anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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amistre64
  • amistre64
HA!! shells, i knew they were comin
amistre64
  • amistre64
since we is going around the y axis; its simpler right now
anonymous
  • anonymous
yup :( i never used this method before ><

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amistre64
  • amistre64
the radius of each shell moves from x=0 to x=1 right?
anonymous
  • anonymous
yup
amistre64
  • amistre64
and the height of each shell is determined by: f(x)
anonymous
  • anonymous
wait, i know why its x=1, but why is it x=0
amistre64
  • amistre64
so the area of each shell, when flattened out is height* width...
amistre64
  • amistre64
umm... y axis means x=0 they are the same thing
anonymous
  • anonymous
ohh okay gotcha
amistre64
  • 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
  • amistre64
2pi [S] x(x^2) dx ; [0,1] 2pi x^4 ------- = (pi x^4)/4 = pi/4 4
anonymous
  • anonymous
how did you get rid of the 4th power in the numerator?
amistre64
  • amistre64
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amistre64
  • amistre64
you dont... why would you?
anonymous
  • anonymous
but you put (pi x^4)/4 = pi/4 where did the 4th power go?
amistre64
  • amistre64
F(0) = 0 F(1) = pi/4 F(1) - F(0) = pi/4
amistre64
  • amistre64
over 2 lol
amistre64
  • amistre64
i forgot i changed 2/4 to 1/2.... doh!
amistre64
  • amistre64
\[\frac{2\pi x^4}{4} = \frac{\pi x^4}{2}\]
anonymous
  • anonymous
much clearer, lol thank you :D
amistre64
  • amistre64
when im wrong; say im wrong...not ask me how im right lol. I just assume im right ;)
anonymous
  • anonymous
lol will do xD
amistre64
  • amistre64
you understand the mechanics of the shell method tho?
anonymous
  • 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
  • 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
  • amistre64
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anonymous
  • 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
  • amistre64
when we flatten out the "shell" we get a flat rectangular piece that the area is eay to determine right?
anonymous
  • anonymous
correct
amistre64
  • amistre64
the width = 2pi x...can you tell me why?
amistre64
  • amistre64
we aint using half the shell, but the whole thing
anonymous
  • anonymous
i think it has to do with the radius, is that why the width is 2 pi x?
anonymous
  • anonymous
ohh is it the circumference of the cylnder?
amistre64
  • amistre64
it does, but i want to make sure you understand this :) its very basic and easy...exactly lol
anonymous
  • anonymous
lol i took a careful look at the cylinder and realized haha xD
amistre64
  • amistre64
you did good :)
amistre64
  • amistre64
the shell method makes things easier for some problems
anonymous
  • 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
  • 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
  • amistre64
{S} 2pi x[f(x)] dx is what we do right?
amistre64
  • amistre64
if there is a constant we can pull it aside right?
anonymous
  • anonymous
mhmm
anonymous
  • anonymous
the 2pi
amistre64
  • amistre64
2 pi goes out and what are we left with inside?
anonymous
  • anonymous
x*f(x)
amistre64
  • amistre64
excatly; so we integrate 'x*f(x)'
amistre64
  • amistre64
do you se how that differes from your first statement of ; we int f(x)?
anonymous
  • anonymous
yesss, thanks for clearing that up for me :) this method seems a bit easier than the others lol
amistre64
  • amistre64
look at the disc method
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anonymous
  • 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
  • amistre64
the disc method take all the areas of a given circle; with a radius of f(x) and adds them up
amistre64
  • amistre64
volume is always a positive number
amistre64
  • amistre64
what is the formula for the the area of a circle ?
anonymous
  • 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
  • anonymous
and area of circle is pi r^2
amistre64
  • 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
  • anonymous
so f(x) is the radius?
amistre64
  • amistre64
look at the drawing i did in this one
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amistre64
  • amistre64
and tell me what the radius of each given circle is?
anonymous
  • anonymous
oh, f(x) xD haha
amistre64
  • amistre64
that is all these volume of rotation problems amopunt to :)
anonymous
  • 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
  • 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
  • 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
  • amistre64
lets say 4 high and 4 wide; the area should be 16 right?
anonymous
  • anonymous
correct
amistre64
  • amistre64
if we integrate f(x) = y = 4; from -2 to 2 we know we should get 16 right?
anonymous
  • anonymous
how come its from -2 and 2?
amistre64
  • 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
  • amistre64
the distance from -2 to 2 = 4 right?
anonymous
  • anonymous
yupp
amistre64
  • amistre64
and a square with sides = 4 would straddle the y axis would sit from -2 to 2 on the x axis right?
anonymous
  • anonymous
correct. i gotcha now :) lol
anonymous
  • 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
  • amistre64
i pic is worth a thousand words :)
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anonymous
  • anonymous
it really is. thank you for that :D
amistre64
  • amistre64
yes, but we just need to make sure we are rotating it properly; make sure the numbers come out correctly
amistre64
  • amistre64
once we can draw a picture, the rest is just intuitive
anonymous
  • 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
  • amistre64
youre welcome :)

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