Find Volumes Using The Washer Method (Step-by-Step)

Video Series: Applications of Integration (Complete In-Depth Tutorials For Calculus)

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

00:00
in this episode you’ll learn how to use python to find the volume of a solid
using the washer method hi everyone what’s up let’s do some math [Music]
hi everyone welcome back we’re going to begin by talking uh about what the
washer method is um and um you know before we start
though i wanted to mention that this episode is part of the series
applications of integration complete in-depth tutorials for calculus
so in a previous episode we talked about the washer method
and this time in this episode i want to emphasize python programming language
but before we do that i want to just briefly recall
or you know talk about what we talked about in the last episode so we’re going
to say a of x is the cross-sectional area of a disk and we’re going to have

00:01
an outer radius and an inter radius so let me sketch the graph and explain to
you what i’m uh what i mean by this right here so let’s say we have a or
axis right here so let’s say here we got this and this and let’s say we got a
function coming through here let’s use blue for the upper function right here
so let’s say this is f of x here and then we’re going to have a lower curve
here g of x coming through here so that would be g of x right there
and then we’re going to be talking about a point right here and so we got a
boundary right here x equals a and x equals b
and what we’re going to do is we’re going to revolve this around the x-axis and
so we’re looking at this region right here which will shade in orange right here
just real quick and now you know when we revolve this region around the
x-axis what we’re going to have is a volume we’re going to have a solid we’re

00:02
going to find that volume of that solid so
when we pick an x value right here let’s just say we pick an x between a and b
and so what we’re what we’re looking at right here is this distance right here
um and so you know we have y equals f of x right here
and y equals g of x right here and so we got these two two heights right here
and we’re looking for this distance right in here
so we’re going to call this r of x right here the distance from the x-axis here
this is called the outer radius right here and this distance right here is
called the inner radius right here and so we want to find that um
you know we want to you know if we think about this right here we can go to a b
pi and then this will be r of x squared and this is uh what we did in the last
episode where we talked about the disk method and this will be dx

00:03
and then we can integrate and you know have the inner radius so we’ll have minus
the integral from a to b pi and now we have the inner radius right here
and so that’ll be squared and so this is similar to
um in that episode where we talked about finding the area between two curves
we had the difference of two functions but now we’re having the difference of
two cross-sectional areas and so basically what we’re doing is
we’re using the disk method to revolve and get the full solid and then we’re
going to also do the the inner one right here and then we’re going to take away
those solids so think about it is this is the solid of the
full object and then we’re going to take away the the solid on the interior here
like we hollowed it out and so that right there is the idea behind why
we have a difference of two squares right here
so this is the cross sectional area here

00:04
which goes all the way from here to here
then and then we have a a representative rectangle and that representative
rectangle is revolving and that gives us a disk when you
revolve it around the x-axis so we get a disk right here
and the radius right here is is the r of x right there
and then we’ll do the same thing right here we have a representative
and then we have a disk coming in here and we have a and this height is r of x
and this full height right here is capital r of x so that’s the capital
r of x so we have the inner radius and the outer radius so this distance right
here in red is r of x minus um yeah so anyways um
so when we revolve this let’s see if we can sketch a diagram of that

00:05
let’s try to put it right i’ll just draw a little little doodle over here
and we got this coming in here f and we got the g coming in here and we got this
f coming in here it’s supposed to be symmetric and we got this coming in here
and this coming in here and then this right here is coming in here like this
and then we got this little part right in here and this part right in here
and so we’re removing this inner um solid right here where we’re moving
that out and that’s why this minus comes in here because when we revolve this
around here uh instead of having disks what we’re
going to be using is the washers so we’re going to have this right here
the capital r of x and then and then we’re going to have the little r of x
here’s inner and outer radiuses okay so i want to kind of help you understand

00:06
that all right and so now let’s look at our first example right here
now when we look at these examples right here
i want to emphasize in this episode using the python method
because in the previous episode we did this by hand here and so you know
we can look at that i want to set up the python first here so let’s go to a
python notebook now if you’ve never used python before
then what i want to say is that there’s a link below in the description and you
can open up your first python notebook just by following that link in the
description now once you open up your python notebook
then what you’ll need to do is to click in a cell and type this up right in here
these are the packages that i’m going to use in this episode so we’ve used all
these packages right here before in previous episodes uh we’re using the
numpy package and the plot package because we’re going to we’re going to

00:07
sketch the region that we’re trying to revolve and we’re going to use the sci
pi and we’re going to do some integration and as i previously said
previously said in all the other episodes hey where i’m going to
customize my axes to make this look nice so let’s execute those two
cells right there after you type in that
right there exactly as it appears on the screen
all right so then now we’re going to go down here to the washer method and
here’s going to be our first example use the washer method to find the volume of
the solid generated by this line and this curve right here
and so yeah we did this one right here in a previous episode so i want to focus
on the python right here so what i’m going to do is i’m going to
say this is 2 times square root of x and that’s going to be my function f so i’m
defining a function f and i’m going to use the numpy package
so i’m going to do mp dot square root so it’ll be 2 times the square root of x
and then for the other function it’ll just be the x the identity function here

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and so i’m going to call that g so g of x just returns an x
and then now we’re going to make a figure we’re going to
make some axes i’m going to customize my axes
then i’m going to create a bunch of x values
and i’m going to be looking on 0 to 4 so i’ll explain y 0 to 4 and i’m going to
make a thousand of them and for each of these x’s i’m going to plug it into the
function right here the square root function i’m going to plot that
and then i’m going to make some y’s from the x function
from the g function and then i’m going to plot that
and then now we’re going to fill between the x the y 1 and the y 2 to get our
region and i’m going to color it a certain way
here put an alpha channel on it so you can see what i’m pointing at here and um
yeah so then after that we’re going to show the plot
and then we’re going to actually do the calculus part right here
so here’s a sketch of the region right here square root of uh two square roots

00:09
of x and here’s the line uh y equals x and we’re shading in between here here’s
our region and now why four four well because that’s where our intersection
points are so where are these two functions equal to each other right
so when x is four um you know you you can see that that
works right there you get uh when x is four you’re going to get 4 equals 4. so
let’s see here we’re going to integrate we’re going to use psi pi and we’re
going to integrate and i’m going to use the quad method
here now if you haven’t seen this lambda before all that means is that is an
anonymous function so here is a function called f here’s a function called g
and we could to go define this function down here but i’m just going to type it
up anonymously so i use the word lambda and my variable is going to be an x and
so here’s the function right here and so this is just pi times f of x squared

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minus the g of x squared and then i’m going to be working on zero to four and
i’m going to be interested in the in the numeric value for that right
there so i’m going to execute that cell right there and we see we get out 33.5
right there and so there is the um value of the
volume right there when you revolve this
around the x-axis right here so we’re in this problem right here we’re revolving
around about the x-axis right there so to set this up and solve this by hand
we would do you know volume equals and then this would be 0 to 4
and then we would have 2 square roots of x and we would have that squared
and so i forgot my pi so pi and then uh two square roots of x and that squared
and then we would have minus and then a pi and then we would have an x and that

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would be squared and then we would have dx so this is according to the um
you know washer method here where we’re going to use washers right here okay so
what we’re going to do here is just simply integrate that
and when you do we’re going to get pi and we’re going to get
2 x squared minus x to the third over 3 and then we’re going to go from 0 to 4
and then when we finish this out we’re going to get 32 pi over 3
and this is going to be the exact value right here
so this is the exact value of the volume that we’re going to get
when this right here this object right here is revolved around the x-axis so i’m
going to be revolving around the x-axis right there
and when we do that we’re going to get an object it’s going to be closed down
here and it’s just going to be kind of opened up here and it’s going to have

00:12
some interior almost like a curved shaped cup or something like that
all right so there’s the difference between the
by hand you know you get the exact value and the numeric value now if you didn’t
use a 0 right here then what you get is a an array
and you’re going to get the first entry is the numeric value which we just had
and then we’re going to get the um you know this is really small this is this is
exponential scientific notation right and so this has got a lot of zeros on it
this is really close to zero and so it’s telling you that this is very accurate
right here all right so let’s look at a second example here
um so let’s go here and look at that right here so on this one right here
we’re going to use the washer method to find the volume of the solid generated

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so we have y equals x right here uh oh sorry that’s the previous one
sorry example 2. so we’re going to look at y equals 2 sine x
and the x axis and we’re going to be working on 0 to pi and we’re going to be
rotating around the line y equals -2 so in the previous example we did the
x-axis which was y equals 0 and so now we’re going to do
a different horizontal axis uh just to see how this works right here
so um let’s do this out by hand here if we can first before we go to python um
and so i’ll just you know go over pretty quickly because we did this in the
previous episode so just want to refresh
your memory how to do that but we have a sine function going right through here
um so this will be y equals 2 sine x and here the height is you know a 2 right
here and then this is the line right here so this is the x axis

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and this is the line right here y equals what minus 2
and we’re going to be revolving around this axis right here this axis right
here so we’re going to be taking this region right here let me shade it in
orange really quick we’re going to be taking this region right here and
revolving around the line y equals minus 2 here and so when we do that
we’re going to get a object that’s been hollowed out
like someone drilled through it or something and so you know this right here is
the origin right here and this is the height of two [Music]
and so you can kind of get a feeling for what this shape right here is it’s just
sine x but it’s been scaled a little bit
in any case we’re going to be looking on 0 to pi so 0 to pi
and we’re going to revolve this and so what happens when it gets revolved so

00:15
this is 0 to 2 and then this is -2 right here um so
you know this is going to get revolved around here so this is going to
look something like this right here we’re going to need to make this be a 4
right and so we’re going to come down here
this distance is going to be a 2 right here so i’m going to need a 2 let’s put
a 2 first and then we’ll put the 4. so let’s put a 2 here and then we’ll put a 4
and so now i’m going to have the shape right here but upside down let’s see if
i can do that it’s just going to kind of go like that
and then this will be the line uh sorry this would be the line here
so this distance is 2 just like this distance is 2. so this will be minus four
and then this will be minus six right here
and so this right here it’s right here at the zero and the pi right there and
so when we get when this revolves around y equals minus two we’re getting this

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object the solid that’s been hollowed out here in the middle right there
so just so that you can kind of kind of visualize it there this will be a minus
4 and that will be a minus 6 right there all right so now we can visualize it a
little better we want to identify the outer radius and the inner radius so the
outer radius is going to be given by 2 plus 2 sine x
and the inner radius is just going to be given by a2 right so the outer radius
is um the distance from here to here let’s go ahead and put it in red right
so we’ll pick an x right here and we’re looking at this part right here right
and so we want to say what the r of x here so the r of x
is this distance all right here so it’s going to be this distance right here 2
and then come up to the height of the function which is 2 sine x so that’s why
that’s the outer radius there now the inner radius is just this

00:17
distance right here so i’ll call this right here if i can make this symbol here
it’s not very good um there that’s close
enough r of x is that distance there how about i just use arrows yeah so this
distance right here is r of x that’s the inner radius
and this distance right here is the capital
that’s the capital of the outer radius right there
all right so let’s find the volume using
the washer method so we’re going to have a
i’m going to say 0 to pi and then we’re going to have a pi and i’ll i’ll say pi
i’ll do brackets and then pi and then we have our
r of x squared so 2 plus 2 sine x so all that squared
minus and then pi and then the inner radius which is just 2. so this will be
a 2 and then a squared and then all of that d of x so

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it’s not so bad when you’re first starting out and doing it like that that
you write it in a form that which you can make sense of it um
so here i have all of this this is everything that i’m integrating and so
this is pi r squared minus pi r squared of course if you’ve done you know 50 of
these you can start factoring out the pi on your first step or whatever but
anyways we’re gonna have a pi uh zero to pi and then we need to expand this out
right here and then we’re going to have a minus 4 pi right there
and so we’re going to or let’s say a 2 squared a 4 and then a minus 4
right in any case when you factor that out and simplify it we get 8 sine x
plus 4 sine squared x and after you do the
complete work you’re going to get 2 pi times pi plus eight
so we filled in the details in the previous episode i recommend checking

00:19
that out but any case there’s the exact value right here
and now what i want to show you is you know how to use python
to get a numeric approximation for that um and to get a plot of the region there
and so yeah let’s go back to python let’s go back to our notebook here
and let’s get rid of all this but you know over here
let’s go ahead and put the exact value so we can just remember what that is
right there 2 pi times pi plus 8. and so yeah let me get rid of
this real quick and we’ll look at the python right so here we go so um we’re
looking at two sine x x axis zero to pi and we’re revolving around minus two
here all right so i’m going to make the function two
uh times sine x so that’s our function right there
and i’m going to make a plot some axes i’m going to customize my axes and then

00:20
i’m going to make up a bunch of x values between 0 and pi
and i’m going to make up a thousand of them i’m going to plug each one of those
into f and then i’m going to plot it and then i’m going to fill between x and y
and there’s my plot right there all right so
there’s my plot right there and it’s just the sine x plot from zero to pi um
and so there’s the region right there now this plot right here doesn’t show
you all the details that we had on the on the when i did it by hand so um
[Music] you know doing this by hand is is a lot perhaps better in this example
but i mean you can we can tack that on here and sketch it out but we already
did that let’s just get on to the calculus part so we’re going to use psi
pi we’re going to integrate we’re going to use the quad method we’re going to do
an anonymous function here what’s the formula for finding the
for the washer method right for finding the volume so the formula for the washer

00:21
method is uh pi times and then we’re going to have 2 plus and
the two plus comes from the fact that we’re revolving around the line y equals
minus two so we’re with the our axis of revolution has been shifted
so we have two plus f of x and then all that squared and then minus
and then here we have the um [Music] the why do we have minus two here um
yeah so minus two or two either way get the same numeric value but
so minus and then we’re going to use a 2 for that distance right there and then
squared and then we’re going to integrate from 0 to pi
and then i’m just going to be looking at the numeric value right there
all right and so there we go and so there’s where we get the
70.005 or something like that pretty close to 70 there

00:22
all right so there’s uh example two um now let’s look at an example three
and let’s go back over here and look at this one right here
so we’re gonna use the washer method again and we’re gonna find the volume
bounded by the graph two sine oh no that’s the same one again my bad all
right so now let’s look at a vertical axis
so this time we’re going to be revolving around a vertical axis so instead of
revolving around horizontal now i’m going to revolve around a vertical axis so
what does that look like here so now i’ll try to sketch the graph
or show you what that would look like over here
and so now let me draw this one pretty straight here
and we’re looking right here and so now i’m going to say i got a function
coming in here now when i say function i don’t mean
a function of x i mean a function of y so it could come in something like this
right here and then come out and this one right here just kind of come in here
like that and sweep through there and let’s call this one here a function

00:23
of y so let’s call it x equals w of y and let’s call this one right here
x equals let’s just call it v of y here and now we’ve got bounds along the y
axis so this right here will be a bound and you know what let’s just bound it up
and just just look at this part right here we don’t need to sketch it all the
way down there and so this this we’re going to say is a c
and this is going to be a d and so that’s going to give us our region right here
that we’re going to be bounded in between and let me shade this in orange right
here as i’ve been doing so far so going to be taking this region right here
that’s defined using these horizontal lines and these functions of y
and we’re going to revolve this around a horizontal axis now sorry a vertical

00:24
axis now and we’re going to choose to do this around the y axis so we can see
that way first all right so there we go and when we uh revolve this around the y
axis now we’re going to get a solid we’re
going to get a solid and we’re going to revolve this around there so what would
the solid kind of look like so let’s call this here x equals
this one right here is just x equals v of y for this curve right here
in any case um yeah so here’s how we do the washer
method when we’re going to be using functions of y
so we’re going to be looking at the cross sectional area which will be r y
uh for the outer radius and our roi here for the um

00:25
yeah so this right here should be um a lower case r and so yeah all right
so i was just fixing that right there in any case yeah so here’s going to be the
r of y right here that distance right there that’s the ry
and this distance right here is the um little r of y the inner radius right
there all right so i like that all right so this is how we’re going to
set it up and what we’re going to end up with is this um
solid um it looks something like this right it comes in here like that so it
comes in here like that and then it has this one right here like this one right
here and then um it has the outer one here let’s maybe
make it a little bit more curvier like that and then like like this right here
something like that and then and so at the bottom here it’s going to have a

00:26
it’s going to have a bottom because it’s chopped off at the bottom right there
and it’s chopped off at the top right there and so let’s make this rounded right
here and looks like a vase except for the fact that
we got some separation right here so what’s going to happen is we’re going to
have a um we can draw a smaller one right here
and that’s the sides of this is going to be given by this function right here v
of y and so we can just say it comes in like that and then it comes in like that
and so we’ll have some smaller thing there and so what we’re looking at here is
this think of it as like a vase or something but then it’s been hollowed out
so it’s really solid but then it has this uh region in here right there

00:27
all right so um yeah so here we go let’s look at an example here we go
so find the volume of the solid that results in the region enclosed by the
curves x equals 1 minus y squared x equals 2
plus y squared y equals 1 and y equals minus 1. so let’s look at that region
right there really quick if i could sketch a graph by hand and
so we have this um one minus y squared so that’s going to
come in here and look like this and so it’s like a parabola but sideways
and it’s got a minus right here so it’s opening up to the left and this right
here is going to be opening up to the right and so it’s going to be looking like
you know something similar but and so we’re going to have a
horizontal line right here at y equals 1 and a horizontal line right here at

00:28
y equals minus 1. and so what we’re looking at here is this region right here
we’re going to shade this in right here and we’re we’re going to take this
region right here and revolve it around the y-axis right there and so this is
going to use the washer method not the disk method because this is not flush
with the axis of rotation so this is the y-axis right here and we’re
going to revolve it around the y axis right here and so how do we do that right
so we need to pick a y and think about what the r of y is right here
what is that distance right there in red and what is this distance right here so
this distance is little r of y and the big distance right here is capital r y

00:29
and so what we’re going to do is we’re going to say r of y is 2 plus y squared
right so we got a function of y right here and our lowercase of y is the
1 minus y squared and so we’re going to use the washer
method here to find the volume so the volume will be pi
and we’re going to integrate here from what to what um so
a little sloppy right here wasn’t i let’s make that a little bit better
right there this is going to come down here and meet us right here there we go
and so we’re going to be integrating here this is a height of minus 1 right
there y equals minus 1 and this is a height of 1 right here so we’re going to
be integrating from minus 1 to 1. and all right so i pulled out my pi
already so i need my outer radius 2 plus y squared and then

00:30
we need to square that and then i have 1 minus y squared and we
need to square that and then all of that right there so
close off the bracket and then we have d y right there
and so there’s how we would set that up um and then one intermediate step for
you i’ll put here will be three plus six y squared d y and then
show some steps and then we get 10 pi so we already did this on a previous
episode so we got 10 pi when we’re finished so let’s look at uh how to do this
using python now so pause that right there if you need to
look at that anymore but anyways let’s look at the python there so here we go so
oh yeah let me get rid of this really quick so example three

00:31
and we’re going to be looking at the python right here
all right so here we go we got the x and the x and the y equals 1 and the y
equals minus 1. it’s revolved around the y axis
now to do this i needed to break the up into pieces because
if we tried to solve this for y you know i mean if you just look at this
right here it’s just a it’s not a function right it’s not a function of x
but we could break it up into two functions of x so we could break up this
piece right here as a function of x which i’m calling f up
and we can look at this function of x right here just this lower branch right
here and we can call that a function of x which i called uh f down
and so that’ll be positive the square root so i’m just moving the y
over and then moving the x over and then taking the square root of that so in
other words just solving this for y right so it’s just like

00:32
y squared equals one minus x and then taking square root
and you get plus or minus and so one for the plus one for the minus
all right and then we’ll do the same thing for the g for this one right here
we’ll do the same thing we’ll move the y squared over here we’ll move the x over
there multiply through by a negative in any case we get the g up and the g
down because this function right here is a parabola that opens up this way
and so we got the g up and the g down function right there
so there’s the four functions that we’re going to use
and for each one of these we’re going to um well before we do each one we’re
going to we’re going to define a figure axes customize axes and set the aspect
ratio so for each one of these the f up the f
down the g up and the g down functions i’m going to make a bunch of x’s
i’m going to plot the x’s and the y’s and then i’m going to fill it and i’m

00:33
going to do that for each of them one two three four
so once we do all four of those then i’m going to fill in between
so those four does this part this part this part in this part and then the last
part is i need to fill in between one and two
and i need to go to a height of one and minus one to fill that in there oops
and so that will give us the whole region right there in yellow that we’re
going to revolve around the y-axis so there’s how i shaded that in
with python and plotted that function right here here’s f up here’s f down and
the same for g over here okay so now let’s integrate this
so we’re going to integrate from minus one to one there’s our bounds right there
and what are we going to be integrating so we’re going to be integrating a
function of y and so here’s the function right here is

00:34
this is the outer right here so it’s 2 plus and then the y squared right here
so that’s the um r of y and then minus and then this is
the lower the lower radius the inner radius um and so this was 1 minus y squared
and then we’re going to square both of those and multiply it by pi
so we’re using the cross-sectional area right there
all right and so then yeah let’s go ahead and execute this cell right here
and then that’s where we get the uh 10 pi or 31.41 so on
all right so very good there’s a good example right there for revolving around
um horizon vertical axis so let’s do one now that is a little bit
different so let’s go back to and look at the next example here
so on this one right here we’re going to be um revolving um y equals natural log

00:35
and the y axis and we’re going to be looking on this interval here
and we’re going to be revolving around the line x equals minus 1. so
this will give us another example where we’re doing the washer method and
revolving around a vertical axis right here
but not the y-axis all right so we’ll get one example of each type
all right so let’s uh sketch the graph here and see what see what’s going to
happen so the first thing i’m going to do is graph the natural log of x so it’s
coming right through there like that and now on the
y-axis we’re only going to be looking at these y values right here so 0 up to
about a one so i’m going to take this part here away we don’t need it and that
and so this is coming from the natural log of x and

00:36
and it’s going to be okay so the region bounded by this and the y-axis so
what’s this height right here so this is coming through here right here
is going to be a and the height right here is going to be so this is like a 2
and then an e right here because natural log of e is 1 right and
so we’re going to get the height of 1 right there all right perfect
and so let’s just chop this off right through here
and the reason why is because we’re bounded less than one and we’re greater
than zero so we’re bounded right here also
so the y’s bounded between zero and one all right so let’s just shade this in
real quick we got this uh region right here that
we’re going to be revolving around the line x equals um one

00:37
and so yeah let’s sketch that in here let’s call this the y-axis the x-axis
and here’s about an x there’s about a one there so let’s call this about a
minus one here and let’s say that this is the vertical line here x
equals minus one and we’re going to be revolving around
this axis here this vertical axis right here x equals minus one
so we’re going to be getting this uh solid right here generated as we rotate
this around and so yeah we’re going to be using
washers when you rotate it around you’re
going to get a disc but it’s going to be
hollowed out so what you really get is a washer so when we revolve this around
here we’re getting washers all right and so we need to know
what is the uh r of y the outer radius so the r of y is 1 plus e to the y so
this right here is y equals natural log of x right
or if you solve for x it’s just x equals

00:38
e to the y right because natural log and e are inverse functions of each other
and so or if you want to think about it like
this e to the y is e to the natural log of x
but e to the natural log of x is just x well however you want to think about it
this is e to the x equals e to the y also so because we’re revolving around a
vertical axis right so this is my uh r of x here let’s put it in red
this is my r of x here so here to here and so this is going to be 1
and then plus the e to the y so this right here is e to the y right here
and so we pick a y value on the y axis and this distance is the
one and then plus the e to the y all right and so then for the lower
radius right here uh r of the y is just one and the reason why is because we’re

00:39
chopped off right here by the y-axis so this distance right here is just a one
all right and so now that we know the inner and outer radiuses we can go find
the volume so the volume will be pi i’m going to pull out my pi already 0 to 1
and then we’re going to have this first function right here the uh outer radius
1 plus e to the y squared minus and then the second one 1 and then
that squared and then d y so that’s a pi pi
all right so yeah we’re gonna integrate this out we’re gonna square this out
minus one we’re gonna expand that and simplify it and so this right here will be
um well i’ll just uh skip to the end here because we did this in a previous
episode so anyways the value that you get at the end is pi over 2 e squared

00:40
plus four e minus five all right so pi over two times e squared
plus four e minus five and so that’s the volume that you get of
that solid right there so i think setting this up by hand sketching the
region and understanding the outer radius and inner radius is
really nice to do by hand when you’re learning this
and then let’s go see how to do this in python now
so here we go let’s look at the python now for example 4 here
and so let me get rid of this real quick right here check this one out
please subscribe to the channel it helps us out a lot or me
to get noticed and recognized anyways find the volume of the solid and by by
the way thank you for all the people who’ve already subscribed really
appreciate it i hope you enjoy the new episodes and look forward to them
all right so find the volume of solid generated
by natural log of x y axis and the interval is revolved about the line x

00:41
equals minus one so first thing i’m going to do is uh define the function
right here natural log of x now when you write in python it’s just logx
that’s already the natural log right there and i’m using the numpy the
you know the numpy package right there to do that
and then we’re going to make some axes of a figure customize and
you can play with the aspect ratio if you want but i commented out
all right and so then i’m going to make a bunch of x’s so i don’t want to pick
exactly 0 so i don’t want to start it at
zero that’ll give me an error so i chose something really really close to zero
and then i go to e so this will be like really close to zero all the way to e
which is about two point seven something so there’s my x’s i’m going to plug in
my x’s into y i’m going to plot that function right

00:42
there you can see it right here and then um now i’m going to plot the
this goes from one to e um and then the y’s again all right so i
did this right here so that i can get um this
let’s see here this is a bunch of x’s from 1 to e
all right so this is to help me shade so i want to fill this in here and shade
this region right here and then i want to shade this region right here so this
is 0 to one and then this is one two to e so i graph i graphed this one again
and i filled it in and i grabbed this one right here and i filled it in um
so that way i can shade it in so it’s a little difficult to not
i don’t want this to shade this end down here so that’s why i did it is to do
those two fill betweens i filling in here and i’m filling in here
and so that’s the region right there that’s filled in

00:43
all right so to integrate this i need to identify
my outer and inner radiuses so i’m going to use sci pi i’m going to integrate
using the quad method now i’m going to have an anonymous
function here it’s going to be a function of y so this will be um
pi times the e to the y plus 1 squared and then minus 1 squared
and then i’m going to integrate along the y-axis 0 to pi
and i’m going to take the numeric value from that all right so let’s click in
here and execute it there we go and so yeah this gives us the
approximation right here 20.832 and if you’re wondering how accurate
that is we can take that off there if you want to and then we’ll get not only
the numeric value but also how accurate it is
and gives you some indication of how accurate it is right there lots of zeros
in there so this is very accurate right here so you know

00:44
speaking for educational purposes that’s very accurate right there
and so that gives us the numeric approximation to the value that we had
if you remember the exact value was pi over two and then it was
e squared plus four e minus five all right and so that was the exact value
and then we found the approximate value right there
so yeah i want to say thank you for watching and look forward to seeing you
in the next episode and i’ll see you then if you enjoyed this video please like
and subscribe to my channel and click the bell icon to get new video update

About The Author
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David A. Smith (Dave)

Mathematics Educator

David A. Smith is the CEO and founder of Dave4Math. His background is in mathematics (B.S. & M.S. in Mathematics), computer science, and undergraduate teaching (15+ years). With extensive experience in higher education and a passion for learning, his professional and academic careers revolve around advancing knowledge for himself and others. His work helps others learn about subjects that can help them in their personal and professional lives.

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