BAM!
Mr. Tarrou.
In this lesson, we are going to learn something
extremely cool, we are going to learn how
to find the volume of 3 dimensional objects
using something called the Disk method.
So we are going to be able to learn how to
find the volume of things that are more complicated
than like just cylinders and cones.
We are going to do that or be able to do that
if we know a function which describes the
outer shape of the region and that is going
to set up, what we have here, plane region,
I have got a region that's bound by the x
axis, the y axis and this parabola is defining
the outside curve of this, going up to an
x value of 3.
Now we are going to do this by expanding our
knowledge of the first fundamental theorem
of calculus, where we estimated the area of
a plane region using rectangles.
And we know that as we let the widths of these
individual rectangles approach zero, the number
of them increased until infinity, thus giving
us a nearly 100% accurate estimation of the
area of that plane region.
Well if we allow that to rotate around an
axis, what we would get is a bunch of cylinders
and we can find the volume of a cylinder,
it is pi r^2 times height.
And if we again, as we reduce the widths of
these cylinders, we are going to get a more
and more accurate estimation of the volume.
And that means, we would be able to find questions
like, if I had the function describing the
outside of this light bulb, how much volume
does it have, how much air, actually it is
in a vacuum right..
But how much volume does this glass hold?
How much glass did it take to make this wine
glass or how much does it hold inside of it...volume
wise?
And this is kind of a boring example, cause
this is just a disk or stack of DVD's, it
is thus a cylinder and we can just do pi r^2
times the height, but we could also find the
volume of all these individual disks and add
them up, and thus get an estimation, in that
case, an exact estimation, if I will, of that
volume of that larger cylinder that all those
small little cylinders make up.
So let's see here, the function is going to
define the radius of each of these disks or
white rectangular cylinders and delta x, if
we have a horizontal axis of rotation, is
going to describe the widths of those disks.
Let’s get this out of the way and read formally
what I have said.
If a region in a plane is revolved about a
line, the result is called a solid revolution.
The line is called the axis of revolution
and in this case the axis of revolution would
be...my x axis or y=0.
The simplest region in a plane is a rectangle,
now I have covered up that diagram here but
again the simplest plane region is a rectangle
and those getting rotated around an axis is
going to set up the..
Move my picture in a picture here, resulting
if we rotate those is a plane a rectangle
which results in the simplest such solid...a
right circular cylinder which we will be calling
disks thus we are finding the volume or estimating
the volume of 3 dimensional objects using
the Disk Method.
Alright..let me get this cleared off and get
to the next page of notes.
BAM!
Now how do you find volume of a disk when
you have a whole bunch of them?
Let's put this in some calculus notation.
Volume of the disk, delta v is equal to pi
times r^2 times delta x, we can see that pi
r^2 times height that we had back in geometry.
Approximating the volume of a solid by n disks
of width delta x and radius r(x sub i).
Now in the next screen I am going to give
you some summaries, I am kind of building
up to what we see here is a formula based
off our first fundamental theorem of calculus
or using definite integration.
And we'll summarize based on whether we have
a horizontal axis of rotation or a vertical
one.
Now if we have a function in terms of x and
we see a delta x here, we are talking about
a horizontal axis of rotation and this disks,
if you will, are kind of vertical, or the
heights of these is determined by delta x.
The volume is approximately equal to the summation
of, where we have i starting at 1 and going
up to n, this is the number of disks that
we are of course using to estimate our volume.
So it is the summation of pi R(x sub i) squared
times delta x, and if we want to go ahead
and go right to using those definite integrals
and allowing delta to approach zero, we have
the limit is exactly equal to the limit as
delta approaches zero, again, pi times the
summation of R(x sub i) squared times delta
x where i starts at 1 and goes to n, Now n
approaches infinity, we have a set difference
of b-a, this would be our lower and upper
limits on the x axis, again, we have, again,
a horizontal axis of symmetry.
So we have a set difference and if we let
n go to infinity that is going to drive the
widths of all those disks down to zero.
So indeed as we let delta x approach zero,
we are actually saying that n approaches infinity
thus giving us an infinite number of disks
and basically an infinitely accurate estimation
of the volume.
The width of the largest subinterval of a
partition, delta is the norm which is denoted
by delta surrounded by what looks like these
double absolute value symbols, so we are saying
that our largest partition is going to approach
zero thus all the other partitions, if they
happen to be at different widths, remember
that back when we were estimating the area
of a plane region using an infinite number
of rectangles, well if the largest sub interval
approaches zero then all of them are going
to as well.
And ultimately we get into, sort of what we
are going to be using through these examples
which is pi times the definite integral from
a to b of, again, R(x sub i) squared dx.
And that's going to be pi, basically, r^2,
times the height.
Alright, let’s get a summary statement up
here for when we have a horizontal and vertical
axis of revolution before we get to our first
example.
Horizontal axis of revolution.
This side of the board kind of matches the
notation so far used in the video, the volume
of a solid using the disk method is equal
to pi times the definite integral from a to
b, that’s the lower value along the x axis
and the upper value, or the lower and upper
limits, of R(x) squared dx.
And if we have a horizontal axis of rotation,
then we have perpendicular representative
rectangles with a width of delta x. SO here's
our horizontal axis of revolution and we'll
be drawing those representative rectangles,
again, like I just read vertically and we'll
have a width of delta x or dx.
So we will be integrating with respect to
x or dx.
Now, if we have a vertical axis of revolution,
then the volume is equal to pi times the definite
integral from c to d, our lower and upper
limits of course, or our lower and upper value
of y or R(y) ^2 dy.
Now you see that we're integrating with respect
to y because we have a vertical axis of rotation
as opposed to over here integrating with respect
to x but we have an R(x) and an R(y).
So when we want to rotate a solid around a
vertical axis, we'll have to make sure that
our function is x in terms of y as opposed
over here, our function is y in terms of x.
Now, it would be a good idea to make sure,
look at your picture, draw or get it in your
calculator and look at that vertical axis
of rotation, draw those representative rectangles
and then once you have those representative
rectangles drawn you can see, well are my
widths dy or are my widths dx just to make
sure that you are setting the problem up correctly.
Now r is the distance from the axis of the
rotation and the function.
Our first couple of examples are going to
have either the x axis or the y axis be the
axis of rotation, so it will look like that
we are just taking the function that's describing
the more regular shape of our solid and squaring
it.
But our third example, I believe, in my notes,
we have a solid with a hole in it, so we are
going to need to remember that, like if this
is the y axis, then my axis of revolution
is going to be x=0, so it's going to be the
value of my function minus zero, so it's going
to just seem like I am taking my function
and squaring it.
But maybe if this red line is the axis of
revolution, then as I rotate this plane region
around, I am going to have a hole.
Well that radius, well this is going to be
more complicated example but, I would have
this overall distance here and then I would
have to subtract this distance so you will
need to remember that r is not just a function
describing the outside of your solid squared,
it is the distance from the axis of rotation
and the function.
So we will get more into that as we get to
our third example, but now we need to get
to our first one!
BAMM!
Our first example.
We are going to set up and evaluate the integral
that gives volume of the solid formed by revolving
this region here about the x axis.
And before we get that started though, on
the previous screen, I misspoke a little bit,
our third example is going to have an axis
of revolution that is not the x or y axis,
but it is not going to have a hole in the
solid, we are going to save that for the fourth
example.
SO, we are still going to have to take special
care that we are setting up that r function,
that radius correctly that we are going to
be squaring in the problem.
But it doesn’t have to have a hole in the
solid for you to need to, maybe, take some
special care in setting up that r function.
So we have a parabola opening down.
We see that we have one squared term, so we
have y= -(x-3)^2 +9.
And that's this parabola here.
We also have our region bound by the lines
y=0 which is the x axis, x=0 which is the
y axis, and x=5.
So we got it drawn.
We recognize that we have a horizontal axis
of revolution so I have drawn 1 representative
rectangle which is perpendicular or vertical,
and we see that the width is equal to dx.
So now I am just making sure that I am going
to integrate with respect to x thus my function
need to be y in terms of x and it already
is.
You also want to make sure that you draw the
picture if the book doesn’t give you one
before you go through this process of finding
the volume of this solid that's being revolved
around an axis.
Because, we don't have the issue with this
one, it is the first example, but maybe the
parabola, what if it had crossed through the
x axis and part of the area was below the
x axis and part of the area was above the
x axis.
Well we would have some negative sums and
some positive sums and we would have to find
the total volume of our revolved solid in
two different parts not just one, so make
sure that you get those pictures drawn and
you pay attention to that before you get started.
We'll have that issue come up as we go through
these more and more complicated examples.
There are going to be a lot in this video
and you will find it footnoted in the description.
So the volume of this solid once we revolve
it around the x axis is going to be found
by doing pi times the definite integral from
0 to 5 of now, our radius, our radius or the
heights of these rectangles are going to be
found by taking the y value from our function
and subtracting it with the y value from the
bottom.
Well that's just going to be y=0 and that's
why it seems like we are taking the function
and squaring it, but it is going to be the
function minus 0 and that height is of course
-(x-3) ^2+9 and that is going to be, our function
giving us the radius of each disk or again,
the length of each of these rectangles that
are going to get revolved, squared dx.
Now we have done a lot of finding definite
integrals.
To speed this video a little bit, I am going
to step off and just reveal the solution and
go over them step by step.
Alright.
So we took the binomial and squared it, we
distributed the negative and combined some
like terms and ended up with 6x-x^2 and now
we can go ahead and apply that outside power
of 2, get (x^4-12x^3+36x^2) dx and just going
through the power rule we have, we are going
to increase each of these powers by 1 and
divide.
so increasing 4 to 5 and dividing by 5, increasing
3 to 4 and dividing by 4 and so on... and
just continue it on finishing up our process
of finding the definite integral.
And the volume of this plane region, rotated,
I can't draw the 3 dimensional figures for
all these, I am not quite that good of an
artist but taking this plane region, revolving
it around the x axis, we would get a volume
of 250 pi or approximately equal to 785 cubic
units.
Now what would happen if we were taking a
function, I am not going to do the same one,
but taking the function and revolving it around
the y axis thus giving us horizontal representative
rectangles?
Let's see what that looks like.
So our second example- find the volume of
the solid formed by revolving the region bounded
by the function y is equal to x raised to
the 2/3 power, x=0 or our y axis and y=3 and
we are going to revolve that about the y axis
or revolve it around the line x=0.
So I have got it drawn, we have an intersection
point over here of (0, 0) and the function
stays to the right of our axis of rotation
through the entire, I guess if you will, the
entire domain or the range of values 0 to
3, so this is just going to look a little
but different that the past example but it
would really not be that much more complicated.
All we need to do is, because we see that
we have drawn a representative rectangle perpendicular
to the axis of rotation, and we have a delta
y so we are going to integrate with respect
to y which means that we need this in terms
of... x in terms of y and not y in terms of
x.
So we are going to take this equation, raise
both sides of it to the power of 3/2 so that
we can have these powers to cancel out giving
us x=y^3/2.
Now the volume is going to be equal to pi
times the definite integral, our lower limit
on y is 0 and our upper limit defined by the
problem is 3 times r(y)^2, r(y) the radius.
Okay so the radius, how long is this orange
rectangle..well it's equal to, we are going
to take the, well we are going to plug in
y and get the x values from the function and
we are going to subtract that with, or the
distance is going to go from that x value
back here to zero.
So it's x minus 0.
And x is defined by y^3/2 power.
So it's just simply going to be y to the 3/2
power squared dy.
And that means we are going to take y to the
3/2 power and square it.
That's going to give us pi times the definite
integral from 0 to 3 of y^3 dy.
Then we are going to just use the power rule
and integrate this and get pi times 1/4 y^4
power again with those limits of 0 and 3 and
applying that we get pi times 1/4 times 3^4
minus 1/4 times 0 ^4 and of course putting
in our lower and upper limits but that's just
going to be 0, just writing this for teaching
purposes and this comes out to be 3 to the
4 power is 3, 9, 27..
81 over 4 pi cubic units.
And
that would be the volume of this solid bounded
again by this function y = x^2/3, y=0 and
y=3. And that would kind of look like a funnel,
if you will, coming down vertically to a point
at the bottom. One last example in this part
of the video, where we have solids that don’t
have holes in them and it will have an axis
of rotation that is not an x or y axis. Third
and last example of finding the volume of
a solid that is created by rotating, basically,
just one function about a line so there's
no hole in it. Find the volume of a solid
formed by revolving a region bounded by y=
square root of x, y=0 and x=4 about the line
x=4. Okay, well I have got our function drawn,
it’s just y= square root of x and the square
root of 0 is 0 and the square root of 4 is
equal to 2. So let's go ahead and fill that
in. And as I draw my representative rectangle
which is perpendicular to the axis of rotation,
I see that my rectangle is actually laying
on its side and so we are going to be integrating
with respect to y and not with respect to
x. So we need to take this equation and turn
it around and get in x in terms of y. So that's
not going to be too difficult, we have y=square
root of x. We are going to square both sides
and get x=y^2, And that means we also need
to, we needed that upper limit of y there,
so we have again, a zero down here and a y
value of 2 so we are going to be finding the
definite integral between 0 and 2 and that
means that the volume is going to be equal
to pi times the definite integral from 0 to
2 of, well I have kind of hinted all along
we need to put some more thought into you
know what is this R(y) , this radius. Well
my axis of rotation is way up here at x=4
not x=0 or y =0, so how long is this rectangle?
Well no matter how far up the y axis I go
and my y is now sort of my independent variable,
I am going to put the y in and find x, well
once I go up the y axis, my largest value
of x, this axis of rotation, sort of the base
of this rectangle or the right side of this
rectangle is 4 units away so R(y) is some
kind of value of 4. But my r doesn’t go
from x=4 all the way back to the y axis, it
stops, Well what’s this distance? Well I
am putting in values of y and getting values
of x so this distance is y^2. So from x=0
to my axis of rotation I have 4 units and
this little piece here is y^2 so the length
of my rectangles are going to be 4-y^2 And
that is my r(y), that is my function that
is describing the length of each of these
rectangles or the radius of each of these
disks that I am going to use to estimate the
volume. So it's going to be 4-y^2, yes squared,
dy. And now again, I have done a lot of examples,
let me just step off and reveal the solution
one step at a time. Alright, well it looks
extremely similar to our previous problems,
I just took this binomial, squared it, used
the power rule to integrate, finish up the
process of finding the definite integral and
the volume of this solid formed by taking
this region and rotating it around the line
x=4 comes out to be approximately 53.62 cubic
units. Now so far 3 examples have not had
any holes in our solids after the revolution
about the axis but let’s just say we wanted
to revolve this region about the line of x=6.
Well we would have a hole in it, it would
kind of look like a doughnut. So let me get
this cleared off, let me give you a set of
notes about how to approach those problems
if the axis of rotation isn’t really sort
of one of the boundaries of our region and
finish up that example. And I am actually
going to give you 2 more, the last one is
going to be quite complicated algebraically,
there will still be some polynomials but quite
the irregular shape. Alright I am looking
for the remote but a different camera, there
isn't one. Alright let’s get to those notes.
So as we can see in our notes here, if you
get your diagram drawn, you look at the bound
region and the axis of rotation. If you notice
there's going to be a hole in that volume
of revolution then we are going to let r represent
the inner radius and the R represent the outer
radius and then I have this drawing set up,
it is not the region that we are about to
find in our first, well not our first, but
our next example. But we do have a delta y
or a dy because we have a vertical axis of
rotation. So if we do have a vertical axis
of rotation, the volume is going to be pi
times the definite integral from c to d of(
[R(y)]^2 - [r(y)]^2) dy and what you can see
with each of these disks..these infinite number
of disks that we are going to use to find
the volume of our solid of rotation..or our
solid..that we are just going to basically
be finding the volume of each of the disk
and coming through and cutting this center
out with the subtraction right there so that’s
what the formula looks like. Let’s see what
it looks like in practice. So now we have
our notes out of the way, let’s get back
to that previous example. Everything is going
to be the same except our axis of revolution
is going to go from x=4 to becoming x=6. And
this time, I went ahead and tried to draw
the revolution of the solid so we can see
that hole going through. Or you are starting
from scratch, you draw the function, you draw
the boundaries that you are trying to encompass,
you try to put the axis of rotation and you
go ‘wait a minute the axis of rotation isn’t
a part of my plane region, so again I am going
to have that hole’. Once you recognize that,
don’t forget the area of a solid using the
disk method is pi times the definite integral
from c to d of [R(y)]^2 - [r(y)]^2 dy in this
particular case cause we have a vertical axis
of rotation. So, we need to figure out..our
board is the same so our lower and upper limit
are still going to be 0 and 2, remember I
just moved the axis of rotation. I can’t
find a disk with a hole in it..I need to find
the volume of the larger disk and then subtract
out the volume of the inner disk. So I can’t
just say well how long is this orange rectangle,
I need to find R.. the function that will
give me R(y), so as I work my way up the y
axis, my axis of rotation is x=6. So initially
this rectangle before you take out the center
area, has to reach from x=6 to wherever the
function is. So it’s going to be from x=6
and then minus, I want to take out this section
right here as I work my way up towards 2 and
that distance is found by well x=y^2 So that's
this little piece right here.. very similar
to the last problem. Except before with the
axis of rotation being at x=4 that radius
was 4-y^2, well now my axis is at x=6. So
it is 6-y^2. And then my inside radius, the
part that we want to subtract out, well that’s
just a constant difference of 2 but again
thinking about working my way up the y axis
and then seeing these distances away from
where x=0 basically how far is that, well
that's a distance of 6 and then I want to
take out this distance of 4 to give me a constant
inside radius of, in this particular case
that we have, of 2. Now that I have my two
radii set up, now again from my axis of rotation,
my large outside radius and my inside radius,
I can finish filling up my formula here which
is going to be R(y) which is going to be [6-y^2]
squared minus my inside radius which is a
constant value of 2^2 dy. And now I am just
going to step off. If you want to try this
on your own, but I am going to reveal it step
by step so that you can too! nanananana....
Alright, so I took that binomial, I squared
it, I combined like terms, I went through
the process of finding the integral I did
the definite integral and came up with the
volume of this solid which is equal to 192/5
pi cubic units or approximately equal to 120.637
cubic units. If you are still watching, I
got one more example, it is going to have
some functions which are intersecting within
the lower and upper limits of which we are
trying to find the volume so we are actually
going to have to find 2 definite integrals
or 2 different volumes and add them together
and that’s going to be the increase of difficulty
in that particular problem if you are still
watching. I certainly hope so. BAM! BAM! Alright
let’s get this last example finished. Find
the volume of the solid generated by revolving
the region bound by the graph of the equations
y=x^2+2 and y=-x^2+4x+8, and x=0 and x=4 about
the x axis. So that means if we are revolving
around the x axis, we are going to have vertical
representative rectangles and we are going
to, again, have a lower and upper bound on
the x axis of 0 and 4. I see that I have a
parabola basically, this is a pretty basic
one, it just has a vertical shift of 2 so,
and I am going to draw a better sketch here
in a second, but we have a parabola opening
up with that one squared term and positive
leading coefficient and this sounds like a
bit more complicated parabola, so if I just
pull this out on a side, I am going to get
that into standard form y-8, moving this over
to the left, is equal to -x^2 +4x. You have
to complete the square to get the parabola
in the standard form, so we are going to factor
that negative out and I am also going to leave
some space because I know that I am going
to need to complete the square. Okay we are
going to take half of that 4 and square it.
Half of 4 is 2… squared is equal to 4. And
now we have added 4 to the right but haven’t
really added 4, there's a negative out front,
so we actually sort of added a 4 but there
is a, it is not really an addition of 4, it
is a subtraction of 4. So we end up with y-12=
negative, now this is a perfect square trinomial
of course, we know that. Well there's a power
of 2 on the outside there. So we have y-12
= -(x-2) ^2. So we have a parabola opening
up that’s got a vertex at (2, 12) and its
opening down, well that means that we are
going to have to find a region here, bound
between 0 and, well is it bound between 0
and 4? Or is there an intersection point somewhere
within this interval and actually going to
give us 2 separate problems because if these
graphs intersect somewhere between 0 and 4,
remember finding area bound between functions?
You have to do the greater function minus
the lesser function and then they swap places,
or they might swap places if they intersect
before we get to x=4. So we need to find out
where these graphs intersect and to speed
up the video a little bit, I am going to set
these equations equal to each other and show
the work to see if they intersect between
0 and 4. So as you can see by our algebra,
we have set the equations equal to each other
and got to a point where they intersect, these
functions intersect at a value of 3 and -1.
And well 3 is the one that is the intersection
value that is between our boundary of 0 and
4. So that's really all I am concerned about.
I have done a better job of drawing our sketch
and indeed we see that our function- this
downward opening parabola -x^2+4x+8 is the
greater function or the outside radius from
0 to 3 and then they switch places. So we
are going to have to find a volume, that is
created by this region rotated around the
line y=0 or the x axis and then we are going
to have to find another volume of this region
and that region is going to be done by seeing
that our function x^2 +2 is the outside radius
or the inside radius is the other function.
So we are going to have to find 2 different
volumes and add them together. So, let’s
get this erased. And we have got volume 1
or v sub 1 is equal to pi times the definite
integral from 0 to 3 of our outside radius,
now again our axis here is y=0, so it's going
to be whatever y value we get from our outer
function or from -x^2+4x+8 minus y=0. So it's
just going to be that function squared. So
we have got (-x^2 +4x+8)^2 minus that inner
radius which is going to be that inner function,
and again we are referencing it from y =0
so it’s going to look like it is just that
function being squared. And this all of course,
integrating with respect to x because we have
a horizontal axis of rotation. So let’s
step off, step out and reveal this solution
one step at a time. Alright, as we saw these
graphs did intersect tat hat value of 3, so
we have a definite integral from 0 to 3 giving
us a volume after the rotation of 848.23 cubic
units approximately and then setting up another
definite integral from 3 to 4 we have another
volume, which is approximately equal to 362.33.
Giving us a total volume of 1210. 56 cubic
units. Hope all these examples helped. I am
Mr. Tarrou. BAM! Go do your homework.
