Shell method around a non-axis line - Mathematics Video Lecture - Engineering Mathematics

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.