# The Disk Method Explained (Using Python)

(D4M) — Here is the video transcript for this video.

00:00
you’ll learn how to use python to find the volume of a solid of revolution
using the disk method we’ll work through several problems so
let’s see how to do this step by step let’s do some math [Music]
hi everyone welcome back i’m dave so we’re going to begin with the
question uh what is a solid of revolution so solid of revolution is a solid that
is generated by revolving a plane region about a line
that lies in the same plane as the region so for example this line is called the
axis of revolution so let’s look at some familiar examples so let’s say the axis
of revolution here is say the x-axis or some horizontal
line and i want to revolve it around and

00:01
i have this object right here this plane region right here and it’s just a
rectangle and so if we if we revolve this around the uh x-axis for example
then uh what we get is a right circular cylinder
so we’re going to end up revolving this area right here this region right here
and when we revolve this around the x-axis then we’re going to get this
right circular cylinder here so right circular cylinder and
you know what if we revolve say a semi-circle so if we have something like
a semi-circle like that and we revolve this around the x-axis
then now we’re going to get a sphere and so i’ll just draw it like that so
now if we revolve this area the semi-circle right here the

00:02
top part of the circle and we revolve it around the x-axis then we’re going to
get a sphere out of all that and then what about if we have a uh
right triangle right here so have like a right triangle right here
and we’re looking at revolving this area reach this region right here around the
x axis right here so this will now be a what what do you think we’ll get here
right so this re this region right here gets revolved i’ll just draw a
circle right there and so we get a right circular cone right there
so this will be a solid cone right here and it’ll be a right
uh right angle cone right there um and so then you know what if we get
something like this which is something that will look in um
in a in the next episode but what about if say your rectangle just isn’t flush
with your axis of revolution what if your rectangle is say floating up here

00:03
right so let’s say that’s a rectangle right there and there’s my region right
there and i’m going to revolve it around the x-axis and so now
what we’re going to get here is a solid and it’s it’s going to
it’s going to be like a hollow pipe so it’s going to because part of this
has been removed so think about it is you know you got that and then you got
an inside that’s been removed there it comes over here and then comes over here
and so we can draw it like this but this part right here has been removed so
just that part in there it’s so it’s been hollowed out the
hollowed right circular cylinder and so there’s some examples there

00:04
of generating some solids by using some of the common shapes that you know of
and so let’s see here uh yeah and so let’s see what’s next
here so next i want to talk about how do we do this what the disk method is
um and so what we’re going to say here is that
we’re going to start with a continuous function non-negative on closed interval
so it’s non-negative right it may touch but it’s not but it’s you know zero
positive right so and let um r be a region that’s bounded above by the curve
and below the x-axis and on both sides right so let’s sketch a you know graph
of that something like this right here so we’re going to be looking like this
right here and we got this function and it’s not negative so it could touch the
x-axis but i’m just going to draw it for
the sake of drawing it here i’m going to
in fact let’s draw it in blue right here so let’s draw this function right here
and we’re going to be stuck between a and b right here so let’s put an a right

00:05
here and let’s put a b right here and chop this off at a b right there
and so you know we’re bounded above um and so we’re bounded above by this below
by the x-axis and on both sides by the line and so
basically this is just the region under the curve us on a to b and we know how
to find this area of this region right here from calculus one right calculus
one shows us how to find the area of that region right there but now what
we’re going to be doing is we’re going to be taking this region right here and
revolving it around the x-axis and we’re going to actually be trying to find the
volume of the solid and not just the area of the region so we’re going to
revolve this around the x-axis here and what we’re going to end up with is
when we revolve this around we’re going to get a solid
so let’s see if we can sketch that so i’m going to put this line here and
then now i’m going to try to sketch this graph right here again
but let’s say over here it’s going to come up a little bit and come down all

00:06
right and so then let’s say this is a right here again and
we have b right here again and revolving around the x-axis and
so what’s going to happen when when we revolve this around it’s going to come
up with the lower part here which is symmetric right here so it’s going to
come up and then it’s going to go about down like this and then it’s going to
come back like that well let’s make it a little bit better
and then come up and go down like that and so what we’re going to get here is
put this in orange i guess so we’re going to you know get this
cylinder right here and this disc right here also and
we have a disc in here that’s going to kind of be used to represent it
and so we’ll just dot this in here and this is an x

00:07
right because this is going from a to b and this will be just an x and this uh
radius right here uh let’s maybe try to put that in red
so this radius right here is the important part to look at so that right
radius right there is f of x that’s the height right you plug in an x and you go
up and you get the height out which is the f of x and so
the area of the cylinder here so the area of the cylinder is what it’s a of x
because it’s going to depend upon x where the x is
right so over here we’ll have smaller cylinders over here we will have a large
cylinder and then it curves in a little bit
anyway so it depends upon the x right the area and it’s just going to be pi
r squared but the r the radius is given by the function height so pi r squared

00:08
and so that will be the area of the uh the cross-sectional area right there and
so that’s what we’ll say here so we’re going to find the volume of the
solid of revolution that’s generated let’s move this f of x down here
so that’s the height at x and that’s generated by revolving this
region right here around the x-axis right there all right so
so observe that the cross-sectional area right here
of the solid taken perpendicular to the x-axis right here um
at the point x is a circular disk and this is the radius of the circular disk
so it’s important to understand that because you know
as your problems uh switch and change you need to be able to adapt these ideas
so this is this is a very nice situation in the sense that
you got one function you got a closed bounded interval

00:09
um if you looked back at the episodes on finding the area between curves often
you have to break your solid or or whatever you’re working with into pieces
into a finite number of pieces and and so you may apply this problem to
different pieces um and so yeah the area of this region here is
pi f of x squared and so the volume is given by by integrating or adding up all
those little small little pieces and so this right here will be the the um
the volume of the solid right here um so actually the volume of the solid b um
this is not the volume of solid this is the area here so the volume of the solid
will be um actually we need to fix this real quick so we’ll just say pi
and then let’s put parentheses in here and raise this up to the power 2.

00:10
so let’s see if we can get this going here so yeah we want to update the
file right here so we want to update this right here there’s the volume of
the solid so it’s pi and then r squared and then so basically what we’re doing
is we’re adding up all the cross-sectional areas and we’re doing
them infinitely small and that and that’s what we mean by using the
integral right there all right so good now some people like to put the square
brackets just to make it a little bit more readable
so we can put the square brackets on that right there
and yeah so let’s put the square brackets on there and there there’s the
nice formula there all right so let’s look at some examples now let’s look at
our first example and so this was what our first example
is going to be here let’s get this out of the way real quick all right and so
now before we do this example right here i’m going to do this by hand but we’re

00:11
also going to do this by python so i want to quickly look at the setup of the
python that we’re going to use real quick before we start doing our first
example here so let’s switch over to a python notebook here now if you’ve never
used a python before there’s a link below in the description
and when you look at the link in the description you’ll get a you’ll be able
to open up a new blank python notebook and you can type up uh what we show what
i show here of course if you already have python set up you know i recommend
using jupyter notebook you can install it install it using anaconda or
something like that but let’s look and see uh what we’re going to do here in
this episode so first off we’re going to
do a setup here and this is the packages that we need to uh import so that our
work will be a lot easier so we’re going to import the numpy package as mp and
we’re going to make some plots so we’re going to import the math plot library as
plt and we’re going to do some integrating so we’re going to integrate the scipy

00:12
package and we’re going to use the integrate method there and i’m also
going to so let’s um hit shift enter on this cell right here so just click in
there and hit shift enter and then i’m also going to customize my
axes so i’m going to make some plots and
if you’ve watched any of the episodes so
far you know that i like to customize my axes so they look like axes that you
would see in a math course and so there’s our setup there and so now we’re
going to start looking at some examples and so here’s our first example right
here and here’s how we’re going to do it in python
but before we do that let’s go see how to do this by hand
and so what we have here is find the volume of the solid that’s obtained in
this region right here so let’s sketch this region right here in fact i’ll
sketch it down here let’s say so i’m gonna i’m gonna have this region
right here and i like to put this in blue so i’m gonna put the square root
function right here in blue it’s just going up like that now i’m looking on
one to four so let’s put here one to four let’s say

00:13
one is here and four is here and i’m going to be looking at this region right
here between one and four and i’m going to be shading this region
in right here this is the region right here that we’re revolving around and
we’re going to be revolving around the x-axis there so when we revolve this
around the x-axis we’re going to get a volume and our goal is to find the
volume of that solid we’re going to get a solid and we’re we’re going to find
the volume of that so we’re going to revolve around the x-axis there
all right so this is just y equals square root of x
and so now let’s do this here so so the volume is going to be
and we’re going to integrate from one to four so one to four
and we need the cross-sectional area so what is the cross-sectional area going
to be if we make a say a disc so we’re going to have a rectangle right here and
we’re going to be revolving that rectangle around here and when we
revolve it around here it’s going to turn into a disc and so we pick an

00:14
x right there and so what is the radius of this disc going to be remember the
area of a circle is going to be pi r squared so this is going to be pi
and then to get the height it’s just going to be square root of x
and then we’re going to square that so pi r squared or pi and then the
height squared and then this will be dx here
and so now we just need to integrate this right here so this will be 1 to 4
and this will be pi x dx and so now integrating this we’re going
to get pi over 2 and then x squared and then evaluate from 1 to 4.
and so we’re going to get 8 pi minus pi over 2
and then we’re going to get here 15 pi over 2 as the exact value for the volume
of the um yeah and so if we were to say make the negative y equals negative

00:15
square root of x right and then we’re going to get you know a solid right there
so that solid uh three-dimensional solid right here and
we could find the volume of that solid it’s just 15 pi over two right there
then you can put your units on that if you’d like all right so now let’s go to
do this in python and so yeah let’s see how to get the region and
get this numeric value here we’re going to get it numerically not symbolically
so here we go we’re going to let’s see if we can zoom
in here all right so again the same problem find the volume of the solid
region under this curve on this interval about the x axis
and so i’m going to define a function it’s going to be the square root of x so
i’m going to use numpy to get a square root of x function
and i’m going to declare a figure and some axes i’m going to customize the

00:16
axes i’m going to lay down a bunch of x values a thousand of them and we’re
going to go from one to four and then i’m going to use the square root function
so for each of these x’s all thousand of them we’re going to put substitute them
into the function and then we’re going to plot the function
and then i’m going to fill between the x and y so our region is shaded
and then i’m going to show the plot and then we’re going to integrate
so we’re going to integrate um so remember we’re using the scipy
package and we’re going to integrate it and we’re going to use this method right
here to integrate it now what is the lambda mean so up here i defined a
function f of x um and that then the name of this function right here was f
so this is how you define a um ambiguous function right it’s just not ambiguous
anonymous anonymous function sorry anonymous function right there and uh yeah so

00:17
we’re going to see here we’re going to get pi and then the radius squared and
the radius is given by the function one to four and so this is exact
integral that we set up before and so this right here will actually
give us the numeric value right here so let’s shift enter that and it sketches
the graph and then it tells us that 15 pi over 2
which was the exact value and here it gives us a numeric value now if you’re
curious about the error how much error it has you can take off this um
0 0 right here that just tells us we want the first entry in the
output and so if i don’t if i don’t have that it then actually gives us the
numeric value and then this gives us the tolerance or the error range right here
all right so there’s our first example there now let’s go and do a second
example so example two here and now in this example right here we’re
going to have um y equals square root of sine x right
here and we’re going to be looking at zero to pi

00:18
and so it’s you know we don’t have square root of x anymore now we have
square root of sine x right so let’s see if we can put it over here actually um
and so let’s see we have it right here and then we have the square root of sine
x and it kind of looks like a flattened sign so it doesn’t look as nice
and round it looks more like it’s been like squashed or something so i’ll just
put it like that and then um the height is here here is still one
and so we’re looking at this region right here this region here r
and then this will be pi right here and we’re going to be revolving this
around the x-axis here so when we get down here we have the same
function right here and it’s just going to look like an egg
it’s just going to be a solid figure right there that’s just generated by

00:19
revolving that around the x-axis there so that looks like a revolving symbol
a little bit better now all right um and so yeah let’s see how to do this uh by
hand so i’m going to say here the volume is um
so we’re going to integrate from what 0 to pi and we’re going to integrate um
pi and then the radius which is given by
the function right here so the radius so here’s like a representative
uh rectangle right here and when you revolve that representative rectangle it
becomes into a representative disk that’s why it calls the disk method and so
the radius of this disk is going to be given by the height of the function
square root of sine x so pi r and then squared and then dx
all right and so then now we just need to integrate this so i’ll bring the pi

00:20
out and then we have square square root so this will just be zero to pi
of sine x dx and then you probably know how to
integrate this already we’re just going to get a 2 out of that right there so
the whole thing will be a 2 pi all right and so that would be the
volume of the solid that was generated from that region right there so
let’s look to see how to do this on python
and here we go so let’s switch this to python now and let’s get this out of the
way here real quick so let’s see here let’s go and appear right here
and yeah so example two here so we got this uh function right here square root
of sine x zero to pi revolved around the x-axis so i’m going
to define a function and it’s going to be the square root of the pi
square root of the sine right so i’m using numpy square root function and i’m

00:21
using the numpy sine function right there so then i declare a axis and a figure
and i’m going to customize axes i’m going to lay down a thousand x values
between zero and between zero and pi i’m going
to lay down a thousand x values i’m going to substitute substitute each of
those into the function f and get the outputs
now here i’m going to set the aspect ratio equal to each other
that way this distance of 1 and the distance along the x-axis
actually looks like three times that one and so then this way you can get a nice
shape uh get a good a nice feeling for the shape of that square root of
square root of sine x all right and then i’m going to fill
between the x’s and y’s so here’s an x and here’s a y so i’m going to fill
between it and then i’m going to show that plot that we just
defined right we defined the plot now we’re going to show it
all right and so here’s we do the calculus part so we’re going to use the

00:22
scipy integrate quad function and we have i’m going to define this
anonymous function of x and it’s going to be so we’re integrating pi times
the function height squared from zero to pi
and i just want the numeric value right there so let’s execute that cell right
there and there we go we get the two pi approximation right there 6.2831
and so on and so there we go so there’s example 2.
so now let’s move on to example 3 right here
so here we go let’s look back at example three now and here we go so [Music]
we’re going to find the volume of the solid and
so we got y equals x to the third this time i’m we’re going to revolve around
the y-axis right there so i thought we should do one of those

00:23
so here i’m going to look at the graph of y equals x to the third here
and let’s do that right here let’s say yeah let’s just put it right here and so
here we go we have y equals x to the third this is coming in there really
nice like that and then we have y equals eight so we’re
coming across here like that so y equals eight and so right here it’s going to
intersect at 2 8. so 2 to the third is 8. so this is a two right here
so find the volume of the solid obtained by rotating this region [Music]
about the y axis so i’m looking at this region right here
and i’m going to revolve this about the y-axis right here
so let me shade that in and then now let’s revolve this about
the y-axis right here so let’s revolve it right here
all right and so then we’re going to get a solid region right here and so that’s
just going to look like we’ll have like this and then it’ll have a

00:24
disk at the top and it’ll have a representative disk
we can draw the representative disk right here like that
so we need to find the area the cross sectional area i’ll give it a little
thickness right here and that comes from revolving a
representative rectangle right here and revolving that around the x-axis or
sorry y-axis and when you revolve that representative rectangle now you’re
getting a disk and so yeah we need to understand the
cross-sectional area and then we need to
integrate to add them up and we’re going to be integrating along the y-axis so
i’m going to be integrating from 0 to 8.
and so yeah let’s try to find the volume so you know what is the cross-sectional
area so it needs to be a function of y now and the reason why is because
i’m revolving around the y axis and so the disk method is
you need your region to be flush with the axis of rotation

00:25
and so i’m revolving around a vertical line now the y-axis and so i need a
something out here to get me the to get me the radius right here so this
will be the radius right here so this will be the point here for example xy and
we can have it input the x and output the y or we can have the
y is the input here um you know we can solve this and say x
you know because this could be the point here x x to the third
in other words just x y or this could be the same point right
here and we can write it as cube root of y y
right and so we could use either way to represent this point right here x f of x
or y f of y in any case this will be the cross sectional area here would be

00:26
pi x squared or pi solving this for x we get cube root of y squared and so
you know this is just pi and then y to the two thirds power
so this will be our cross-sectional area here and then notice the disks are
horizontal this time all right so anyways let’s find the volume so the
volume is we’re going to integrate 0 to 8 of the cross sectional area
and then times with respect to y and so this is just 0 to 8
and this is going to be pi y to the third two thirds and then we’re going to
[Music] say d y so we can integrate this this will be pi and then we’ll get a
uh what power here y to the two thirds plus one so that’ll be

00:27
uh five thirds and then we’ll divide by five thirds and then zero to eight
and so now we’re going to come in here with an eight
uh sorry and this will be a y right here sorry y to the five thirds yeah um and
so we’ll come in here with an eight now when we come here with an eight
eight to the you know the cube root of eight is two
so that’d be two to the fifth so that’s a 32
so this will be three fifths times 32 pi or said differently 96 pi over five
and so this would be how we would be able to find the exact
value of that volume exact volume using you know just integrating out by
hand and so let’s see how to do this in python now

00:28
here we go let’s erase this real quick all right so find the uh volume of the
solid [Music] region bounded uh rotated about the
y-axis right so what i’m going to do is i’m going to define a function right
here f of x is x to the third um again we’re going to do the same
thing we’re going to declare axes and we’re going to customize
and now i’m going to um go here zero to two um
so yeah we’re just looking zero to two right here and so
here when i did the lean spay uh linear space i did a thousand of them
so let’s try that here let’s do a thousand right there
all right so between zero and two i’m going to do a thousand values for x
a thousand x’s substitute them into my function i’m going to keep keep the ax

00:29
aspect ratio equal to each other so that way this distance of 2
is the same distance between 2 here and here
all right and so then i’m going to fill between the x y and the 8
so we get this part right here and then we’re going to show that plot
and then now we’re going to integrate here so we’re integrating um
so we’re going to use some science pi and then we’ll integrate use the quad
method anonymous function of x so now notice i use [Applause]
anonymous function of x because i used um so [Music] this right here is
the squared so this will be x to the fifth here

00:30
so let’s see here what should this be here so um if i’m going to use the f here
um so i’m going to define a new function f of y but not f i’m going to use
g of y and i’m going to return the um so i’m going to return so we’re looking
at uh y equals x to the third right so i’m going to say x is cube root of y so
input a y y and take cube root of it so here i’m going to input a y and i’m
going to take cube root of that so i believe that’s m p dot cube root cube root
of y so this will be cube root of y just like that

00:31
and now we want to use this function so this right here will be a function of y
and we want to do this function right here g of y
and then i’ll square that so that’s pi r squared and the r is given by this
function right here so yeah we want to measure
this height right here horizontally um and so yeah let’s execute that
and so we’re going to get here almost a 6 out
and we’re going to and we’re going to do this from 0 to my bad 0 to 8.
there we go and so now we’re getting uh a 60 right there

00:32
and let’s see here the 996 pi over 5 which was the exact volume so this is um
[Music] 96 pi so yeah there we go so um you know using this python right here
is only going to give us the approximation
not the exact value so if i want to do an example like this derive the formula
for the volume of the sphere of radius r well for this one here example 4
i’m not going to use the python to do that so let’s just do that
by hand here so let’s go back over here and look at that
and let me get rid of this and so yeah this last example is derive the formula
for the volume of a sphere of radius r so to do that what i’m going to do is
i’m going to say okay we have a sphere of radius r so we have x squared plus y

00:33
squared equals r squared and so that i’m just going to be looking
at the top part and so i’m going to say this will be square root of the r
squared minus the x squared in other words move the x squared over and then
take square root of both sides and just look at the positive root
and that will give us the top part of the circle or a circle of radius r
centered at the origin and so let’s just make a good top part
of the circle right there and so this is
radius r so that distance right there is r and this distance right here is r and
we’re going to revolve this around the x-axis
and when we do we’re going to get the sphere
and so the volume of the sphere will be and we’re going to integrate from 0 to r
and we’re going to have pi and how do we get the radius here
so the radius at a given x and the height will be given by all this right here

00:34
so pi and then the radius squared and then dx
so this will give us the volume of the sphere when we revolve this we’ll get
the sphere so we just need to integrate this so this will be
zero to r and then pi and then r squared minus x squared
and then dx and remember r here is just a constant right so a sphere of radius 4
a sphere of radius 16. and so what we can get here is just um actually um
using the symmetry right two and two because i’m integrating zero to r
and and and this is symmetric right or we can integrate minus r to r if we want
minus r to r but i’m going to multiply by 2 to get the full hemisphere here
or if you want you can just do what i was doing and then multiply by

00:35
4 at the end if you want to just actually no that never mind so yeah i
want to just integrate zero to r but i’m gonna multiply it by two and use
symmetry um and so yeah if we integrate this here
this is a dx so this right here is in fact let’s just pull out the
pi right here so 0 to r and then we’re going to get r squared minus x squared
yeah and then we can just integrate this
so 2 pi and then r squared will get an x and this will be minus x to the third
over 3 and then all this will get evaluated from 0 to r right there
and so then if we come in here with an r we’re going to get 2 pi
and then we’re going to get an r to the third right there so r to the
third and then here we’re going to get minus r to the third over three
and then when we come in with a zero that all vanishes so i don’t need to say
minus or i could say minus zero if i want but in the end what do we get here

00:36
so think of this as three over three so this will be three r cubed minus r cubed
so that’ll be two r cubed so this will be two pi
times two r to the third over three and so then lastly we’re going to get
the 2 times 2 so we’re going to get 4 pi thirds r to the third
and so then there’s the volume of a sphere of radius r you give me the
radius i’ll find the the volume of it uh yeah so that’s a lot
of fun uh so you know this is a numerical uh answer here at the end so
we would just do this by hand and you know um and it’s a lot of fun and so
yeah let’s uh i’m gonna say i’ll see you in the next
episode and uh hey have a great day if you enjoyed this video please like
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