(AB/BC) I don't understand Volume by Slicing, Disk, Washer, Cylindrical Shells

[silent pasture]

So I know how to do them, but i'm just doing them without really understanding what's going on. I'd like an explanation for each method, visualization, and please specify how depth of an object is introduced when working with each method's formula.

Thanks!

[fair hollow]

definite integration and volumes of revolution?

slicing(purple): y or f(x) is your radius of each slice, add up all the (pi)r^2 slices.
disk(yellow): x and y flipped cuz your object rotated about y axis(hence integrate wrt dy) and x is your disk radius
washer(blue): bun(outer, f(x)) - core(inner, g(x)) = donut
cylinder(red): flipped washer for same reasons as disk

depth is specified by the upper and lower limits of your integral(b and a), without them youd be doing indefinite integration and your depth would be x (or y if you int wrt dy)

[silent pasture]

@fair hollow
Ok. So here's what I understand:

when trying to find the volume in the region bounded of a symmetrical function about some axis--and touching-- i can use the slide method bc the rotation creates a circular shape in the y-z direction, thus pi*r^2 for each instance (let's say function was about x, points of integration will be front the x-axis that bound the region).

disk method is used when a part of the function is identifiable in some region bounded by some axis (and touching). it will be rotated creating a circular shape in the y-z direction, thus pi*r^2 applies again like in disk.

washer method is used when the region is not bounded by some axis, creating a donut shape when rotated about that axis bc the closest region, when there are two regions, to the axis is missing. Pi*r^2, blah, blah (same reason as disk and slice)

let's say i was talking about depth regarding slicing method. the purple object depth would be the Xs, right? some a & b value picked from the x-axis bc i'm doing the problem in terms of x, i.e., dx
and it's getting the depth of each slice, finding that slice's volume, then adding it, right?
...
but the red object for the washer method's depth, after subtracting to find just the region, would find the depth of the, in terms of y (going up), find the volume of that, then add it, right?

i still don't understand cylindrical method's purpose though

I think i'm understanding everthing else rotary. just cylindrical method's purpose. Oh! And maybe cross-sections. still don't know why i draw that line connecting graphs then finding the point on the ends creating some kind of function out of that to integrate

🙂

[silent pasture]

But i'm still lacking critical understanding

[magic dust]

Do you understand how normal integration works?

In terms of a Riemann sum

[silent pasture]

yup

[silent pasture]

it's just that i don't understand why we use cylindrical shells to find volume?

when do we prioritize it over washer or disk?

[magic dust]

I think it's just the same thing as washer tbh

I never learnt volumes of revolution as "methods" though

[silent pasture]

honestly, i understand the formulas for the most part, but it's mostly just memorization. Like, I don't understand how we can assume that another axis--z-- can be considered. That's Multi-var Calc

[magic dust]

You don't need to consider another axis

[silent pasture]

when creating a 3d shape there are 3 axes

[magic dust]

Sure

But the 3rd one is not important

Look at the diagram I've drawn

[silent pasture]

i see area

bounded by a to b

[magic dust]

Let A be the area from a to b under the curve

We have
A = sum of rectangles

Each rectangle has an area of base x height

The base is dx

Which means a very tiny change in x

And the height is f(x)

So the area of each rectangle is f(x) dx

And then we sum them

A = ∫[a, b] f(x) dx

That's what this means

∫ means the limit of the sum when dx goes to 0

[silent pasture]

I understand. But how can a 3d shape has formed? is it when the integral of area is taken?

[magic dust]

One second

So imagine we now take the same rectangles

And rotate them around the x-axis

What shape do we get?

[silent pasture]

some disk

[magic dust]

Yeah

We can now add those disks together to get a volume

The volume of a disk is:
V = π r^2 h

In our case, r = f(x)

h = dx

[silent pasture]

ok

[magic dust]

And we need to sum and take the limit again

So we have V = ∫[a, b] π y^2 dx

Does that make sense?

[silent pasture]

i'm following

[magic dust]

Do you see how that formula arises?

[silent pasture]

dx is infinitesimally small meaning we can't determine the actual depth of each rectangle, right?

[silent pasture]

yes

[magic dust]

Well we're taking the limit so the depth is essentially 0

What we're saying is that the approximation gets closer to the real value as you use more rectangles

And if you extend that process infinitely, you get the exact area

[silent pasture]

ok

[magic dust]

Okay so now let's imagine we rotate them around the y-axis instead

What shape do we get?

[silent pasture]

some disk shape again

[magic dust]

Nope

[silent pasture]

?

a donut

bc it's not touching

[magic dust]

= a cylindrical shell

[silent pasture]

oh ok

[magic dust]

Let's just say a = 0 for now

So it is touching

Actually nvm

So again, let's work out the volume of a cylindrical shell

V = π (r2^2 - r1^2) h

[silent pasture]

ok

[magic dust]

Now imagine taking the cylindrical shell and unwrapping it

To make a cuboid

[silent pasture]

a very thin cuboid

[magic dust]

Mmhmm

[silent pasture]

toilet paper for an example

?

[magic dust]

It doesn't even need to be very thin at this point

Although it will be infinitely thin when we take the limit

But sure

[silent pasture]

wait, why is that?

[magic dust]

Because we're just setting up an approximation to the volume

[silent pasture]

oh, right

[magic dust]

Then we're going to make them smaller (infinitely so)

[silent pasture]

i misunderstood the unwrapping

[magic dust]

Like this

[silent pasture]

so you've cut it, then bent it to form a cuboid

?

[magic dust]

Yep

The inner radius will be slightly smaller than the outer radius but the difference becomes smaller and smaller as the shell gets thinner, so a cuboid is a valid approximation

Happy so far?

[silent pasture]

yes

[magic dust]

The thickness of the cuboid is dx

The height is y

And the other side is the circumference of the cylinder

Which is 2πx

So we have 2πxy dx

And then we just have to sum them and take the limit

V = ∫[a, b] 2πxy dx

[silent pasture]

sorry, i'm lost again. what is the length of the cuboid if the thickness is dx? and what do you mean the "other side" is circumference?

[magic dust]

So our cuboid has 3 side lengths right?

[silent pasture]

i don't understand. 6 sides, no?

oh

sorry

i see, i see

lol

[magic dust]

Like this

And our volume is trivially abc

[silent pasture]

yes

[magic dust]

c is the thickness of the cylinder

Which is the thickness of our rectangle from before

So c = dx

The height b is the value of our function

So b = y

And a is the circumference of our cylinder

From when we unwrapped it

[silent pasture]

ok

i get that

[silent pasture]

but i don't get this

@magic dust nevermind

i do

[silent pasture]

yeah

[magic dust]

So we have
dV = 2πxy dx

[silent pasture]

ok

[magic dust]

dV = infinitesimal chunk of volume

[silent pasture]

wait

pi*y(b^2-a^2), no?

chain rule

[magic dust]

?

[silent pasture]

dV = change in volume

[silent pasture]

.

[magic dust]

You can think of dV as an infinitesimal chunk of volume

[silent pasture]

ok

[magic dust]

V = ∫ dV

What that means is that the total volume V is equal to the sum of all the infinitesimal volumes

1 infinitesimal volume in our case is the volume of this cylindrical shell

Which we're approximating as the volume of a cuboid with the correct dimensions

[silent pasture]

yes

[magic dust]

(this approximation becomes perfect in the limit so we're fine)

So dV = height * base * thickness

dV = (y) (2πx) (dx)

dV = 2πxy dx

V = ∫ 2πxy dx

You don't really need a specific washer method tbh

[silent pasture]

ok

[magic dust]

The disk method is sufficient if you understand how to use it

[silent pasture]

yes

i understand this now. thank you

lastly, cross-sections

i understand how to use them, but not why. Is it basically using one rectangle as an approximation for other rectangles as x changes, then when added together find the area of the region?

[magic dust]

Specifically, to find the volume between two functions, you can just take the volume for the outer function and the inner function separately, and subtract the volumes

[magic dust]

What do you mean by "using one rectangle as an approximation for other rectangles"?

[silent pasture]

i'll draw it out

@magic dust

[magic dust]

What rectangle are we approximating?

[silent pasture]

the orange one

and the yellow ones

f(x)-g(x) changes when x does

[magic dust]

How are we approximating it?

[silent pasture]

the orange, f(x)-g(x), is a radius. when x changes, we add the next value that comes from the radius. Nevermind, it's not approximating. Bad word choice.

When adding all these values up then area is found

Is my understanding solid for this?

[magic dust]

Can you rephrase this?

[silent pasture]

two functions, f and g, on the interval [a,b], are used to find radius (r(x)). since the radius changes with every change in x, taking the definite integral will find the area of the bounded region by adding up all the values r(x)

sound?

[magic dust]

f(x) - g(x) is not the radius

f(x) is one radius and g(x) is a different one

[silent pasture]

oh?

[magic dust]

The difference between f(x) and g(x) is not approaching 0

So we can't do it that way

[silent pasture]

but the difference between f(x) and g(x) are finding one y value

if we add up all the y values we find area

[magic dust]

For this example, yes

But for the cylinder version, it doesn't work

[silent pasture]

yes, yes. area vs. volume

[magic dust]

Yellow area = blue area - green area

[silent pasture]

but i guess this is also a more complex way of thinking of things rather than just taking the area of the top function and subtracting it by the area of the bottom one

[magic dust]

blue area = ∫ b(x) dx

green area = ∫ g(x) dx

yellow area = ∫ [b(x) - g(x)] dx

yellow area = ∫ b(x) dx - ∫ g(x) dx

And this isn't surprising since
∫ [b(x) - g(x)] dx = ∫ b(x) dx - ∫ g(x) dx

But if we are rotating them to get a solid of revolution, we have:

blue volume = ∫ π b(x)^2 dx
green volume = ∫ π g(x)^2 dx
yellow volume = blue volume - green volume = ∫ π b(x)^2 dx - ∫ π g(x)^2 dx = ∫ π [b(x)^2 - g(x)^2] dx

=/= ∫ π [b(x) - g(x)]^2 dx

Does that all make sense now?

[silent pasture]

It does. Thank you very much, Green

[magic dust]

Has it solved your issue?

[silent pasture]

i won't know until i do a problem.

i'll get back to you on that

[magic dust]

Hopefully this has given you some understanding of the structure of the formulae at least

[silent pasture]

it has. Usually when doing these problems I feel an empty space in my mind, but just regurgitation.. well, memorization

so thank you

[magic dust]

Yeah so hopefully you're now able to create the formulae yourself rather than memorising them

Btw, it's generally possible to use either method for any given problem

Let's say we want to find the yellow rotated volume (around the y-axis)

We can either do cylindrical shells like this

Or disks like this (and then we need to subtract from the green rotated volume to get it)

Does that make sense?

[silent pasture]

yes

[magic dust]

You can use that to practice and test your answers as well. Try doing it both ways and you should get the same thing

[fair hollow]

theres no "depth of each slice", depth of each slice is infinitesimally small cuz thats what calculus is capable of. the depth is the height of your object.

"but the red object for the washer method's depth..." youre not FIRST subtracting small ring from big ring THEN integrate, thatd be ∫(f(x)-g(x))^2 dx. youre FIRST finding big volume and small volume THEN subtracting inner from outer, which is ∫f(x)^2 dx - ∫g(x)^2 dx. if i assume that by "then add it" you mean integrating, then no theres no such thing as adding slices after finding some volume(of slices?), since each slice is infinitesimally small unless youre using riemann sum or trapezoidal rule

cylinder is when the washer is upright, all it mathematically means to you would be to int wrt dy. if you can never see an upright washer you can always inverse its function, changing y into x and x into y, making it a washer, so you can int wrt dx

[silent pasture]

thanks

also, when washer method, why is it that you can't subtract to find area, then rotate that area to form a solid? why must it be rotate existing area to form solid and subtract that from other formed solid?

[fair hollow]

1. a^2 - b^2 = (a+b)(a-b) != (a-b)^2, so if a^2 - b^2 works, (a-b)^2 doesnt work
2. these two circles have the same bounded area between the 2 functions (top/bottom semicircle) but make donuts of different sizes

[silent pasture]

i see it now. ok, thanks!

[fair hollow]

np

[silent pasture]

hey, just ran into this problem

[fair hollow]

essentially its a half spun diamond

[silent pasture]

i understand cross-sections make-up a solid. so when it's symmetric about the x-axis, or whatever axis you're rotating about, we should cut it in half then rotate--I mean that's how we should think of it before taking the "slices"?

[fair hollow]

yeah thatd work

you mean cutting in half like this?

[silent pasture]

yeah

[fair hollow]

more accurately, overlap the area of whats below the axis to that on top of the axis

such that blue is the area rotating

[silent pasture]

so it foldslike a sandwich? when it's revolved it fills up that space though

[fair hollow]

yes it would

but since this isnt “cutting in half” i thought id clarify