Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about the y-axis.
y = x^2
y=0
x=1

- anonymous

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- amistre64

HA!! shells, i knew they were comin

- amistre64

since we is going around the y axis; its simpler right now

- anonymous

yup :( i never used this method before ><

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## More answers

- amistre64

the radius of each shell moves from x=0 to x=1 right?

- anonymous

yup

- amistre64

and the height of each shell is determined by: f(x)

- anonymous

wait, i know why its x=1, but why is it x=0

- amistre64

so the area of each shell, when flattened out is height* width...

- amistre64

umm... y axis means x=0 they are the same thing

- anonymous

ohh okay gotcha

- amistre64

the height = f(x)
the width = 2pi [x]
the area of any given shell =
2pi x[f(x)]
so integrate that from [a,b] = [0,1] in this case

- amistre64

2pi [S] x(x^2) dx ; [0,1]
2pi x^4
------- = (pi x^4)/4 = pi/4
4

- anonymous

how did you get rid of the 4th power in the numerator?

- amistre64

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- amistre64

you dont... why would you?

- anonymous

but you put (pi x^4)/4 = pi/4
where did the 4th power go?

- amistre64

F(0) = 0
F(1) = pi/4
F(1) - F(0) = pi/4

- amistre64

over 2 lol

- amistre64

i forgot i changed 2/4 to 1/2.... doh!

- amistre64

\[\frac{2\pi x^4}{4} = \frac{\pi x^4}{2}\]

- anonymous

much clearer, lol thank you :D

- amistre64

when im wrong; say im wrong...not ask me how im right lol. I just assume im right ;)

- anonymous

lol will do xD

- amistre64

you understand the mechanics of the shell method tho?

- anonymous

all the methods like the washer, disk, etc all those methods seem so similiar to eachother and i get them all confused. because in the end youre always taking the antiderivative and plugging in the boundaries right?

- anonymous

like i understand theres certain formulas you have to follow for each method, which i have to memerize lol all of them seem really similiar

- amistre64

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- anonymous

so for the shell method, basically youre taking half of it, and spreading it out which is why it turns to be a rectangle shape?

- amistre64

when we flatten out the "shell" we get a flat rectangular piece that the area is eay to determine right?

- anonymous

correct

- amistre64

the width = 2pi x...can you tell me why?

- amistre64

we aint using half the shell, but the whole thing

- anonymous

i think it has to do with the radius, is that why the width is 2 pi x?

- anonymous

ohh is it the circumference of the cylnder?

- amistre64

it does, but i want to make sure you understand this :) its very basic and easy...exactly lol

- anonymous

lol i took a careful look at the cylinder and realized haha xD

- amistre64

you did good :)

- amistre64

the shell method makes things easier for some problems

- anonymous

thanks ;) so basically for the shell method we're plugging in f(x) into the formula and taking the antiderivative of it? then plug in the boundaries and solve?

- amistre64

lets think that thruough; we want to add up all the areas...which is what intagrating does;
each area = 2pi x[f(x)]

- amistre64

{S} 2pi x[f(x)] dx is what we do right?

- amistre64

if there is a constant we can pull it aside right?

- anonymous

mhmm

- anonymous

the 2pi

- amistre64

2 pi goes out and what are we left with inside?

- anonymous

x*f(x)

- amistre64

excatly; so we integrate 'x*f(x)'

- amistre64

do you se how that differes from your first statement of ; we int f(x)?

- anonymous

yesss, thanks for clearing that up for me :) this method seems a bit easier than the others lol

- amistre64

look at the disc method

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- anonymous

oh, i have one question. when you solve for an integral, is it possible for it to be negative? or does it always have to be positive

- amistre64

the disc method take all the areas of a given circle; with a radius of f(x) and adds them up

- amistre64

volume is always a positive number

- amistre64

what is the formula for the the area of a circle ?

- anonymous

uh oh. darn i should of took the absolute value. i was so sure i did it right too. lol but thank you :)

- anonymous

and area of circle is pi r^2

- amistre64

so when we add up all the areas of the cicles thru integration we do:
{S} pi [f(x)]^2 dx right? adding up the areas of circles

- anonymous

so f(x) is the radius?

- amistre64

look at the drawing i did in this one

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- amistre64

and tell me what the radius of each given circle is?

- anonymous

oh, f(x) xD haha

- amistre64

that is all these volume of rotation problems amopunt to :)

- anonymous

i remember there was this one method i think where you had to divide by two, is there a method that involves that? or am i making stuff up lol :p

- amistre64

i dont think thats a volume of rotation one :)
in some area propblems we are asked to find the area under a curve that hopes the y axis... and if the symmetry is the same from left to right side of the y axis; we get a skewed result

- amistre64

lets try something simple and see; suppose we want to find the area of a square that is sitting halfway between the sides

- amistre64

lets say 4 high and 4 wide; the area should be 16 right?

- anonymous

correct

- amistre64

if we integrate f(x) = y = 4; from -2 to 2 we know we should get 16 right?

- anonymous

how come its from -2 and 2?

- amistre64

{S} 4 dx -> 4x
4(2) - 4(-2) = 8 -- 8 = 16, so that one doesnt apply to the /2 thing you discussed lol

- amistre64

the distance from -2 to 2 = 4 right?

- anonymous

yupp

- amistre64

and a square with sides = 4 would straddle the y axis would sit from -2 to 2 on the x axis right?

- anonymous

correct. i gotcha now :) lol

- anonymous

if the prob asked to rotate around the x-axis, its generally the same idea of solving it like we did with the y axis right?

- amistre64

i pic is worth a thousand words :)

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- anonymous

it really is. thank you for that :D

- amistre64

yes, but we just need to make sure we are rotating it properly; make sure the numbers come out correctly

- amistre64

once we can draw a picture, the rest is just intuitive

- anonymous

gotcha. i shall try and attempt the rest of these shell method questions. thank you so so much for all your help. :D you make it possible for me to understand calculus lol xD thank you times infinity !

- amistre64

youre welcome :)

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