Question

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

Answer 1

HA!! shells, i knew they were comin

Answer 2

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

Answer 3

yup :( i never used this method before ><

Answer 4

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

Answer 5

yup

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

ohh okay gotcha

Answer 11

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

Answer 12

2pi [S] x(x^2) dx ; [0,1]

2pi x^4
------- = (pi x^4)/4 = pi/4
4

Answer 13

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

Answer 14

you dont... why would you?

Answer 15

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

Answer 16

F(0) = 0

F(1) = pi/4

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

Answer 17

over 2 lol

Answer 18

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

Answer 19

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

Answer 20

much clearer, lol thank you :D

Answer 21

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

Answer 22

lol will do xD

Answer 23

you understand the mechanics of the shell method tho?

Answer 24

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?

Answer 25

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

Answer 26

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?

Answer 27

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

Answer 28

correct

Answer 29

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

Answer 30

we aint using half the shell, but the whole thing

Answer 31

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

Answer 32

ohh is it the circumference of the cylnder?

Answer 33

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

Answer 34

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

Answer 35

you did good :)

Answer 36

the shell method makes things easier for some problems

Answer 37

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?

Answer 38

lets think that thruough; we want to add up all the areas...which is what intagrating does;

each area = 2pi x[f(x)]

Answer 39

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

Answer 40

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

Answer 41

mhmm

Answer 42

the 2pi

Answer 43

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

Answer 44

x*f(x)

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

look at the disc method

Answer 49

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

Answer 50

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

Answer 51

volume is always a positive number

Answer 52

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

Answer 53

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

Answer 54

and area of circle is pi r^2

Answer 55

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

Answer 56

so f(x) is the radius?

Answer 57

look at the drawing i did in this one

Answer 58

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

Answer 59

oh, f(x) xD haha

Answer 60

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

Answer 61

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

Answer 62

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

Answer 63

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

Answer 64

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

Answer 65

correct

Answer 66

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

Answer 67

how come its from -2 and 2?

Answer 68

{S} 4 dx -> 4x

4(2) - 4(-2) = 8 -- 8 = 16, so that one doesnt apply to the /2 thing you discussed lol

Answer 69

the distance from -2 to 2 = 4 right?

Answer 70

yupp

Answer 71

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

Answer 72

correct. i gotcha now :) lol

Answer 73

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?

Answer 74

i pic is worth a thousand words :)

Answer 75

it really is. thank you for that :D

Answer 76

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

Answer 77

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

Answer 78

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 !

Answer 79

youre welcome :)