# How to Find the Area of a Surface of Revolution (Explained)

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

00:00
[Music] hi everyone uh welcome to uh day for
math live this is our first live stream um in this episode we’re going to talk
about finding the area of a surface of revolution this episode’s part of the
series applications of integration complete in-depth tutorials for beginners
and i’m excited to get things started for us let’s do some math
so um the goal of this video is to understand how to find the surface area
and we’re going to be revolving a region around an axis
we’ll start off by talking about the x-axis and so the goal of this episode
is to understand how this formula right here
will allow us to find the surface area and by surface area
we’re going to talk about the lateral surface area so i’ll explain what i mean
by that here in a minute so this is the goal and so the way we’re going to
approach that goal is we’re going to talk about first finding the surface

00:01
area of a cone and then we’re going to talk about
finding the surface area of a frustum and then we’re going to write out an
approximating sum where we can find the surface area
where we have a non-negative smooth function
and then we’ll pass the limit and then we’ll have this formula right here and
so we’ll see how to do that and as we as we go along we’ll work out lots of
examples here so let’s uh go ahead and get started here and
so uh uh first um i want to talk about uh finding the surface area of a comb now
as we talked about surface area of a cone let’s actually talk about cylinder
first just very briefly here so if i look at the surface here or this
cylinder here so i’m going to have this radius right
here let’s put the radius in blue if we can so we’ll call that r right there

00:02
and we’ll call the height h and if we’re trying to find the surface
area so i’m not going to count the top surface area here or the bottom surface
area underneath there i’m just going to talk about the lateral surface area
so the surface area you know if we uh unroll this right here then we’ll just
get a rectangle and the area will be the width times the
height you know w times h and this w here what will the w be it’ll be
the circumference right we unroll it and the circumference is 2 pi r
and so this area right here will be 2 pi r times h or
if we want to use an l then we would have an l and an l and an l
and i’m going to use a cone now let’s look at a cone now let me use an l for
that so let’s say we have a cone here and let’s say this height here is an l

00:03
make a nice l here and let’s say here this right here is an r
and so what we got going on here is to try to find the surface area here and
again i’m going to talk about the lateral surface here i’m not going to
talk about the area underneath or anything like that okay so
now if we unroll this and to try to find the surface area and should try to make
it flat we could try to find the surface area so if we enroll this unwrap it
we we got this curved distance here and this will be some angle theta
and this right here will be the 2 pi r and this right here will be some l
um and yeah so l is this distance right here from right here to right here and
we can find the area of a sector there um of course you may remember also like
like if this was called um you know um you know s equals r theta right where s

00:04
is the arc length r is the radius so s equals r theta
but in our case here we have s which is 2 pi r
and r is the radius which is l in our example here and so theta is
you know going to be 2 pi r over l um let’s write it over here so theta is
you know 2 pi r over l 2 pi r over l so we’ll use that here in a second
so any case remember how to find the area of the sector right here so the
area of the sector will be one half and then the radius which is l
l squared theta and it would be nice if we didn’t have
theta in here if we just had the r and the l
so that’s why i solved for theta over here so this will be one half l squared
and then this will be 2 pi r over l for the theta 2 2 pi r over over l

00:05
and so the area here will be the twos cancel you know we’re
going to get the l’s to cancel so we’re going to get here pi r
and then one more l left so this will be the area right here of a cone
and when we have a radius r and a and a and then
that length there right there is the l right so
now uh what we want to do though is we want to use the frustum um to help us
find the surface area and uh we’ll ex we’ll see why here in a
second uh why we’re going to use the fresh term first of all what is that um
how to find the surface area of your frustum so i’m going to redraw the cone
again up here and this time i’m going to chop it off here
right about here parallel to the bottom and so we’re going to end up with two
r’s now so let me put them in um yeah let me put

00:06
them in blue again so i’m going to call this one here r2 so that one’s r2
and we’ll call this one right here r1 so i’ll just put an r1 here
and i’m going to call this uh distance right here l1 [Music]
and then an l for this one right here and the fresh drum is this part right here
it’s this part right in here so it’s like the whole cone but then we took
away the bottom cone and we’re left with this um solid down in here
and that is uh that’s what that’s what i’m calling the frustum and i’d like to
find the area of that frustum right there and then we’ll see why i’m looking
at the area of a frustum in a moment here what would be the area of the
frustum so the area would be you know using what we just did down here
we would look at the area of the whole uh cone which will be um you know just
it’s going to be just pi r l so pi and then the radius so the radius of the
whole cone is going to be r2 and then the l will be the length right

00:07
there so it’ll be l1 plus l because i’m calling this distance i’m
calling that length right there l1 and this one right here l
and in the end we would like to have a nice formula that only had r1 r2 and l
in it it didn’t have l1 in it all right so anyways the area of the
frustum will be the area of the cone the surface area of the cone
take away the area of the smaller cone so that’ll be pi
and that radius will be r1 and that length right there is l1
so this will be the area of the surface area of the thrust drum right there that
we’re going to be looking for here but like i said we would like to have all
this out without an l1 so that when we talked about the area of
a frustum we would just have r1 r2 and l in it so can we get rid of the l1 here

00:08
so let’s just factor out the pi here for a moment and say okay so we have r2 l1
and we have an l there and i’m going to leave that r2l alone
and i’m going to say minus r1 l1 and then don’t forget the r2 l plus r2 l here
and we’ll see why write it like that in just a moment uh i want to try to get
rid of this part right here if i can now to do that i’m going to use a similar
triangles so let me get rid of this part right here
let’s look at similar triangles right here so my similar triangles right here
in this diagram right here i’m going to say l1 over r1
is going to be equal to so that’s the small triangle right there and now i’m
going to be looking at the larger triangle which will be l1 plus l
and then that’ll be the full radius right there so r2
so i’m going to try to get rid of the l1 i’m going to try to solve this for l1

00:09
and i’m going to try to substitute in and get and get a get out from under
this right here so this was a r2 right here r2
all right um so let’s see here what can we do with this so this will be
i’ll multiply both sides by r1 times r2 and multiply by r1 times r2 so the r1s
over here cancel so we get r2 l1 equals and then r1 l1 plus r1 l1
and so if we move this over here we get r1 l1 equals r1 l1 um
actually that should be just an l right there
right because i’m multiplying r1 times this one and then r1 times this one all
right so good so this part right here is actually this part up in here isn’t it

00:10
so now we’ll come back here and write the a the area is pi so area is pi
and then times all of this right here i’m going to rewrite it as an r1l
so r1l plus and then an r2l and i’m going to write it out even nicer
by saying pi and then we already have an l right here
and so now i’m going to say r1 plus r2 and so now we have the
area of the frustum right here only in terms of r1 and r2 and l
and that’s pretty nice but actually we’re going to do even more what i’m
going to do is i’m going to put a 2 here
and i’m going to put a 2 underneath here and now i’m just going to rename this
right here as an r so what the r is going to represent is
the average of the two radiuses r1 plus r2 if you add them up and divide by two
that’s just the average so the area of the frustum is two pi l

00:11
times r where r is the average r is the average of the radii is the average
and so this is a nice uh formula right here to think uh to use right here to
find the um area of a frustum now imagine this that
i tipped it over and put it on its side and so now we have something like this
something like that and we’d have this part in this part and we would have
a radius right here and we’d have a radius right here and we would have a
length which we’ll just call l again right so this is like an r2 this is like
an r1 so i can find the surface area of that object right there
um and that’s a frustum and we would use this formula right here to do that
so we’re on our path now to understanding
how we can find the surface area now by using these little frustums as
approximating an approximating sum and so let’s do that now

00:12
uh so first part of what is the surface area so let’s uh get rid of this here
real quick so what i’m going to do is i want to make a sketch for us here a
diagram to help guide us so we’re going to need a function f is
has a continuous derivative continuous derivative
on a closed bounded interval a b and so some people call a function that has a
continuous derivative in other words the derivative exists and is continuous so
people just say a smooth a smooth curve i’ll just uh leave it like that but
anyways um so let’s say we have something like this right here and here’s an a
and here’s a b and here’s my f and we got the x axis coming out here and
what we’re going to do is we’re going to take this region right here

00:13
and we’re going to revolve it around the x-axis and we’re going to create a
three-dimensional solid so if you don’t know what that is
please take heart and what i was saying at the beginning
which is um this episode is part of the series applications of integration
complete in-depth tutorials for for calculus so in the uh previous
episodes we talked about what a surface of revolution is
and also pre previous episode we talked about how to find the arc length and
that was an incredibly important video for this one right here
because what we did in that video is we chop this length up right here this
function right here into little tiny segments
and we’re trying to approximate the arc length and then we pass the limit and
we’re able to find the exact value when we had a continuous derivative on a
closed bounded interval also and so for that formula and for that method you
know check that out but in this episode what we’re going to

00:14
do is find the surface area and so we’re going to do is we’re going to revolve
this region around the x-axis and we’re going to get a solid of revolution
and our goal is to find the lateral surface area of that
and so bilateral surface area i mean i’m not going to try to find the surface
area right here or the surface area of that part right there but anyways when i
look at the solid here it’s going to look something like that and then it’s
going to look something like that and over here it’s going to have you
know some round part to it some round part to it right there
but we’re going to have here coming through here these a b and the y axis and
you know we’re going to be revolving around the the x axis so we’re going to
get some kind of solid and this region right here this you know this area in
here right here is going to get revolved and we’re going to generate the solid
and so we want to try to find the surface area of that solid and the way
we’re going to do that is we’re going to
put down a partition of a b so let’s put that over here partition

00:15
um interval a b using subintervals and i’m going to go ahead and use uh
regular sub intervals so let’s do a regular partition and so we’re going to have
um the delta x is b minus a over n and so i’m going to have n of them and
they’re going to be equal with and so i’m just going to draw one of
them an arbitrary one of them so we’re going to divide them up in a whole bunch
of sub intervals here just imagine a whole bunch of tick marks for our partition
and so for the left endpoint i’m going to use an x i
uh minus 1 and then for this tick mark right here this will be the i x i so
this is an arbitrary sub interval we made a whole bunch of
them and then this is just an arbitrary one and so this width right here will be
delta x i and we can find a formula for that right

00:16
there but what i want to do here is i’m going to come up to the graph here
and come up to the graph right here and i want to make a triangle right here
a right triangle and i want to try to find the distance
between these two points right here so this will be f of x i minus one and
this height right here will be f x i but this lateral distance of this
horizontal distance right here uh we’re going to call it delta x i
and this right here will be delta y i so for each one of these intervals i’m
going to do this process right here for each one of them i’ll have a tiny little
line segment going across here and we can find the length of that it’ll be delta
l i and so delta l i just either distance between two points
or if you want to think about pythagorean theorem we got a right angle
right in there so this will be equal to the square root of the delta x i’s

00:17
squares plus the delta y i squares and this will give us those little links
right there and we want those links because remember
when we take one of these sub intervals right here so i’m going to chop this
down into a straight edge right there and when we revolve this little piece
right here around the x-axis we’re revolving the whole thing so what we’re
going to get here is a frustum and so this will have some kind of circular
and this will be so this will be a frustum right here and we know how to
find the area of this little uh p segment of it
of this little piece of the whole solid we know how to find the area of this
because we can find the average of the radii and we can find this length right
here so i need to know this length right here this little distance this little
straight straight edge right here it’ll be delta l i
so for each one of these we’re going to do it to do this okay now

00:18
in order to get the average radius so i’m going to say let x
our ri be the average radius average radius um of this frustum right here
of the i thrust strum so remember we’re doing this from i equals 1 to
we’re doing this for n sub intervals right here so this is just an arbitrary one
so we’re going to use the intermediate value theorem
the intermediate value theorem and say ri equals f of d i um so
where d i exists because of the inter intermediate value theorem where d i is
in the i sub interval i sub interval so where d i is in the i sub interval so
in other words we have some height here on this one we have some height here for
this radius and because this is a continuous

00:19
function right here there has to be some input di
where we hit that uh output right there so this will be the um ri will be the
average radius and i can get some input that hits that output right there all
right and so now the lateral surface area of the frustum delta s i
and let’s move it let’s bump it over here delta s i here
will be given by 2 pi r i and then delta l i
and i have to keep putting an i here as an index because we’re we’re doing a
whole bunch of these to find the whole surface area we can’t just do one one in
sub interval there we have to do all of them
and so we’re gonna um in two pi and then here’s the average radius and then
here’s the uh delta l i this distance right there so this will be two pi
and then the ri will have f of d i and then delta l i right here
okay and so what we can do is we can bring in what our delta l i is right

00:20
here now which came about by using the pythagorean theorem
you know the hypotenuse squared is equal to the sum of the leg squared
and so what we end up with is two pi and then f of d i f of d i here
and then we have square root of delta x i squared plus delta y i squared
and now we can do some algebra to manipulate this here a little bit so
what i’m going to do is i’m going to factor out a delta x i squared here
and we’re going to [Music] say 1 plus i’ll just uh do 1 plus delta y i squared
and then over the delta x i but we’re factoring it out and so we
actually just have delta x i squared and we’re going to take the square root

00:21
of that so we’ll have delta x i squared and the square of the square root so
i’ll just put delta x i and there we go right there so there’s
the approximation right there and so this is for the eighth one
and so then what what what we’ll do is we’ll sum up all those uh frustrums
right there the surface area of all those frustums i just drew one of them
but we’re we’re going to draw a whole bunch of them we’re going to have i of
them uh uh sorry is the index and so we’re going to add some from 1 equals 1
to n we have a finite number of them let me uh write it a little bit
let’s put it let’s put it right here so sum i equals 1 to n of the delta s i’s
is going to be equal something from 1 to n of of this part right here so 2 pi
f of d i and then we’re going to have the square root of all this right here

00:22
so we’ll have i’ll just erase that right there square root 1 plus
delta y i over delta x i and all that will be squared and then delta x i
and so we’re summing up all of them so we did the arbitrary one right here and
then we’re going to sum up all of them all right and so now as you can imagine
you’ve seen this argument before in the previous
episodes in this series we’re going to now pass the limit
and so we’ll have limit on both sides we’ll have the limit as n goes to
infinity so we won’t have a finite number of them um what we’ll have is a
process we’ll have a finite number of them and then more of them and more of
them and more of them and because there it’s a regular partition as n goes to
infinity right here the delta x is going to zero
and so what we’ll have is the width of each one of them is going to zero
and so what we’ll have is the definite integral there

00:23
and so if we write that out as a limit which i’ll do right here now
here we don’t have the derivative so what we have here is the delta y i over
the delta x i and what we did in the previous episode
where we change this uh these slopes right here to derivatives using the mean
value theorem so i’ll just put here by the mean value theorem here i highly
recommend you go and check out the previous video where we did that process
right there very diligently so the by the mean value theorem here what we’re
going to have here is the limit as n goes to infinity of the sum from 1 to n
of 2 pi so all this right here we’re putting
limit on both sides so we’re going to have the limit as n goes to infinity 2
pi f d i and we’re going to um just call that x i
star as the representative in that interval right there
and then we’ll have 1 plus and then the derivative at delta x i star

00:24
squared and then we’ll have here a delta x delta x i and here um
we can take off the x i here if we want they’re all the same
um but in any case what we end up with here is the uh integral here so the area
the surface area the lateral surface area sometimes i’ll use an s or an a or
or both we’ll have the definite integral here from a to b
and then we’ll have here this 2 pi and it’ll be 2 pi
and then we’ll have our function here f of x
and square root of 1 plus the derivative at x squared and then dx here here so
um let me make myself a little smaller there we go so
this will be the formula that we talked about at the beginning of the episode

00:25
here so it basically comes from thinking about f of x as the average radius
and this is the distance or the um the distance along this edge right here
and we’re going to be letting that limit go to infinity
and so we end up with a definite integral right here and under these
conditions right here this will give us the exact value and this will give us
the lateral surface area and so what we can do now is actually try to
work out some examples and try to generalize this [Music]
a little bit further so let me erase this real quick and see how to do that
sometimes you’re not always revolving around the x-axis or the y-axis
sometimes you’re revolving around just a horizontal or vertical axis so i’ll use
s here for surface area and we can say 2 pi integral from a to b

00:26
and we can have an r of x and then one plus uh f of x the derivative squared dx
and so again we have the same conditions on f f is um [Music] continuous
derivative on a closed bounded interval a b
and here y is a function of x so y is a function of x
so i’ll just say y equals f of x and here r of x is the distance is the distance
between so in the previous screen we used
f of x here and so now i’m changing that to r of x it’s going to be the average
radius so it’s the distance between the graph of f and the axis of revolution

00:27
so if you’re using the x-axis as your axis of revolution the graph of f and the
um axis of revolution axis of revolution
but if you’re using the x-axis then this r of x will just be the regular f of x
but if you’re using an axis of revolution okay and so now we’ll have s equals
2 pi and we’re going to integrate from c to d
and then we’ll have r of y and then we’ll have 1 plus g prime y squared d y
and so now we’ll say g is uh can has a continuous derivative
on closed bounded interval c d along the y axis and now we’ll say that
um x equals g of y so we have a function of y and here we have r of y is the

00:28
i’ll just put quotes here um the axis of revolution here
and this is going to be a horizontal axis so i usually like to call this the
horizontal version and so r of y is the distance between
is the distance between the graph of g and the axis of revolution
and here we’ll have horizontal axis and here we’ll have vertical axis so
horizontal axis and here we’ll have vertical axis
all right so we have a horizontal version and a vertical version
if we’re if we’re going to generate our solid
horizontal uh using horizontal rotation or vertical rotation right here
and so these r y’s right here represent i use r’s to represent the radius
but here’s what happens if you can solve for y in terms of x
or if you can solve x in terms of y so we’ll see some examples of this right

00:29
here so let’s do one um let’s go to the first example here
so let me get this out of the way here real quick
all right so let’s find the surface area generated by revolving this curve right
here on this interval right here and we’re going to revolve about the x axis
so here we have y as a function of x and so we need to find d y d x
so d y d x will be 1 minus x over square root of 2x minus x squared
so there’s our derivative and we’re looking on this interval right
here 0.5 to 1.5 and on that interval right here this
function right here this derivative is continuous
and so we have a continuous derivative on this interval right here
obviously this derivative is not doesn’t exist everywhere it’s not continuous
everywhere but on this interval right here we have a continuous derivative
and so we’re going to use the surface area that we just came up

00:30
with here so we’re going to integrate from 0.5 to 1.5
and we’re going to have a 2 pi and we’re going to have our function
2x minus x squared and now we’re going to have the square root of 1 plus
and then we’re going to square our derivative right here so we’re going to
have 1 minus x squared over 2x minus x squared and that’s a
square root but we’re squaring it so i’ll just leave off the square root down
there and then we have dx right there so we can try to integrate this right here
uh pull out the 2 pi and we’ll have the integral from 0.5 to 1.5
and so now we’re going to say 2x minus x squared
and let’s just go ahead and expand all of this out here so let’s put this
square root 2x minus x squared and then plus 1 and then minus 2x

00:31
and then a plus an x squared and then all of this over the square
root so you know we need to get a common denominator and put all that together
just like that right there all right and so here we go 2 pi
now these cancel out and what does all this become here two x’s cancel out or
add up to zero x squares add up to zero we got square root of one so in the end
we just have the integral from zero point five to one point five of one times dx
and that difference right there is one so we just end up with 2 pi right there
and so that’ll be our surface area lateral surface area
for this example here so let’s do one more
let’s find the surface area of a sphere now so let’s try this real quick so
now in this example we’re going to let y be square root of r squared minus x

00:32
squared right so we have the surface area of a sphere
and so think about r as being a constant so we have 4 pi r squared
and we want to evaluate the double integral in other words we want to use
our formula right here so we have y prime is so we have minus x over
r squared minus x squared and so we can take a look at what is
going to be 1 plus the derivative squared
because we’re going to need to integrate all that so what y prime squared
so if we simplify all this down we get r over r squared minus x squared
so doing these computations over here uh you know sort of helps us out let’s go
find our surface area now so it’s going to be 2 pi
and we’re going to be integrating from minus r to r
right minus r to r and then we’re going to be integrating y and

00:33
yeah so this will be the just the y and then we have here the r over
the square root of r squared minus x squared dx all right perfect
um you know because that’s just what we found right there
all right so then we have 2 pi r r is constant just um
so we’re going to integrate from minus r to r can’t have that r there
um and so we’re going to have dx here and this will come out to be 2 pi
from minus r to r and so this will be yeah there is an r here uh
we’re integrating with respect to x yeah
so this is just 2 pi and then 2 pi r and then we’re going to get another 2 pi r

00:34
out of this right here or sorry 2 r so in the end we get 4 pi r squared
which is you know pretty much as we expected that to be okay perfect um
so let’s do one more but first let’s just recall here this theorem here um
if we’re going to be revolving around the y-axis
we still use frustums we still use the same technique we’re just going to be
integrating along the y-axis and our function is solve for x we have
x equals g of y and we have the derivative right here
and so let’s go see an example of that now so we have x equals and we have a
function of y and we’re going to be integrating from 1 to 2 along the y axis
so first step is let’s find x prime so here’s x prime we’re taking the
derivative of x with respect to y and after we simplify that we’ll get y

00:35
to the fourth minus one over two y squared
oops can’t see that sorry so x prime is y to the fourth minus 1 over 2y squared
and so we can say what 1 plus x prime squared will be
so let’s just do 1 plus and then square this and so that’ll be 1 plus
and this will be y to the 4th minus 1 squared over
4y squared and let’s just break that down a little bit y to the fourth plus 1
will all be squared and then we still have 4y to the fourth
so expand that out get a common denominator combine it all back together
and so yeah i just find this to be a little bit easier if we
just kind of bring that out on the side right there
all right so here we go let’s find the surface area

00:36
so we have pi integral from one to two um so one to two on the about the y axis
and so we’re going to do the uh function right here um y to the third over six
plus one over 2y and then we’re going to have the um y squared if we yeah
so we’re going to have y squared over 1 plus y squared here so um
you know because we’re going to have the
square root of all this right here so we need to take square root of that
in any case [Applause] this will be pi and then integral from one to two
and then just expanding this out so i get y to the fifth over six plus

00:37
two y over three and then plus one over two y to the third and then d y
all right and then without making this boring we just integrate piece by piece
find the antiderivative and substitute it in
and after all the work you get 47 pi over 6 is the exact surface area there so
that does it for this video here so i want to say uh thank you for
watching and i’ll see you in the next series take it easy
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