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

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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

Mathematics
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HA!! shells, i knew they were comin
since we is going around the y axis; its simpler right now
yup :( i never used this method before ><

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the radius of each shell moves from x=0 to x=1 right?
yup
and the height of each shell is determined by: f(x)
wait, i know why its x=1, but why is it x=0
so the area of each shell, when flattened out is height* width...
umm... y axis means x=0 they are the same thing
ohh okay gotcha
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
2pi [S] x(x^2) dx ; [0,1] 2pi x^4 ------- = (pi x^4)/4 = pi/4 4
how did you get rid of the 4th power in the numerator?
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you dont... why would you?
but you put (pi x^4)/4 = pi/4 where did the 4th power go?
F(0) = 0 F(1) = pi/4 F(1) - F(0) = pi/4
over 2 lol
i forgot i changed 2/4 to 1/2.... doh!
\[\frac{2\pi x^4}{4} = \frac{\pi x^4}{2}\]
much clearer, lol thank you :D
when im wrong; say im wrong...not ask me how im right lol. I just assume im right ;)
lol will do xD
you understand the mechanics of the shell method tho?
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?
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
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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?
when we flatten out the "shell" we get a flat rectangular piece that the area is eay to determine right?
correct
the width = 2pi x...can you tell me why?
we aint using half the shell, but the whole thing
i think it has to do with the radius, is that why the width is 2 pi x?
ohh is it the circumference of the cylnder?
it does, but i want to make sure you understand this :) its very basic and easy...exactly lol
lol i took a careful look at the cylinder and realized haha xD
you did good :)
the shell method makes things easier for some problems
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?
lets think that thruough; we want to add up all the areas...which is what intagrating does; each area = 2pi x[f(x)]
{S} 2pi x[f(x)] dx is what we do right?
if there is a constant we can pull it aside right?
mhmm
the 2pi
2 pi goes out and what are we left with inside?
x*f(x)
excatly; so we integrate 'x*f(x)'
do you se how that differes from your first statement of ; we int f(x)?
yesss, thanks for clearing that up for me :) this method seems a bit easier than the others lol
look at the disc method
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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
the disc method take all the areas of a given circle; with a radius of f(x) and adds them up
volume is always a positive number
what is the formula for the the area of a circle ?
uh oh. darn i should of took the absolute value. i was so sure i did it right too. lol but thank you :)
and area of circle is pi r^2
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
so f(x) is the radius?
look at the drawing i did in this one
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and tell me what the radius of each given circle is?
oh, f(x) xD haha
that is all these volume of rotation problems amopunt to :)
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
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
lets try something simple and see; suppose we want to find the area of a square that is sitting halfway between the sides
lets say 4 high and 4 wide; the area should be 16 right?
correct
if we integrate f(x) = y = 4; from -2 to 2 we know we should get 16 right?
how come its from -2 and 2?
{S} 4 dx -> 4x 4(2) - 4(-2) = 8 -- 8 = 16, so that one doesnt apply to the /2 thing you discussed lol
the distance from -2 to 2 = 4 right?
yupp
and a square with sides = 4 would straddle the y axis would sit from -2 to 2 on the x axis right?
correct. i gotcha now :) lol
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?
i pic is worth a thousand words :)
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it really is. thank you for that :D
yes, but we just need to make sure we are rotating it properly; make sure the numbers come out correctly
once we can draw a picture, the rest is just intuitive
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 !
youre welcome :)

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