Question

Consider the given curves to do the following.
y=5sqrt(x)
y=0
x=1
Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about x = -3.

Answer 1

wha!?!

Answer 2

:) just kiddin' this ones easy too

Answer 3

yup :( i went on this online tutor thing from the hw website i use, they got me set up but i think i have the boundaries wrong

Answer 4

lol xD
the integral they told me to use is

Answer 5

\[2\pi \int\limits5x^(3/2) + 15x^(1/2)\]

Answer 6

sorry it should be 5x^(3/2) and 15x^(1/2)

Answer 7

use {} when using exponent

Answer 8

kay i will

Answer 9

i thought the boundaries for the integral would be 1 and -3, but when i plug in -3 to the integral itll give me trouble lol

Answer 10

this is a rough idea of what you got and since we are spinning around an x; we integrate with respect to y.....lets move that x line; and I got the rotation axis in the wrong spot

Answer 11

so if we integrate with respect to y, that means we have to get x = in terms of y ?

Answer 12

like this

Answer 13

the shells have a radius along the x; so lets stick to that.... i had to draw the pic for clarity in me head :)

Answer 14

lets +3 to all the 'x' values; to move this thing; NOT f(x) + 3; but f(x+3)

Answer 15

how come were adding 3?

Answer 16

y=5sqrt(x+3)
y=0 = x axis

x=1+3;

you see that instruction to spin about x=-3? get that to zero dear lol

Answer 17

LOL oh. haha thank youu

Answer 18

our inteval is then ; [0+3,1+3], or simply; [3,4] right?

Answer 19

yes

Answer 20

Our farthest Raduis =4 and our inner radius = 3 so we integrate from r to R roght?

Answer 21

so like lets say the instruction told us to spin about x = 1, would we subtract 1 from the x values?

Answer 22

and correct

Answer 23

and our height only changes as 5sqrt(x+3) changes now; so lets do this :)

Answer 24

make that 5sqrt(x-3) ; you know why?

Answer 25

mmm, im not too sure why

Answer 26

we want 5sqrt(0) to equal 5sqrt(3+___) so remember that we moved it but we dont want to change its value; just its position

Answer 27

the value at 3 away from center of rotation nedds to be the same regardless of where we put this thing right?

Answer 28

correct

Answer 29

so; lets integrate this thing like this:

2pi {S} x*5sqrt(x-3) dx ; [3,4]

Answer 30

you see the 5sqrt(x-3)

Answer 31

http://www.wolframalpha.com/input/?i=int%282pi*x*5sqrt%28x-3%29%29+from+3+to+4

Answer 32

can we put it as 5(x-3)^1/2

Answer 33

i see it :) we can pull out that 5 right? nothing is stopping that from not being a constant

Answer 34

ooh that makes it easier than what i was about to do lol

Answer 35

we can put that lonely x inside the sqrt

Answer 36

x*sqrt(x-3) = sqrt(x^2(x-3))

(x^3 -3x^2)^(1/2) right?

Answer 37

can i ask one thing though. can we use that integral of 2pi x f(x) dx for any cylindrical shell problem? because when i went to the hw tutor from the hw website, they told me to use the 2pi r h or something like that

Answer 38

r = x or y depending on if you spin it up or sideways

h = f(x) regardless

Answer 39

well; maybe h can = f(y) too lol

Answer 40

ohh so they're basically the same thing, just different way of putting it

Answer 41

x*sqrt(x-3) = sqrt(x^2(x-3))

(x^3 -3x^2)^(1/2) right? and yes lol xD

Answer 42

yes; its like calling your dad by his given name; hell respond to either right? the person is the same, the name can change

Answer 43

haha nice comparison xD

Answer 44

we could integrate this numercally instead; with like the simpsons rule or trap rule if trying to find a morph that works proves too difficult

Answer 45

i havent learned those rules yet :(

Answer 46

you have, you just havent learned how to apply them yet lol

its like this shell methid; you already knew how to find the area of a sheet of paper; but dint realize how that fact could aid you

Answer 47

lol oh xD im sorry, im still a little confused on why we added 3. i understand to get x=0 but im still confused on why we did it.

Answer 48

becasue in analysis it is always easier to bring things to a standard working table that is right in front of you.

If we dont move it, it is stuck up high on a shelf and that is no way to examine it right?

so, we bring it to the table by moving it off the shelf

Answer 49

we change its position ; but retain its value

Answer 50

this is the basic of the trapazoid rule

Answer 51

you mean everything is the same, the value is the same, its just on a different point on the graph?

Answer 52

exactly :)

Answer 53

so the answer we receive from it will be the same;

if you move your breakfast form the stove to your plate to the table; what has changed?

Answer 54

the place of where its at lol xD i gotcha, thank you for clarifying that for me. i was so confused before lol :p

Answer 55

lets see if we can int this function :)

x*sqrt(x-3) then :) any tricks up your sleeve thatll be useful for us?

Answer 56

ohohoh...maybe int by parts lol

Answer 57

u = x and v = sqrt(x-3) ??

Answer 58

are we using substitution right now?

Answer 59

its a kind of subsitution based on the product rule for derivatives

Answer 60

remember the product rule?

Dx(uv) = uv' + vu' right?

Answer 61

yesss i remember that :)

Answer 62

so it would be equal to x/2 (x-3)^-1/2 + sqrt of x-3 ?

Answer 63

{S} (uv)' = {S} u v' + {S} v u' right?

uv = {S} u dv + {S} v du

Answer 64

uv - {S} v du = {S} u dv

Answer 65

we want to integrate " u = x sqrt(x-3)" and it has 2 parts; so we use this trick to find it lol

Answer 66

how did you get from
{S} (uv)' = {S} u v' + {S} v u' right?
to
uv = {S} u dv + {S} v du

Answer 67

the {S} of a derivate (y') = y;

{S} y' = y; right?

Answer 68

correct

Answer 69

v' = dv; just different notations

Answer 70

lets focus on that part of my "u" column...

we start with:

(x-3)^(1/2) ; and that ints up to:

2/3 (x-3)^(3/2) ; in tthat again... pull out the 2/3 and do:

(x-3)^(3/2) goes to what?

Answer 71

(2/5) (x-3)^(5/2) right?

Answer 72

so we take the antiderivative twice for it?

Answer 73

\[\int\limits_{}(x.\sqrt{x-3}).dx = [\frac{2x}{3}.\sqrt{(x-3)^3}] - [\frac{4}{15}.\sqrt{(x-3)^{5}}]\]

Answer 74

right?

Answer 75

we integrate it till we can stop pretty much

Answer 76

if we pick a value for u that derives to 0 then it doesnt matter what v suits up to cause when it gets to u=0 its done

Answer 77

okay i see what you mean now

Answer 78

u =x

u' = 1

u'' = 0 right?

Answer 79

correct

Answer 80

v = sqrt(...)

{S} v = {S} sqrt(...)

{S} [{S} v] = {S} [{S} sqrt(...)]

Answer 81

+u * {S} v

- u' * {S} [{S} v]

+ 0 * ...doesnt matter lol

Answer 82

this is the method known as integration by parts; and it can get tricky but the key is to pick smart values for u and v to begin with and itll work itself out in the end :)

Answer 83

int by parts is just using the product rule as a basis; and then using it again, and again, till you get to 0 lol

Answer 84

seems so hard :( lol but i think i got a jist of what you mean. thank you for all your help and for staying and explaining all of it to me. :))

Answer 85

youre welcome :) the more i explain it the better i understand it meself ;)

Answer 86

its says your typing.... its like waiting for the waffle iron