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.

- anonymous

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- amistre64

wha!?!

- amistre64

:) just kiddin' this ones easy too

- anonymous

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

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## More answers

- anonymous

lol xD
the integral they told me to use is

- anonymous

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

- anonymous

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

- anonymous

use {} when using exponent

- anonymous

kay i will

- anonymous

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

- amistre64

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- amistre64

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

- anonymous

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

- amistre64

like this

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- amistre64

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

- amistre64

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

- anonymous

how come were adding 3?

- amistre64

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

- anonymous

LOL oh. haha thank youu

- amistre64

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

- anonymous

yes

- amistre64

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

- anonymous

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

- anonymous

and correct

- amistre64

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

- amistre64

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

- anonymous

mmm, im not too sure why

- amistre64

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

- amistre64

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

- anonymous

correct

- amistre64

so; lets integrate this thing like this:
2pi {S} x*5sqrt(x-3) dx ; [3,4]

- anonymous

you see the 5sqrt(x-3)

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

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

- amistre64

we can put that lonely x inside the sqrt

- amistre64

x*sqrt(x-3) = sqrt(x^2(x-3))
(x^3 -3x^2)^(1/2) right?

- anonymous

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

- amistre64

r = x or y depending on if you spin it up or sideways
h = f(x) regardless

- amistre64

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

- anonymous

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

- anonymous

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

- amistre64

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

- anonymous

haha nice comparison xD

- amistre64

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

- anonymous

i havent learned those rules yet :(

- amistre64

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

- anonymous

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.

- amistre64

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

- amistre64

we change its position ; but retain its value

- amistre64

this is the basic of the trapazoid rule

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- anonymous

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

- amistre64

exactly :)

- amistre64

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?

- anonymous

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

- amistre64

lets see if we can int this function :)
x*sqrt(x-3) then :) any tricks up your sleeve thatll be useful for us?

- amistre64

ohohoh...maybe int by parts lol

- amistre64

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

- anonymous

are we using substitution right now?

- amistre64

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

- amistre64

remember the product rule?
Dx(uv) = uv' + vu' right?

- anonymous

yesss i remember that :)

- anonymous

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

- amistre64

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

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

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- amistre64

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

- anonymous

correct

- amistre64

v' = dv; just different notations

- amistre64

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?

- amistre64

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

- anonymous

so we take the antiderivative twice for it?

- amistre64

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

- amistre64

right?

- amistre64

we integrate it till we can stop pretty much

- amistre64

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

- anonymous

okay i see what you mean now

- amistre64

u =x
u' = 1
u'' = 0 right?

- anonymous

correct

- amistre64

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

- amistre64

+u * {S} v
- u' * {S} [{S} v]
+ 0 * ...doesnt matter lol

- amistre64

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 :)

- amistre64

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

- anonymous

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. :))

- amistre64

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

- amistre64

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

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