help-13

[finite badger]

differential systems is the topic

[cedar kiln]

@finite badger Has your question been resolved?

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@finite badger Has your question been resolved?

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[crimson sedge]

why does ln^(-1) go to e?

[crimson sedge]

its not the derivative yet, i just got confused how that happened

[autumn fox]

It thinks you mean the inverse of ln(x) which is eˣ

[crimson sedge]

oh

so 1/lnx does not equal e^x?

[autumn fox]

No

[crimson sedge]

i see

thank you for clarifying

.close

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[autumn fox]

[crimson sedge]

.close

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[frank bronze]

i dont undetstand

[frank bronze]

right = positive and left = negative yes?

[cyan eagle]

correct

[frank bronze]

that does not make sense

we are saying if function is above x axis it is goign right?

and vise versa

[rigid lava]

it's not moving up and down

It's moving left and right

[frank bronze]

well positive means above x axis and negative means under

[rigid lava]

I mean

positive just means greater than 0

and negative means less than 0

On the real number line

the positives are to the right of 0

Analogous for neg

[frank bronze]

like in what sens

[rigid lava]

every

[dusk goblet]

there is no above or below the x axis

the particle is moving ON the x axis

going back and forth to the right and left

[frank bronze]

hm

[dusk goblet]

do you mean v(t)?

[frank bronze]

yes

[dusk goblet]

if v(t)>0 then yes it means it’s moving to the right

and if v(t)<0 then it’s moving to the left

it’s just a standard convention

[frank bronze]

how would i solve this]

[dusk goblet]

set v(t)=0

to find tR

[frank bronze]

but i dont get it

like

[dusk goblet]

is this not your work?

[frank bronze]

Tr = 1.426, but i dont understand y we have to set vt = 0 to find that

[frank bronze]

its the answer key

[dusk goblet]

so ln(t^2-4t+5)-0.2t=0

ln(t^2-4t+5)=0.2t

[lone plaza]

Its calc medic i assume

[dusk goblet]

then just graph it with a calculator

and calculate the intersection point

[lone plaza]

This was AB frq 2

[dusk goblet]

ahh

[dusk goblet]

do you know why you set v=0

[frank bronze]

no

[dusk goblet]

ok

so

for something to be at rest

[frank bronze]

ah

[dusk goblet]

what must its speed/velocity be

[frank bronze]

needs speed of 0

[dusk goblet]

yes

[lone plaza]

mhm

since its not moving

[dusk goblet]

because it’s not moving

just graph the function

and find the zero

[lone plaza]

For the second part of the question I just tried a point within the interval and checked if it was positive or negative

[dusk goblet]

and you can see from the graph if its greater or less than zero before the zero

[frank bronze]

so what we are really doing is just solving for t when v(t) = 0

[dusk goblet]

mhm

you have a graphing calculator yes

[frank bronze]

well then where does the Tr come from

[lone plaza]

Tr is that point

[dusk goblet]

tR is the time that satisfies that equation being equal to zero

[lone plaza]

where v(t) = 0, tR is the t value which makes this equation true

basically tR just means Time when the particle is at Rest

[frank bronze]

okay and because we solved for it and got a positive number as a t value, that is why we are saying it is positive (aka moving right)

?

[lone plaza]

nono

at tR the object isnt moving

so when we solve for v(t) = 0, the t value just means the time when the object isnt moving

that is the answer to the first part of the question

the second part asks within the interval from time t=0 and tR when the object is at rest, what direction is the particle moving

so the value you get for v(t) = 0 doesnt correspond to left or right, it is just a time value when the object isnt moving

[dusk goblet]

yea t is just the time it tells you nothing about the objects motion

and t is always going to be restricted >0 for these motion problems because you can’t have negative time

[frank bronze]

how can you say the particle is moving a direction when its at rest

[dusk goblet]

it was asking for the time before it came to rest

[frank bronze]

i just dont undetsand this part

[dusk goblet]

like which direction was the object moving in before it came to rest

[frank bronze]

wat

[lone plaza]

ok

theres two different parts to the question

part 1

[dusk goblet]

it’s just asking if it was moving to the right before it stopped

[dusk goblet]

or left before it stopped

[lone plaza]

part 2

[frank bronze]

BUT HOW CAN WE FIND THAT

it does not make sense logically

if you track the object from the left or rigth side, either way you will come to the point where vt = 0

[lone plaza]

have you tried graphing the function

[frank bronze]

no

[lone plaza]

ok so part 1

over the interval 0<t<2 where is the particle at rest

we have already stated it is at rest then v(t) = 0

so we can see that v(t) = 0 at t = 1.426

then the point that is at rest is also known as tR according to the problem so 1.426 = tR

then for part 2 we are finding whether or not the function is moving left or right between 0< t < tR which we can rewrite as 0 < t < 1.426

since the graph is of velocity, we know that when the function is greater than 0, it is moving to the right, and when it is less than 0, it is moving to the left

since from 0 < t < 1.426 the graph v(t) is greater than 0, we know the particle is moving to the right

[frank bronze]

hm

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[dusky hare]

Please help me math masters

<@&286206848099549185> ...

P(x) = axⁿ+¹ + bxⁿ + 1

divisible by x² - 2x + 1

a = ?

Its looks easy but i dont know how to do that

@lone plaza @modern sparrow can u help me?

[wraith dagger]

misha

misha

[native rain]

solve the system of equations to get your answer

[cedar kiln]

@dusky hare Has your question been resolved?

[dusky hare]

Thank you so much

I got it

[cedar kiln]

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[vapid otter]

not math, but can anyone help me with a chemistry problem?

[vapid otter]

the bond S--P

what is the dipole movement

[fringe moth]

No

[vapid otter]

is it ^δ-S --> P^δ+, or P^δ+ --> ^δ-S?

@cerulean sail

[cerulean sail]

Oh no, I don't know, just amused that there's a chem problem that isn't even anywhere close to math here
(also re. the below, #old-network literally has a chem server )

[vapid otter]

theres no chemistry server though

HELPP

thanks i found it

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[crimson sedge]

2.36

whats going on

ive read a dozen stackex posts, some saying its a contradiction others saying its a contrapositive

some saying its all about perspective

they are showing the negation of the result implies the negation of the assumption?

[crystal raptor]

this is a common mistake many people do, a lot of contradiction arguments are actually redundant and are really contrapositive

That proof is directly showing that $\cap K_\alpha$ empty $\implies$ there is some subcollection that is empty

[wraith dagger]

ΣAC

[crystal raptor]

which is exactly the contrapositive

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@crimson sedge Has your question been resolved?

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[balmy yoke]

how do I solve a)

[balmy yoke]

I'm not sure how to start

I need help with any question and I can do the rest on my own

[shy widget]

im assuming you haven't taken differentiation?

[balmy yoke]

no idea whats that

[shy widget]

ok

[shy widget]

its a way to find the exact slope at each point of a curve

but dont worry about that yet

you can estimate the slope

by taking two points very close to each other

and using the slope formula

$m=\frac{f(a)-f(b)}{a-b}$

[wraith dagger]

The د

[balmy yoke]

oh I see

[shy widget]

so to estimate the slope of the first function at 3...

use 3 and a point very close to 3

like 3.01

makes sense?

[balmy yoke]

yeah I used 3 and 3.01, and got 0.01 as the slope

is that right?

[shy widget]

looks about right

[balmy yoke]

coool thank you

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[balmy yoke]

hey uh one more question please

number 2

I get the slope as -1.52

which doesn't seem right

.reopen

[cedar kiln]

✅

[balmy yoke]

this

I chose sin(90) and sin(90.01)

[shy widget]

yeah not quite right

[balmy yoke]

$\frac{\sin90.01 - \sin90}{0.01}$

[wraith dagger]

rynite

[shy widget]

yeah

make sure you're on degree mode

[balmy yoke]

p sure im on deg mode

[shy widget]

its giving me a right answer

[balmy yoke]

wait I I meant the answer it gave me is
-1.52 * 10^-6

forgot the last part

does that make a difference

[shy widget]

ahh yess

that means that its about -0.00000152

[balmy yoke]

oh okay, thanks for confirming

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[meager flare]

vec(b) - vec(a) = vec(AB)

apparently

I can show this by doing the following equation: vec(a) + vec(AB) = vec(b)

however visually I dont see how vec(AB) is equal to vec(b) - vec(a)

[mortal minnow]

Say this is your graph

I just connect A and B because that is the vector AB

To get from A to B, you must go backwards first.

You travel down OA towards the origin

This is the vector a, but in reverse

Hence, you have travelled -a

Now you are at the origin, you have to get to B, well you xan travel along OB, and OB is just the vector b

So you have AB = -a + b

Looks neater to say b - a

Hence, AB = b - a

[neon raven]

you can also look at it this way

when you subtracting vec(a) from vec(b), you're subtracting their heights and their lengths

when you subtract their heights, you get the height between point B and point A

when you subtract their lengths, you get the length between point B and point A

and vec(AB) contains them both

the grey line there is parallel to the x-axis

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[raw parrot]

Quick question, but what does

P(x,y) represent in regards to permutations? Specifically the x and y.

[mental trail]

context?

[austere ember]

this looks like slightly different notation but without more context, i'd assume this represents the number of ways to select a collection of y objects from a set of x objects where the order of the y objects matters

[raw parrot]

Context is you have two jobs, given this many people how many ways can you fill the job

I'm assuming x would represent the total people and y would represent the number of jobs?

[neon raven]

that's partial permutation

P(n,k) = n!/(n-k)!

in this case

it's got many different notations

these are the ones from wikipedia

k-permutations of n

[raw parrot]

So as an example let's say we have 30 people and two offices, and 2 people can get an office

Would that be P(30,2)?

[neon raven]

do you mean that 2 people can be in 1 office or that only 2 people can get an office?

[raw parrot]

one person gets an office from the 30

it can only be used by one person at a time

[neon raven]

then in that case yeah, you use P(30,2) which gives you 30!/28! = 30*29

[raw parrot]

sweet, ty!

.clsoe

/close

[neon raven]

.close

[raw parrot]

.close

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[pseudo trellis]

How would I setup this problem if the about is x=2?

Find the folume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.

y=x^3, y=0, x=1, about x=2

[pseudo trellis]

I'm doing the integral with respect to y, so would I have to change $y=x^3$ to ${3}\sqrt{y}=x?$

[wraith dagger]

Someone

[pseudo trellis]

then convert y to y to be y-2 as the about is x=2?

[timid crypt]

not necessarily

[dusk goblet]

x=y^(1/3)

[timid crypt]

you can do shell method

where h = x^3 and r = 2 - x

i believe that would work

[carmine bronze]

The problem states disk or washer.

[timid crypt]

oh oops

yea then do in terms of y

[pseudo trellis]

what do I do about the x=2?

In this problem, I did -1

or is that only if there's two curves?

Would I also use a disk?

[dreamy void]

Didn't you ask that recently?

[pseudo trellis]

probably lol

all my notes were wiped for some reason 😔

im integrating with respect to y and using a washer

[dreamy void]

#help-15 message

[pseudo trellis]

we're on question 15 now tho

[dreamy void]

ah ok

[dreamy void]

yes

[pseudo trellis]

so, the question i have is

if there's only one curve (y=x^3) and not a second curve as it's just a line (x=1), we don't need to subtract the about x=2 from the equation?

[wraith dagger]

Someone

[dreamy void]

Well since we are rotating around x = 2

the radius moved

[dreamy void]

just like in that

previous example

[pseudo trellis]

right, what I did in the previous example is just subtract the 1, so would this problem be the same with subtracting 2?

[cerulean sail]

Shouldn't your little r be squared too?

[dreamy void]

yes i forgor thanks

[pseudo trellis]

ohhh, so since there's a difference of 1 between the x=2 and x=1, we're subtracting 1?

[cerulean sail]

And that's the way I think about it, whatever the "radius" ends up looking like, in that case the distance between x = 1 and x = 2 is 1, so subtract 1^2 = 1

[pseudo trellis]

I see, i see

[torn sorrel]

Hey if you smart at math can you guys pls help me out

[pseudo trellis]

so i'm at this point

now, we need to find the bounds, right?

[torn sorrel]

I gotta project that if I fail I fail the whole 10th grade

[cerulean sail]

Also note that you want the [horizontal] distance from the curve x = y^{1/3} to the line x = 2 as the big R

(also @solid juniper I see you )

[pseudo trellis]

so we would set them to each other?

y^{1/3} = 2?

that would give us the upper bound, no?

lower bound is 0

[dreamy void]

yes

[pseudo trellis]

we would not use y^{1/3}-1 = 2?

[dreamy void]

that's equivalent to 2^3 = y

[cerulean sail]

Wait hang on a moment, I might be a few steps ahead of where I think we are

[dreamy void]

you are looking at the horizontal axis

[pseudo trellis]

right

[dreamy void]

subtracting - 2 is like imagine we shift our problem by 2 units down

the integral bounds remain

[pseudo trellis]

wait what, it's -2 now??

that's what i had

i do not understand any of this

i'm getting contradicting information i feel

[cerulean sail]

Yea I think I was ahead of where I thought

I thought you already had the big R and needed to work out the little r

[pseudo trellis]

i have nothing

[cerulean sail]

The point here was that you want to have the horizontal distance from x= y^{1/3} and x = 2, that's 2 - y^{1/3}, and becomes the big R

[pseudo trellis]

why is it -2????

someone said -1????

i'm so lost

holy shit

[cerulean sail]

What I was saying was that you subtract, from that one, the little r^2, which is the 1^2

So like-

[pseudo trellis]

so subtract 2 from big R?

which is y^{1/3}-2

[cerulean sail]

That is your big R

[pseudo trellis]

little r is y^{1/3}-1?

that doesn't make sense

[dreamy void]

r is the distance between axis of revolution and the function

[pseudo trellis]

i still have no damn idea why we need big R and little r

[pseudo trellis]

little r is 2-1 then?

[dreamy void]

So in order to get this distance r you do 2 - y^(1/3) or y^(1/3) - 2 (it's getting squared so it doesnt matter which way)

[pseudo trellis]

1/3 then 1/2

what

[dreamy void]

😂

1/2?

i did a typo lmao

i am sorry

relax

[pseudo trellis]

okay so that is little r?

[dreamy void]

yea

[pseudo trellis]

isn't big R the same thing?

y^{1/3}-2

[cerulean sail]

As in this one's the little r

[pseudo trellis]

what?

little r is y^{1/3}-1 makes sense to me

[cerulean sail]

We're saying that big R is like the "outer radius" (as per here), but the little r is the "inner radius"

[pseudo trellis]

big R is the pink, little r is the red, right?

[cerulean sail]

I messed up the highlighting, one second again (couldn't find colours either )

Both in those are big R

[pseudo trellis]

oh

[dreamy void]

I think we need this to do

[pseudo trellis]

idk if this helps, but this is what i have

R and r surely cannot be the same

[cerulean sail]

Argh my fingers are too fat to draw it out

[pseudo trellis]

so little r is indeed -1?

[dreamy void]

We have two integrals

we want the volume between 0 and 1 to be gone

[pseudo trellis]

what

why do we have 2?

when and how?

[dreamy void]

We want to get the thing below

[pseudo trellis]

the sideways graph may be confusing me

[pseudo trellis]

,r 90

idk how to do it with the bot

[pseudo trellis]

,r

[dreamy void]

it's y to x

not x to y coordinate system

because we do now everything in terms of y

x = y^(1/3) is a root function

[pseudo trellis]

right

[dreamy void]

So

Look upper corner left

This is what we would actually get

but we want to subtract that tiny volume (see upper right corner)

that's why we get 2 different integrals

We are also rotating around 2 different axis

[pseudo trellis]

this is making 0 sense to me im sorry

like i'm so lost you have no idea

can we go back hella far?

[dreamy void]

upper left corner is around x = 2 hence why we do 2 - f(y) or f(y) - 2
upper right corner is around x = 1 hence why we do or 1 - f(y) or f(y) - 1

[pseudo trellis]

that's all i understand

[dreamy void]

ok where

[pseudo trellis]

nothing else

[dreamy void]

I suggest to draw it in terms of y and x not x and y

[pseudo trellis]

it would cause more harm than good

[dreamy void]

it's basically 90 degree rotation counter clockwise and then reflection at y-axis

[pseudo trellis]

my brain wont work if i draw the basic graph differently

i've never drawn it in terms of y so it'll screw me over

[dreamy void]

this is your bread

actually maybe washer method is for you complicated

[pseudo trellis]

i hate the washer method, yes

[dreamy void]

I got 31pi/10 btw let's see with the other one

shell method was it?

[pseudo trellis]

all of these are disc/washer

no shell yet

shell is 6.3, we're 6.2

been doing 6.2 for 4 days

i'm actually fucked

[dreamy void]

well disc/washer is the same, isn't it?

[pseudo trellis]

shell has a 2pi or something

2pi integral bounded between a and b of x f(x) dx

[dreamy void]

,, 2 \pi \int_a^b x \cdot f(x) : \dd x

[wraith dagger]

𝔸dωn𝓲²s

[pseudo trellis]

yes

i cant use that on these problems

[dreamy void]

well then you have no choice

but to understand

[pseudo trellis]

i know, we just have to back up

[dreamy void]

shell method would be better for this

but anyway

[pseudo trellis]

i'm here

i erased stuff idk

[dreamy void]

this it right

[pseudo trellis]

so we need to find that volume

[dreamy void]

the yellow

[pseudo trellis]

yes

[dreamy void]

it's the same what I sent

If we integrate from 0 to 8 (y-wise)

you will take that little piece with you

[pseudo trellis]

which is our second integral, right?

[dreamy void]

yes

the red one is the second integral

[pseudo trellis]

oh

[dreamy void]

which we want to subtract

that is it

[pseudo trellis]

okay

[dreamy void]

so we have two different volumes

we get the red by rotating around x = 1

we get the yellow + red by rotating around x = 2

because we can't find a way to integrate between x=1 and x=2

we need to subtract

lol #help-9 message

your counterpart

[pseudo trellis]

😭

okay so i kinda understand the need for two integrals now

[dreamy void]

yes

take a look again

the reason for subtractin 2 and 1 is because of the radius

radius is the distance between axis of revolution and function

you can see

[pseudo trellis]

like this?

[dreamy void]

yes

you cannot achieve the red one

by rotation around x = 2

[pseudo trellis]

right

[dreamy void]

if you close your eyes and imagine what's happening like einstein haha

charbit left ;-;

[pseudo trellis]

dont blame him

im actually stupid as hell

it's like teaching math to a 2 year old

this concept is so god damn hard for no reason

it really should not be this hard

[dreamy void]

⚔️ one more time and you're done

dont be hard on yourself

it's a new concept for you

that's normal

that's why this server exists

[pseudo trellis]

i got 16pi/5 for the first integral

[dreamy void]

nice

the other one?

[pseudo trellis]

1pi/10

31pi/10 is the final answer

[dreamy void]

⚔️

I got that too

[pseudo trellis]

yayayayayayay

[dreamy void]

and don't you ever compare yourself to a 2 year old ⚔️

😡

[pseudo trellis]

💀i have the brain capacity of one

lmao

[dreamy void]

⚔️

[pseudo trellis]

okay i have one more problem before a short break because i totally didn't spend a solid 4 hours on only question 15

[dreamy void]

you said solid

is this a pun or some

[pseudo trellis]

😭

[dreamy void]

😂

[pseudo trellis]

we need two integrals this problem, right?

[dreamy void]

The closed region around x = 3?

[pseudo trellis]

i think closed region

[dreamy void]

I suggest again to draw it in terms of y not x

[pseudo trellis]

it says:

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.

y=x^2, x=1-y^2, about x=3

i dont think my brain will work if i draw the graph in terms of y

[dreamy void]

ok

wait

[pseudo trellis]

the first step i need to do is find the bounds of y?

[dreamy void]

your sketch is wrong

[pseudo trellis]

wot

[dreamy void]

,w plot y =x^2 and x = 1-y^2

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by adoniszgks

[dreamy void]

It's x = 1-y²

not y = 1-x²

ah actually no

x =y^2

[dreamy void]

no y = x^2

[dreamy void]

It's because the y-axis is the x-axis

[pseudo trellis]

oh yikes

[dreamy void]

READING THIS IT'S

y=x^2,
x=1-y^2,
about x=3

ok?

[pseudo trellis]

yeah, lemme fix one sec

[dreamy void]

damn that is uglly

[pseudo trellis]

what did you call me?

jkjkjk

[dreamy void]

hahaha

you got nerves

[pseudo trellis]

we still gotta find the bounds with respect to y?

[dreamy void]

so you did it now in terms of y haha

after all

[pseudo trellis]

oh god dang it

[dreamy void]

it's fine

[pseudo trellis]

i suck at graphs lol

i'll try it

[dreamy void]

Let us think

[dreamy void]

are you kidding

x = y²?

[dreamy void]

Then y =x² is wrong

and I my gut tells me it is

[pseudo trellis]

wot

[dreamy void]

read what you wrote

explain what you drew

x = y²+

[pseudo trellis]

GHFUWIJAF okay i'm doing it in terms of x

lmao

[dreamy void]

pls send the original

[pseudo trellis]

the original?

[dreamy void]

yea

[pseudo trellis]

problem?

[dreamy void]

yes

I got a problem

⚔️

[pseudo trellis]

x=y^2, x=1-y^2, about x=3

that's it

[dreamy void]

ahhhhhhh

man

ok

[pseudo trellis]

https://archive.org/details/calculus-early-transcendentals-8th-edition-2015/page/446/mode/2up

17

lmao why did i buy this $50 book when i could've just used this website

smh

[dreamy void]

rich

looks like a spider

[pseudo trellis]

okay okay

[dreamy void]

yes

[pseudo trellis]

that's in terms of x

[dreamy void]

So

[dreamy void]

y*

We use washer

[pseudo trellis]

yeah, like the graph is drawn in terms of x

[dreamy void]

Can you tell me the bigger and smaller radius?

[pseudo trellis]

we need the bounds in y

[pseudo trellis]

we just plug in a random number, like 2 to find that, right?

[dreamy void]

no

you can see it

which function is farther

from the axis of rebv

[pseudo trellis]

the green is overall

x=1-y^2

[dreamy void]

no

the blue

[pseudo trellis]

wot

oh i see

yeah

[dreamy void]

the blue at (0,0) is farthest

[pseudo trellis]

yeah that makes sense

so that will be our first integral

gotta find the bounds on it though

which will be 0 to 1, right?

[dreamy void]

find the intersections

[pseudo trellis]

wait nononononono

im dumb

[dreamy void]

1-y²=y²

[pseudo trellis]

yeah

[wraith dagger]

Someone

[dreamy void]

nice

[pseudo trellis]

we need the plus only

[dreamy void]

depending on what you do

I suggested two ways

[dreamy void]

remember we integrate wrt y so the bounds are in terms of y!

[pseudo trellis]

oh i see it now

im stupid

[dreamy void]

stop

[pseudo trellis]

i was thinking there was a y=0 for some reason

lol

[dreamy void]

stop calling yourself stupid

😭

else

⚔️

[pseudo trellis]

tada

okay now we need big R and little r, right?

[dreamy void]

yes

be careful

Which is farther away

from x = 3

[pseudo trellis]

the blue, no?

[dreamy void]

yes

So R(y) = ?

[pseudo trellis]

i'm not entirely sure what to do with the x=3 as i think we have to change x=y^2 to something else?

x=y^2-3?

[dreamy void]

no

radius is the distance between function and axis of rev.

So

[pseudo trellis]

so we leave it as R(y) = y^2

[dreamy void]

and do something

The distante at (0,0)

between blue and x = 3

is 3

[pseudo trellis]

right

[dreamy void]

If you only do y^2

0^2 = 0 = R(y)

But it needs to be

3

So

[pseudo trellis]

yeah, so y^2-3

[dreamy void]

or 3 - y^2

doesnt matter

since it gets squared

[pseudo trellis]

me likey y^2-3

lol

[dreamy void]

because you see

ok likey

R(0) = 0^2 - 3 = -3

so techncally "3"

[pseudo trellis]

yeah

[dreamy void]

the same with the green

[pseudo trellis]

2-y^2

[dreamy void]

yes

[pseudo trellis]

or y^2-2

ima do y^2-2

[dreamy void]

not quier

[pseudo trellis]

oh?

[dreamy void]

green is 1 -y^2

so r(y) = 1-y^2 - 3

right?

or confused

[pseudo trellis]

like that?

[dreamy void]

AH

ONE IS y^2 - 3

and the other 3-(1-y^2)

hahahaha

but it doesnt matter

i though you likey

function - radius

[pseudo trellis]

bfiuwahfiwaj

i thought i wrote it down right

i did wrong

[dreamy void]

you mixed it also good

[dreamy void]

no

it's not wrong

but funny

[pseudo trellis]

lol

[dreamy void]

one is

f(y) - 3

the other

3 - g(y)

that's my point

anyway

you wanna keep it that way?

so I can adjust quickly?

[pseudo trellis]

like that?

[dreamy void]

ok

,,\pi \int_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} \underbrace{\left ( y^2 - 3 \right )^2}{= R^2(y)} - \underbrace{\left ( (1-y^2) - 3 \right )^2}{= r^2(y)} : \dd y

[wraith dagger]

𝔸dωn𝓲²s

[dreamy void]

right

[pseudo trellis]

yeah

[dreamy void]

notice

R(y) and r(y)

are even

R(-y) = R(y)
r(-y) = r(y)

ok?

[pseudo trellis]

wot

they aren't even tho

[dreamy void]

EVEN

they

are

because of y²

[dreamy void]

look

we are doing it in terms of y dont forget

[pseudo trellis]

oh then yes

[dreamy void]

So

[pseudo trellis]

i thought you were calling the math equal to each other

[dreamy void]

to make things easier

We can let it go from 0 to

sqrt(2)/2

and the the area twice

that should be familiar calc 1

[wraith dagger]

𝔸dωn𝓲²s

[pseudo trellis]

huh

i don't recall that

[dreamy void]

yes you do

,,\int_{-a}^a x^2 : \dd x = 2 \cdot \int_0^a x^2 : \dd x

[wraith dagger]

𝔸dωn𝓲²s

[pseudo trellis]

oh

[dreamy void]

generally for any even function

[pseudo trellis]

i think

[dreamy void]

look again

whether we go from -sqrt(2)/2 to sqrt(2)/2

or we go from 0 to sqrt(2)/2 but double it

: )

[pseudo trellis]

yeah that makes sense

[dreamy void]

good

,, 2\pi \int_{0}^{\frac{\sqrt{2}}{2}} \underbrace{\left ( y^2 - 3 \right )^2}{= R^2(y)} - \underbrace{\left ( (1-y^2) - 3 \right )^2}{= r^2(y)} : \dd y = 2\pi \int_{0}^{\frac{\sqrt{2}}{2}} \left ( y^2 - 3 \right )^2-\left ( 2 + y^2 \right )^2 : \dd y

[wraith dagger]

𝔸dωn𝓲²s

[dreamy void]

We evaluated the inner

1-y^2 - 3

Notice

[wraith dagger]

Someone

[dreamy void]

((1-y²)-3)² = (1-y²-3)² = (-y²-2)² = ((-1)(y²+2))² = (-1)²(y²+2)² = (2+y²)²

[pseudo trellis]

yeah

[dreamy void]

ok good

[dreamy void]

yea

that's what I got too

[pseudo trellis]

:D:D:D

[dreamy void]

damn

sometimes my soul leave me

[pseudo trellis]

https://archive.org/details/calculus-early-transcendentals-8th-edition-2015/page/446/mode/2up the next problem is 25, which is a bit different, how do you recommend i start? xD

just not sure what it's really asking me to do

[dreamy void]

wow that's nice bookl

$\mathscr R_2$ about $AB$

[wraith dagger]

𝔸dωn𝓲²s

[dreamy void]

haha

[pseudo trellis]

lmao fancy R

[dreamy void]

so

basically between y = 1 and y = x^(1/4)

green

rotating around AB meaning x = 1

I would do it again in terms of y

[pseudo trellis]

okay, yeah

wouldn't it be easier to do it in terms of x though?

[dreamy void]

we are using washer

dont forget

[pseudo trellis]

i still dunno how to tell if i use a washer or disc ngl

washer because it's about x=1?

[dreamy void]

disc is just when we have two radius

[pseudo trellis]

ah, but we have a lot of radius

[dreamy void]

or the other way around

anyway

seems disc is one function

washer involves many

[pseudo trellis]

yeah

[dreamy void]

https://courses.lumenlearning.com/calculus1/chapter/disk-and-washer-methods/

[dreamy void]

That would be the idea

you can still draw it backwards

I would always draw it in terms of what variable we integrate with

It's more difficult if it is r(y)

but your function is in terms of x

or not difficult

but rather confusing

[pseudo trellis]

so i would need to convert y=x^(1/4) so it's x=y something?

x=y^4

[dreamy void]

Yea for y >= 0

This is what we doing

[dreamy void]

to get this

does this make sense?

[pseudo trellis]

yeah

[dreamy void]

good

can you tell me the bounds

upper and lower radius?

[pseudo trellis]

it's in terms of y, so it would be 0 and 1?

[dreamy void]

yes

[pseudo trellis]

i might've cooked

is that the intergral?

-1 because x=1 is where we're rotating it around

🙏

[dreamy void]

something is missing

[pseudo trellis]

hm

[dreamy void]

ok

what is R(y)?

[pseudo trellis]

y^4

[dreamy void]

no

[dreamy void]

look at this

[pseudo trellis]

y=1?

[dreamy void]

almost

[pseudo trellis]

x=1

💀

[dreamy void]

it's technically R(y) = 1 - 0

or your way

R(y) = 0 - 1

[pseudo trellis]

right

[dreamy void]

the left square

ok!

yes

[pseudo trellis]

make your own

[dreamy void]

someone else is using this

lmaooo

[pseudo trellis]

#help-36

okay so R(y) is -1

-1^2

[dreamy void]

yes

that is actually farther

the y-axis

than the y^4

[dreamy void]

radius wise

[pseudo trellis]

is that right?

[dreamy void]

almost

you forgot

[pseudo trellis]

i forgot -1

[dreamy void]

to subtract 1

yes

[pseudo trellis]

on the second term

right

[dreamy void]

ok

i am not good at math

maybe someone

[pseudo trellis]

💀

unless it's addition/subtract you certainly don't want my help

[dreamy void]

💀

[pseudo trellis]

oh yeah then definietly not

this is also kinda geometry 😭

[dreamy void]

this is hyper geomerty

stereometry

the reason i prefer radius - function is because it's cleaner

but ok

so

now we evaluate it

[pseudo trellis]

13pi/45 is the answer i got

[dreamy void]

me too

see

relax

you get the premium bread

[pseudo trellis]

that problem was pretty easy though lol

i guess 15-17 were pretty difficult ones

okay last one for about 30 mins

https://archive.org/details/calculus-early-transcendentals-8th-edition-2015/page/446/mode/2up 26

so we're just flipping the same problem around y=1

instead of x=1

[dreamy void]

awwwwwwwww

[dreamy void]

what's the last one

[pseudo trellis]

26 is the one im doing now

i have to do 39, 41, 48, 49, and 50

[dreamy void]

damn

wtf

I thought you were a software developer

[pseudo trellis]

💀

this is nothing like software

[dreamy void]

this is for you HARDware

[pseudo trellis]

😭 facts

okayyyy so

[dreamy void]

$\mathscr R_2$ about $BC$

[wraith dagger]

𝔸dωn𝓲²s

[pseudo trellis]

we want to do 26 in terms of x now?

[dreamy void]

yes

[pseudo trellis]

okay okay

[dreamy void]

because it's around a horizontal axis

[pseudo trellis]

y=x^4 is that how that works?

[dreamy void]

doesnt?

it's y = x^(1/4)

[pseudo trellis]

oh yes

that yeah

we're rotating it around y=1

so the bounds are in terms of x, so 0 and 1

[dreamy void]

yes

yes

you are getting the idea slowly but surely

[pseudo trellis]

now we need to find both R(x) and r(x)?

[dreamy void]

yes

Have you drawn

[pseudo trellis]

nope

[dreamy void]

like a 2 year old?

do it

always

[pseudo trellis]

damn but we already kinda have a drawing

[dreamy void]

damn

from where that confidence

[pseudo trellis]

lol

[dreamy void]

well okay

we want the green area rotated

actually there is one radius only

do you see why?

[pseudo trellis]

are we using discs here?

is that why?

[dreamy void]

haha yea you could say it like that

we dont have to subtract anything from something

[pseudo trellis]

yeah

[dreamy void]

it's not like a donut or something

where some inner part is being missed or taken away

[dreamy void]

the green rotated is like a

volcano

[pseudo trellis]

yeah

[dreamy void]

haha

yea

[pseudo trellis]

:o

forgot pi

[dreamy void]

the blue function is kinda off but yea

,w plot y = x^(1/4)

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by adoniszgks

[pseudo trellis]

drawing is not my best trait lol

[dreamy void]

as a 2 year old it should

[pseudo trellis]

frfr

[dreamy void]

😂

[dreamy void]

the reason I do 1 - x^(1/4) because radius is always positive

and x^(1/4) is between 0 and 1

it kinda triggers me seeing like that haha

but you can also interpret it as shifting the problem by one unit down

and rotation around x = 0

anyway

[pseudo trellis]

1pi/15

:o

[dreamy void]

,w pi * Integrate[(1-x^(1/4))^2, {x,0,1}]

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by adoniszgks

[dreamy void]

yea

[pseudo trellis]

yessss

[dreamy void]

🥱

[pseudo trellis]

eh 39 now lol

[dreamy void]

describe the solid?

[pseudo trellis]

no idea what that means ngl

[dreamy void]

,,\pi \int_0^{\pi} \sin(x) : \dd x

[wraith dagger]

𝔸dωn𝓲²s

[dreamy void]

describe it

,w plot y = sin(x) between 0 and pi

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by adoniszgks

[dreamy void]

how would you call that

you are american

you should know that

[pseudo trellis]

x=-y^2+1?

is that describing it?

[dreamy void]

a football

https://tenor.com/view/football-football-fan-watch-football-are-you-ready-for-some-football-nfl-game-day-gif-13208639217159682531

[pseudo trellis]

wot

[dreamy void]

close your eyes

and rotate it

[pseudo trellis]

i would've said a mountain

or something

[dreamy void]

what

you are not an American

⚔️

[pseudo trellis]

lol

[dreamy void]

https://tenor.com/view/football-gif-23338780

[pseudo trellis]

i literally just put football as the answer? 💀

[dreamy void]

mountain 😭

do what you want

😭

[pseudo trellis]

surely that can't be what it's asking though

[dreamy void]

well 50 $ for 2 year old tasks

[pseudo trellis]

if i see this question on a test, i'm not allowed a graping calculator, so how would I figure it out?

[dreamy void]

it's basic sine 💀

[pseudo trellis]

oh

[dreamy void]

😂

it's just sin(x)

[pseudo trellis]

im a genius

[dreamy void]

calc 1

the other ones are also managable

oh wait

wait