Calculus 3

[crimson ginkgo]

Can anyone tell me what methods I should be using for this - I am completely lost as to what to do despite t his being a relatively easy task according to peers

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[crimson ginkgo]

any help

T-T

?

[alpine berry]

integrate

what's the radius of the section at height z?

$$V(z_1) = \int _0^{z_1} A(z) dz$$

[uneven ridge]

aL

[alpine berry]

@crimson ginkgo

[crimson ginkgo]

The radius at height z?

[alpine berry]

assuming the liquids mix, you are to find the combined density of the mixture and determine its center of mass (what is the height, basically)

[crimson ginkgo]

root(z)?

[alpine berry]

correct

hence the area is pi z

yes?

[crimson ginkgo]

Yep

then I integrate throught 0 to root(z) of pi(z)

?

[alpine berry]

0 to z_1

[crimson ginkgo]

ofc yeah

oops

Do I times that by pi?

I remember in the lecture

[alpine berry]

$$ \int _0^{z_1} \pi z dz$$

[crimson ginkgo]

I do pi * integral

[uneven ridge]

aL

[crimson ginkgo]

yeah

This is the exact same mathematically right

I'm writing my script on word so

[alpine berry]

yes because pi is a constant

[crimson ginkgo]

Yeah ofc just checking

Part b still makes absolutely NO sense to me though

[alpine berry]

you are adding some unknown volume of liquid

but you do know it goes from z_1 to 2z_1

hence

$$ \int _{z_1}^{2z_1} \pi z dz$$

[uneven ridge]

aL

[alpine berry]

is the amount of liquid you add

[crimson ginkgo]

This looks good?

[alpine berry]

I denoted V(a) as integral from 0 to a before

the amount of new liquid you add would then be V(2z1) - V(z1)

[crimson ginkgo]

So (3pi(z1)^2)/2 is the Volume of new liquid added

[alpine berry]

that is correct

[crimson ginkgo]

In my earlier attempt (no idea if its correct) I used this formula to compute the centre of mass

[alpine berry]

I have no idea about formulae

i cba to learn them

[crimson ginkgo]

Fair

[alpine berry]

waht does center of mass mean to you?

[crimson ginkgo]

Well it is the point through which a bodies weight can be said to act

or like the mean of all the individual point masses in that object

Is how I would see it

[alpine berry]

the paraboloid is symmetric around the z axis

so your center of mass always satisfies x=y=0

your only task is to figure out what the z coordinate is

what's the center of mass for just the first liquid?

[crimson ginkgo]

it was root z right

[alpine berry]

that's the radius, irrelevant

the first liquid is filled up to z1

let's say z1 = 10

do you think the center of mass lies at z=5?

or higher/lower and why?

[crimson ginkgo]

Higher

since the majority of the liquid is still above that point

[alpine berry]

that is correct

[crimson ginkgo]

You'd want the point where an equal amount of liquid is above and below

[alpine berry]

very good

so you're looking for a value of z such that

$$ \int _0^{z^} \pi z dz = \int _{z^}^{z_1}\pi z dz$$

[uneven ridge]

aL

[alpine berry]

make sense?

[crimson ginkgo]

yes, so (pi(z)^2)/2 = 2pi(z)^2 - pi(z)^2/2

Idk how to use the notation thing sorry

And thus

[alpine berry]

let's call it z* = b

[crimson ginkgo]

Yeah okay

[alpine berry]

$$b^2 = z_1^2 - b^2$$

[uneven ridge]

aL

[alpine berry]

where does this come from

you are integrating up to z1

not 2z1

[crimson ginkgo]

ah yeah

I fucked up the integral

mb

so it would be

[crimson ginkgo]

b = 2so

so

b^2 = 2z1^2

b = sqrt(2z1)

[alpine berry]

b^2 = z_1^2 / 2

[alpine berry]

if z_1 = 1, then this would be higher than the total height of the liquid

makes no sense

[crimson ginkgo]

okay yeah idk what is wrong wiht my head

so b = z1/sqrt(2)

[alpine berry]

correct

between z1 and 2z1 is the center of mass for the other liquid

or you assume the two liquids can be mixed

it doesn't matter what the density of the mixture is

[crimson ginkgo]

Yeah it says the liquids can be mixed for part b

[alpine berry]

all that matters is the center coordinate from 0 to 2z1

which is what you attempted before

[crimson ginkgo]

this?

[alpine berry]

yes

cancel the pi so is easier

for problem c you do more of the same

but this time you can't mix them

[crimson ginkgo]

nvm

[alpine berry]

so you find separate center coordinates and take weighted average of the two

and that's all there is to it, no fancy formula required

[crimson ginkgo]

Like this

?

or literally just (a + b)/2

[dense lintel]

@crimson ginkgo

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[crimson ginkgo]

With part c could I assume that A liesbelow C even though in reality this wont be the case as liquid c is denser?

[alpine berry]

that depends on the masses of the two liquids