integration help

[lean condor]

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[twilit hedge]

how did you do it for a curve and a straight line then?

[charred merlin]

well I would add the line and the curve

[charred merlin]

or subtract it depending on what portion they want

[twilit hedge]

How would you you find the volume of revolution of this around the x axis, over the interval [-2,2]?

ie what is the integrand and why?

it's y=x^2+2 and y=1 for the blue and red respectively

[charred merlin]

hold on the curve isn't touching the x axis

[twilit hedge]

yeah

so?

you said you can do the problem if it's a curve and a straight line

so I gave you a curve and a straight line

[charred merlin]

yeah if it's touching the x axis

for example

[twilit hedge]

ok....

after you do the rotation, describe the shape you get

is it a cylinder (esque)? a cylinder with a hole? other?

[charred merlin]

close enough

[twilit hedge]

ok, how would you do that...

[charred merlin]

well if that is rotated 360 degree

I would add the straight line and the curve

I will take my range as 2 to 0 for the straight line

and I will take my range 3 to 2 for the curve

[twilit hedge]

ok yeah, no clue what you're on about

[twilit hedge]

the cross section of the volume will be an annulus

hence $\dd{V}=A(x)\dd{x}=\pi[R(x)^2-r(x)^2]\dd{x}$

[keen flower]

Omegabet_

[twilit hedge]

where $R(x)$ is the big radius, and $r(x)$ is the small radius as usual

[keen flower]

Omegabet_

[twilit hedge]

since area of an annulus is (big circle area) - (small circle area)

[charred merlin]

wait are you doing the dish washer method?

[twilit hedge]

no clue, im doing logic

vertical slices of the volume are annuli

[charred merlin]

never heard of that but ok

[twilit hedge]

you've never heard of logic?

[charred merlin]

"annuli"

[twilit hedge]

,w annulus

[keen flower]

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[charred merlin]

is it like radius?

[twilit hedge]

no, shapes are not radii

shapes have area, radii have no area, they're lengths

an annulus, as you can see from having eyes, is a circle with a smaller circle taken out concentrically

[charred merlin]

hm

and

[twilit hedge]

and........?

[charred merlin]

my question exactly

[twilit hedge]

well it's obvious what an annulus is, I just posted a picture of one

but anyway, as I said already

[charred merlin]

so what am I gonna do with annulus?

[twilit hedge]

the vertical cross sections of the volume are annuli

hence, as was already written, $\dd{V}=\pi[R(x)^2-r(x)^2]\dd{x}$ is the infinitesimal of volume using such cross sections

[keen flower]

Omegabet_

[twilit hedge]

so $V=\int\dd{V}=\int_{a}^b\pi[R(x)^2-r(x)^2]\dd{x}$, where $a$ and $b$ are the values from a

[keen flower]

Omegabet_

[charred merlin]

what should be my input?

[twilit hedge]

as I said....

R is the big radius, and r is the small radius

in the region you're rotating, it's obvious which function is bigger than the other

[charred merlin]

wait if y=5

[twilit hedge]

im talking about your initial question

[charred merlin]

then it's the small radius

[charred merlin]

we don't have a radius there

[twilit hedge]

Im telling you, you do

cause it's annuli...

[charred merlin]

what's the value ?

[twilit hedge]

the value... of what?

[charred merlin]

the annuli

[twilit hedge]

what's the value of a square?

or the value of a hexagonal pyramid?

[charred merlin]

.

[twilit hedge]

.

I can type pointless stuff too

annuli are shapes, they obviously dont have values