually, let me
do something different.
Let me draw here.
So that's my y-axis.
This is my x-axis.
Now let me draw the function y is equal to x squared.
And we know that that could be written as y is equal to x
squared, or we could write that as x is equal to square root of
y, depending on what we want to be a function of what.
This is the y-axis.
This is the x-axis.
Let's say I also have the line y equals 2.
Goes over what I just wrote.
y equals 2.
Now this problem is going to be slightly different then
what we've done so far.
I'm going to take a rotation, but instead of taking a
rotation around the y or the x-axis, I'm going to take a
rotation around another line.
So let's say I want to take a rotation between
x is equal to 0.
Actually let me do something arbitrary.
Let me say between x is equal to 1, so that's that point,
and x is equal to 2.
That's where they intersect.
This is the point right here, this is 2,2.
I'm sorry, no, 2,4.
Because y is equal to x squared.
So this is 2,4.
This is the point 4.
So what point is this.
This is the point 1, right?
So our y values go from 4 to 1, our x values go from 1 to 2.
And that makes sense, because y is x squared.
And so if we were to kind of take the area that we're going
to rotate, and I haven't told you what we're going to rotate
it around yet, and this might prove to be shocking to you.
So this is the area we're going to rotate.
Instead of rotating it around the y-axis, I want to rotate
it around the line y is equal to minus 2.
So if that's 2, y equals minus 2 should be roughly here.
So I'm going to rotate this area around this line.
So what's it going to look like?
It's going to be a fairly big ring.
Like if I were to try to draw it-- let me see if I can
even make an attempt.
Once again this is always the hardest part, just drawing
what I'm trying to rotate.
I'll try to do it from an upward perspective.
So that's kind of the inner loop, and then there
will be an outer loop.
The top is flat right, because it's defined by y is equal
to 4, so that's the top.
And the inside is also going to be a hard edge.
But then the outside is going to curve inward.
I don't know if you see what I'm saying, because this is the
outside, it's curving inward.
So it's going to be a big ring.
So if I were to draw the axis, this would be the y-axis access
coming in-- no, no, sorry.
Whoops.
The y-axis is actually going to be closer to this hand side.
The y-axis is going to be in the middle of kind of-- so
this is going to be the y-axis coming up here.
And then the x-axis is going to come below that.
I'm drawing everything at an angle as best as I can.
The x-axis is going to come a below that.
And then this line we're rotating it around,
that's going to be someplace over here.
That's going to be something like that.
It's going to go behind there and come back over there.
Hopefully that makes sense.
We're just getting a big ring.
So how are we going to do this?
Well actually there's a couple of things we can do.
First we could just use the shell method,
using the x value.
So how do we do that?
The important thing is to always visualize
the shell or the disk.
So the shell method we're going to take slivers like this,
where the width of that sliver is dx.
I could draw it really big.
So that's our direct angle, is going to be dx.
What's the height going to be of the sliver?
Well it's going to be the top function minus
the bottom function.
It's going to be y equals 4 minus x squared.
So this is going to be 4 minus x squared, the height at
any point right here.
And then if I were to do the shell just like we did
before-- let me see if I can draw a decent shell.
I think I'm getting better at this.
This is one edge of the shell, that's the other shell.
We already figured out that the width of the shell is dx.
The height is 4 minus x squared, the top function minus
the bottom function because of the distance between the two.
And then what's the radius going to be?
What's the radius of that shell going to be?
Well, is it going to be just x?
Is it just going to be x value?
No, the x value will tell you the distance from the y-axis to
that shell, and it's going to be from minus 2 to the e value.
So it's going to be essentially 2 plus x.
That's going to be the radius at any point.
And this is where we diverge from what we've done before.
Before the radius was just x, but now it's 2 plus x.
So what's the circumference of each shell going to be?
Well circumference is equal to 2 pi r, and
our radius is 2 plus x.
So it's 2 pi times 2 plus x, which equals 4 pi
plus 4 pi plus 2 pi x.
That's the circumference.
And then what's the surface area of this?
Well it's going to be the circumference times the height.
So surface area is equal to that.
The circumference 4 pi plus 2 pi x, all of that times the
height-- times 4 minus x squared.
And let's see if we can foil this out or
distribute this out.
So 4 pi times 4 is 16 pi.
4 pi minus x squared minus 4 pi x squared.
2 pi x times 4 plus 8 pi x, and then 2 pi x times minus x
squared, so that's minus 2 pi x to the third.
So that's the surface area of each ring.
And then if we want the volume of each shell, essentially
we multiply it times the width, the dx.
So that's the volume of each shell, and if we want
the volume of all the shells, we sum them up.
So we take the integral, that's an integral sign, and I'm
running out of space like I always do.
And where did I take the integral from?
I take the integral from x is equal to 1 to x is equal to 2.
From 1 to 2.
That's probably too small for you to read.
So let's see if we can take the antiderivative of this.
Let me make some space free just so I don't
have to write so small.
So I'll keep this down here, because that's the set
up of the problem.
I think I can get rid of a lot of this.
I think that is pretty good.
OK.
And let me switch to another color.
And we're going to take the antiderivative.
So what's the antiderivative of this?
So the antiderivative of 16 pi is 16 pi x.
And then what's the antiderivative of
4 pi x squared?
It's going to be x to the third over 3, so it'll be minus 4
pi over 3-- sorry-- 4 pi over 3 x to the third.
Well this will be x squared over 2, so it'll be
plus 4 pi x squared.
And then minus, this will be x to the fourth over
4, so minus pi over 2.
Just divided by 4.
x to the fourth.
I'm going to evaluate that.
This is a much hairrier problem then what we've been doing.
2 and at 1.
So what is it evaluated at 2?
It is 32 pi minus 4 pi over 3 times 8 plus 4 pi times 4 plus
16 pi minus 2 to the fourth is 16 divided by 2 minus 8 pi.
And then I just realized I'm running out time, so I will
continue this in the next video.