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in this episode you’ll learn the disc method by revolving around a vertical

line [Music] hi everyone welcome back in this episode

the the disc method around a vertical axis

um is part of the series applications of integration complete in depth tutorials

for calculus so hope you find the link for the series below in the description

let’s go ahead and get started so we’re going to start off by um

understanding this theorem here which is

what we’re going to use this is the disc method when we’re revolving around

a vertical line so we’re going to state in terms of the y-axis though

so here we’re integrating from a to b and a to the y

is the cross-sectional area of a disk of radius r y

so um while we’re looking at this theorem here let’s go ahead and get a good um

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illustration going up here so here we’re going to draw the x and y axis right

here and now we’re going to be looking at functions of y so i’m going to come in

here through here with some function of y something like that

so this will be our g of y and we’re going to be revolving here around the

y axis this time so i’m going to put that down here and say we’re revolving

around the y axis here and now when we do that

um you know we’re going to have some boundary right here so let’s call this here

c to d and so that’s going to give us some

boundary right here some boundary right here

and so what we’re going to have here is we’re going to have this area right here

and we’re going to use this area like i said we’re going to revolve

around the y axis and we do that that’s going to give us a volume of a solid

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so let’s just go ahead and shade this in here and to to know what we’re

revolving around or or what we’re using to generate the

volume here is this area right here now when when we look at this area right

here we can look at representative so i’m gonna look at

a representative right here so cut this right here and we’re gonna say

that this right here is a representative rectangle

so we know we use rectangles they’re easy to work with we know width times

height right and so this width right here will be delta y

so i’ll put that right here like this that’s the

that’s the width or the height in this case

and then this right here right here is going to be given by the r y here and

when we rotate this uh this area in blue

here around the y-axis so this rectangle now becomes a disk and so let’s see how

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that would look so let’s try to sketch it right here

and i’m going to be coming in here with this

g of y it’s going to look something like this and then we got the same area or

something like that over here and we’re revolving around the

the y-axis down here and we still have this c to d and so we’ve got this

right here um and we’re taking this region right here which we shaded in

blue up here and when we revolve it what

we’re going to get right here instead of this rectangle right here

this is our representative rectangle i’ll just call it representative rectangle

so now over here this uh right here is going to give us a disk so

let’s say if we can see if we can draw like a disk coming through here like

this and it’s going to have a little bit of a width to it

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and let’s just draw this like here like that and so this is going to give us a

actually it’s probably better to draw it above

and let’s just so this rectangle right here when you revolve it now becomes a

disk it generates a disk and so this right here is going to be given [Music]

and so we can even maybe see one right here so we can have this disc right here

coming through here and that’ll be the lower boundary and

then we have the upper one also and so this a to the y here that we’re talking

about right here this a to the y is the cross sectional area right here

so that’s the cross sectional area and that is going to be given by

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so here we have the uh r to the y and that’s the radius right there that’s

going to give us the radius given a y a y value

and the r y here gives us the radius right there from the um

from where you’re revolving around so there’s the r right there and that’s

going to be given as right all right so there’s the r y so

the the cross sectional area then will be the um you know the area of that

right there we’re generating circles so we got pi and then

r of y squared right there so that’ll be the cross sectional area

and so that’s how we’re generating this formula right here from a to b

and in fact this should be c to d a to the y d y

from my illustration over here i use a c and a d so pi r y

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uh they only doesn’t matter what you call them what you name them

as long as you stay consistent so i use the c and the d over here so writing the

theorem is a c to d but the point is is that they’re fixed numbers right there

and they’re defining the boundary of the

object that you’re that you’re revolving right there

okay so let’s see lots of examples here and make sure that we understand this

and here we go so our first one here is um and we need

to erase this so let’s do that okay so in our first example here uh

find the volume of saw that results from

the region enclosed by these curves here is you know we have y equals zero y

equals natural log of x and then we got two and we got and we’re

revolved around the y axis so let’s see what all this looks like draw a sketch

right here and so we’re going to be using we’re

looking at this natural log right here it’s going to come in here at 1 right

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here and it’s going to be coming up here and

we’re chopped off at the top here by y equals 2 so let’s dash that in here

and so there’s y equals 2 and here is y equals natural log of x

and then here we have the x axis now we’re revolving around the y axis let’s

put that down here and so let’s see here we’re we’re

stopped at y equals zero or better known as the x-axis right here

um and so let’s just go ahead and shade this area in here in blue

that uh we’re bounded and the x-axis right here or sorry the

y-axis or x equals zero right so we’re we’re stuck in this region right in here

and we’re going to take this and we’re going to revolve it around the y-axis

and we do that we’re going to be generating a solid and we want to find

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the volume of that solid so let’s draw a um a representative so let me just take

this out right there and then now let’s uh we need to make the the sketch over

here let’s say a 1 is about right here when we revolve this it’s going to

generate another region over here and so let’s say that’s about right here

and then let’s make it nice and curvy and so what we’re going to get here is a

sub sub disk representative touching right there

all right and so something like that right there will be our representative

right there and now we need to know what the cross-sectional area is in here

so what is the cross-sectional area in here it’s a of y

in order to find that though of course we’re going to need to know the

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r of y here so the r of y is given as well here we can solve for the x we can

say x is e to the y these are inverse functions of each other so

we have the function of y now here and so now we’re ready to find the

volume here so um you know because the r of y is just e to the y here

right just from here up to the curve right there so e to the y right there

so the volume is um and we’re going to be integrating

here from 0 to 2 that’s our a b or or if you prefer cd

and then that’s going to be here 0 to 2 and then we have pi

and then our r of y squared so e to the y and then squared and then d y

and so this right here will be our volume right here

so let’s go ahead and integrate that let’s pull out the pi

so pi zero to two and then e to the two y and then d y

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and then now you might recognize that we need a u substitution we can say

something like u is two y so d u would then be 2 d y

and so we’ll need a 2 here so i’m going to chop it down by a half and say pi

over 2 0 to 2 e to the 2 y with the 2 out here now

so making the change of variable here we can say this is e to the u d u

and when y is 2 u is 4 and when y is 0 all right so we’ve got 0 still

and so now we can integrate this e to the u d u which is just e to the u

and then zero to four and so then this would just be pi over two

and then this will be e to the fourth minus one all right and so this right here

represents the exact volume of the solid

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that’s generated by revolving around the y-axis in this example here

so that looks like a lot of fun there um

yeah and so uh you know let’s do another one let’s see what happens when we’re

revolving around um you know the y-axis um and let’s just look at

another example here okay so in this one we’re looking at y

equals x to the third y equals eight x equals zero

and this is going to be revolved around the y axis again okay so here we go

let’s look at y equals x to the third first so when we look at y equals x to the

third it’s going to come through here like this

and we’re going to be chopped off at the top by y equals eight

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and so let’s just draw that y equals eight through here x-axis y-axis

and we’re revolving around the y-axis all right

and so what’s the region here so we got this curve we got y equals eight x

equals zero so we’re looking at this region right in here

and so let’s go ahead and sketch that in blue a little bit real quick and

this is remember this is where i remember i

liked coloring books when i was a kid in any case um

we need our i usually like to draw the um sub disk representative

let’s get the other side when this is revolved over here we’re going to get

something shaped like this and so now i’m going to slice through here and try to

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make a sub disk representative so something like this

and then it’s got a little bit of depth to it so there’s my [Music] all right

so what is our r y now so we’re looking we’re revolving around here to the curve

so the r of y is just given by we need to look at this right here

now this is a function of x so we need to be looking at the cube root here

so i’m going to be looking at the cube root of y

in other words solve this for x take the cube root of both sides

and so there there’s our r of x right there and so now we’re ready to find the

volume um this is a 2 right here but we’re going to be integrating from 0

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to 8. that’s all we need all right so here we go so the volume will be

zero to eight pi and then our r of y which is cube root of y squared d y

so let’s integrate this out this will just be pi

and then 0 to 8 of what y to the what two thirds so pi

and then we’re going to be looking at you know three-fifths y to the five-thirds

and then zero to eight and so then this is just three over five pi

and then times the 32 or said differently 96 pi over five

right and so there’s the volume right there excellent

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and i thought we would just do one more um this time we won’t revolve around the

y-axis let’s do one where we revolve around something else but uh this video

is for uh vertical lines so let’s look at another one

all right so here we go we’re looking at the square root of x

y equals zero x equals nine and we’re going to revolve around the line uh x

equals nine uh now before i do this video let me

just mention though that what happens if they’re not flush in other words

we’re this is part of our boundary is part of what we’re revolving around and

so far the disc method and in the last video in this video

um er what we’re revolving around was flush with with our um object that we’re

revolving and so when we look at the washer method of coming in the next

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video we’ll answer that question but here we go let’s look at

the the uh the region here and see what’s happening

so we’re looking at square root of x somebody coming up through here like

square root of x and then i’m going to be revolving

around the line x equals 9 and i’m also bounded by x equals 9. so x equals 9 is

just a vertical line that’s coming right through here

and i’m going to be revolving around it the x-axis

and the y-axis now when we revolve around this right here we’re going to

get this other part over over here let’s say that’s let’s say it’s right there

and so we’re going to take this region right here

and that’s a lot of shading to do i’ll just try to outline it and then

fill it in briefly but we’re going to be taking this region

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here and revolving it around x equals nine and i usually like to draw my um

[Music] the representative disk so let’s say we have here

this disk coming through here then i’ll give it some depth something like that

now we have to be careful here when we figure out what is the radius function

because remember we’re we’re measuring the r of y

relative to the axis of rotation so here what i’m trying to find is what is

the r y here and so know this is the cross-sectional area here a to the y a of y

that’s the cross-sectional area right there to find the a of y we need to know

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the the r of y so the r of y is going to be given as 9 minus y squared

right and so how do we get that right so this is the

y equals x squared so in other words x equals y squared y is greater than zero

and so that that’s going to give this right here this part right here

but y is greater than 0 so we just have the upper part right there

so in other words i took this solve for y i’m going to solve it for x

and we get x equals y squared right there and so we’re going to use that and so

our r y here is going to be the full nine

the full nine right here because that’s nine right there white x equals nine

nine and then take away this part in here and this right here is from the y

axis up to the curve which is the y square right here so there’s how we get

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the r of y right there all right so now we’re ready to find the

the volume so the volume will be we’re going to be integrating from zero

up to the height here so where’s this intersect right here this is where x

equals nine x equals nine y is three so we’re going

to be integrating here from zero to three

let’s actually go ahead and put that on the sketch right here zero to three

and then we’re going to be integrating pi and then our r of y squared so nine

minus y squared squared d y so there we go let’s integrate this so let’s pi

and then zero to three and then we’re going to expand this out

let’s say y to the fourth minus 18 y squared plus 81 d y and then pi

and now let’s go ahead and integrate this out y to the fifth over five

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minus 18 y to the third over three plus 81y and then zero to three

all right so very good and substituting into the three to all of that

we’re going to get here pi and we’re going to get 243 over 5 minus 162 plus 243

and so in the end we get 648 over 5 pi and that will be the exact area of that

solid that we get from revolving around x equals nine all right so

i hope that you enjoyed this video um and i’ll see you in the next one have a

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