The Disk Method Around a Horizontal Axis (Simplified)

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 the disk method by revolving around a
horizontal line [Music] hi everyone welcome back this is dave
in this video the disc method around a horizontal axis simplified is part of
the series applications of integration complete in-depth tutorials for calculus
the link below is in the description um for the playlist for the series and i
hope you find it and i hope you enjoy the series let’s get started
so we’re going to be looking at a the disc method here and this is for the
around a horizontal line in this video so we’re going to be looking at a of x
here which is the cross sectional area and we’re going to be looking at a disk
of radius uh r of x so both of these are functions of x’s

00:01
and and we’re going to be looking at to find the volume of the solid of
revolution around the x-axis so here we’re integrating from a to b of a of x
and so really it’s very useful to look at an illustration
so let’s get going here um let’s sketch this um
set up here and try to understand it better so let’s hear let’s go with um
right about here and we’re going to be looking at some
function here so let’s come in here with some functions let’s say f of x here
and we’re looking on from a to b so let’s put up a boundary here let’s say
this is a and this is b right here and let’s say this is the x-axis
y-axis up here and so let’s slice through here
and let’s look at this area right here so we’re looking at this area right here

00:02
i’ll shade it a little bit of it in blue and we’re going to be taking this area
right here and we’re going to be revolving it around the x-axis so how do we um
put that in symbols so we’ll just take this and we’ll say
here we’re going to be revolving it around the x axis there and
now remember when we’re looking for the area under the graph of a function
we would often take uh representative rectangles
so let’s let’s put this in orange here so let’s put this right here in orange
this is going to be a representative rectangle right here and we’re going to
be revolving this uh this region here in blue
around the x-axis and when when we do that that’s going to generate this
rectangle here is going to turn into a disk
so here the width of the rectangle is delta x
and this height right here is given as r of x

00:03
so we’ll call this the height right here so height times the width
and this is our representative rectangle and
so when we start when we start revolving this around the x-axis here we’re going
to be generating a three-dimensional object and we’re going to try to find the
volume of that so let’s try to see what that looks like um so here we go
we’re going to come through here with this function right here
and a little bit like something like that and let’s go from a to b x axis
and we’re going to revolve here and so what is this region here so we
still have this um region here that’s generating the the volume
and now so so when we gen when we uh revolve this rectangle around the x-axis
it’s going to turn into a a disk so we’re going to have this disk

00:04
here let’s see if we can sketch a disk going through here
and let’s just chop this in through here like this
and so this though with the width of this disk is the same width this is
the delta x here let’s go and shade in this width right here
so this is delta x here and this right here is given the height
here is given by r of x still so r of x here and this right here is the cross
sectional area this is the aid of a of x here so a of x is equal to the
let’s see here what is what is a of x here
right so we need the height right here and so which is going to be the radius
of the disc right here the the cross sectional area so it’s going to be pi r
square so pi and then the radius squared so that’s how we get this part right

00:05
here and so this right here will be the volume
and so this will be our representative disk and this right here is going to
represent the approximate volume using let’s say in disks
and here’s one of the representatives and this right here is going to
represent the exact volume so we’ll say exact volume right here
because we’re going to be integrating we’re going to let these the widths of
these discs right here go to zero so we’re going to be adding up
infinitely many of these and so that that’s going to be integrating there
so um let’s look at some examples here of this right here in action and make
sure that we are understanding everything um

00:06
and yeah so this looks fun let’s let’s look at some examples let’s erase this
so in our first example here we’re going to find the volume
and we’re going to take this right here as our function right here our f of x
let’s go ahead and make a sketch here we’re looking at something like this
the height here comes in at nine and we’re looking at plus or minus three
right here so let’s say three and minus three right here
and so we’re using this uh area in here i’ll shade this blue
and we’re looking at this area right here and we’re going to be revolving
around the x-axis so we’re going to need to uh draw a representative here
so here we go x axis let’s revolve around here y axis

00:07
and now i’m going to draw a representative rectangle right here and so
but this rectangle when we revolve it it’s going to become a disk
so i’ll try to sketch a disk in here and so let’s find the the volume of this
so in this case here our r of x here our r of x right here will be the height
from the x axis up to the curve so it’ll be given by nine minus x squared
for this example here 9 minus x squared and so
we can integrate so the volume will be and we’re going to be integrating from

00:08
minus 3 to 3. so we have pi and then r of x squared so 9 minus x squared
and then dx and so we can integrate this we’ll just
square this out and bring out the pi so we’ll say pi integral from -3 to 3
and we’re going to have x to the fourth minus 18 x squared plus 81 dx
good so just expanding that out and then now let’s integrate term by
term so we’ll have here x to the fifth over five
minus eighteen x to the third over three plus eighty-one x
and from minus three to 3 and so when we simplify all of this we’re going to get
2 pi and then one-fifth x minus 6x to the third plus 81x

00:09
and so basically what i did was i use symmetry i’m going to go from 0 to 3 now
that will cut down your the amount of arithmetic you have to do there
now we just need to substitute in three but basically i use symmetry right here
this is symmetric about the y axis here so i went from minus three to three from
zero to three and you can do that at any point you want you could have set this
up two and then integrate it from 0 to 3. in any case we get 1296 over 5 pi
as the exact volume for that object that
you get when you revolve this around the x-axis there
alright so there’s our first example there
and let’s go ahead and look at another one so let’s erase this here
so here we go um we’re going to be looking at this line this function right
here square root of x plus 1 and now we’re going to be looking at

00:10
revolving around the line y equals 2 um and and this is going to be part of
our boundary here and we’re going to be looking on 0 to 1 here
so let’s look at and sketch this first so here we go
now we’re going to be revolving around the line y equals 2
and someone has put that about right here let’s say
and let’s say this is the x-axis here and we’re going to be integrating from
zero to one so let’s put a one about right here and and we’re looking at this uh
you know square root of x um but it’s been shifted up one
so let’s make it look like you know um so there’s y equals two so it’s been
shifted up one so let’s go with something like this
so a one what’s the height here is two so it’s going to hit right there and
it’s going to be nice and curvy now when we revolve it around the line y

00:11
equals two it’s going to have a part of it up here
right so that’s the lower part there and let’s say this right here is so we’ve
got one we got two and we got three and so we’re going to come in here
through like something like that and so what we’re going to have here are
um representatives they’re going to be cylinder sorry disks
and we’re going to be there’s going to be our like our representative
we can draw another one if we want and then so we’re revolving around y equals 2
there let’s put that right there so we’re revolving around y equals two nice
and smooth and we’re just going to find the volume of this solid in here
all right so in this in this example here what is

00:12
giving us the height the r of x right here so the r of x is
well let’s put it right here let’s put it right here so let’s think
about the r of x right there and so what’s going to give us the r of x here
so the r of x will be so the height here the full height is 2
so i’m going to say two minus and then take away this height right here
so two minus square root of x plus one so that will be our r of x here in this
in this example here and perhaps it’s better if i write it down here
so hopefully you can see how we got that right there that’s this right in here
this this height right here because remember we need the the volume is zero
to one of the cross sectional area and how do you find the cross sectional area

00:13
it’s going to be pi r of x squared where r of x is going to give us the radius
and so that’s why i need to find that radius function right there it’s going
to be fixed at a height full height of 2 and then minus the height that we need
to get up to the function right so there’s the function right there
so in this case right here we’re going to be getting uh pi and then zero to one
and then our r of x is two minus square root of x and then i’ll
distribute the minus one there so in fact that just becomes a one
and so let’s see what we get here we’re going to get here um zero to one
we have to square this out right here think of that as a one so one minus
square root of x right so we’re going to get an x minus two square roots of x
plus one [Music] and then now we can go ahead and

00:14
integrate this right here so we have x squared over two
minus two x to the three halves over three halves and then plus an x
and from zero to one [Music] so looking good so we have pi
and then we’re going to come in here with the one and we’re going to get a
one half minus four thirds plus one and so all of that it turns out is just
pi over 16 pi over 15 16 over 15 pi’s all right very good oh sorry pi over six
just pi over six just put them all over six
all right so there we go that that will be the volume of that solid that’s
generated by revolving this area right in here around y equals two
excellent so let’s work out one more example

00:15
and let’s see how that let’s see how the next one’s going to go so here we go
so this last one here we’re looking at 2 minus x squared
and we’re going to be revolving around the line y equals 1 in this case right
here so here we go so our height here is 2 and so i’m just going to
draw this right in here we have x-axis and we have y-axis
but we’re revolving around the line y equals one and we’re bounded by one also
so let’s just come through here with a one this height here is two
um maybe i should draw that a little bit more in the middle because it is

00:16
and let’s call this y equals one and now we don’t need all this part down here
because we’re revolving this part right here around the y equals one
revolving around here so we’re going to take this little top part of this right
here and let’s see what what are these right
here what are we integrating from here to here so that’s a one
and then the height will be one two minus one and this is a minus one
and now so i’m going to take a representative here and [Music]
i’m going to get a disk right here and so what is this height right here
um what is this r of x right here going to be so we need to find the r of x here

00:17
so the r of x will be it’ll be a height of one and then you know well
so the r of x here will be i mean it’ll be one minus x squared right
because this is going to be 2 minus the x squared minus 1
which just turns out to be 1 right because this is 2 minus x squared that’s
the function right there but then we’re taking away this one part right in here
so that that’s how we get that little height right in there that little radius
height right there it’ll be the whole height the whole
function right there two minus x squared and then take away this one height of
one right here so in other words it’s just one minus the x squared right there
so there’s our r of x right there so now to find the volume um and

00:18
so to find the volume i’m also going to use symmetry again this is symmetric
about the y-axis so i’m just going to be finding the
volume of half of it and so i’m going to double it to find the volume of all of
it so the volume will be 2 and then i’m going to integrate from 0 to 1
and then pi and then the r of x squared so 1 minus x squared squared
and so this right here will represent the volume that’s pi so i’ll say two pi
zero to one and then now let’s just expand this out real quick
and now let’s integrate term by term so x to the fifth over five
minus two x to the third over three plus x
and then now let’s go from zero to one and then the last step will be to

00:19
substitute into one so we get one-fifth minus two-thirds plus one
and then now when we combine these fractions together it’s
going to be over 15 and in fact we get 16 over 15 pi right excellent so
there’s uh three good examples there on how to find how to use the disc method
by revolving around a horizontal line and in this case uh in one example we
did the x-axis but for here’s a here’s another one where we did um y equals one
all right so i hope that you enjoyed this video of course the next question
is how do you revolve how do you find the volume of a solid that you’re
revolving around a vertical line so we’ll check that out in the next video
and i want to say thank you for watching and i’ll see you next time
if you like this video please press this button and subscribe to my channel
now i want to turn it over to you math can be difficult because it requires

00:20
time and energy to become skills i want you to tell everyone what you do to
succeed in your studies either way let us know what you think in the comments

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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