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anonymous
 5 years ago
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
 5 years ago
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.

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0:) just kiddin' this ones easy too

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yup :( 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol xD the integral they told me to use is

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[2\pi \int\limits5x^(3/2) + 15x^(1/2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry it should be 5x^(3/2) and 15x^(1/2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0use {} when using exponent

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0this 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
 5 years ago
Best ResponseYou've already chosen the best response.0so if we integrate with respect to y, that means we have to get x = in terms of y ?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the shells have a radius along the x; so lets stick to that.... i had to draw the pic for clarity in me head :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lets +3 to all the 'x' values; to move this thing; NOT f(x) + 3; but f(x+3)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how come were adding 3?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y=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
 5 years ago
Best ResponseYou've already chosen the best response.0LOL oh. haha thank youu

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0our inteval is then ; [0+3,1+3], or simply; [3,4] right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Our farthest Raduis =4 and our inner radius = 3 so we integrate from r to R roght?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so like lets say the instruction told us to spin about x = 1, would we subtract 1 from the x values?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0and our height only changes as 5sqrt(x+3) changes now; so lets do this :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0make that 5sqrt(x3) ; you know why?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0mmm, im not too sure why

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we 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
 5 years ago
Best ResponseYou've already chosen the best response.0the value at 3 away from center of rotation nedds to be the same regardless of where we put this thing right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so; lets integrate this thing like this: 2pi {S} x*5sqrt(x3) dx ; [3,4]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you see the 5sqrt(x3)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=int%282pi*x*5sqrt%28x3%29%29+from+3+to+4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can we put it as 5(x3)^1/2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i see it :) we can pull out that 5 right? nothing is stopping that from not being a constant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ooh that makes it easier than what i was about to do lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we can put that lonely x inside the sqrt

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x*sqrt(x3) = sqrt(x^2(x3)) (x^3 3x^2)^(1/2) right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can 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
 5 years ago
Best ResponseYou've already chosen the best response.0r = x or y depending on if you spin it up or sideways h = f(x) regardless

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0well; maybe h can = f(y) too lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohh so they're basically the same thing, just different way of putting it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x*sqrt(x3) = sqrt(x^2(x3)) (x^3 3x^2)^(1/2) right? and yes lol xD

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yes; its like calling your dad by his given name; hell respond to either right? the person is the same, the name can change

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha nice comparison xD

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we 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
 5 years ago
Best ResponseYou've already chosen the best response.0i havent learned those rules yet :(

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you 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
 5 years ago
Best ResponseYou've already chosen the best response.0lol 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
 5 years ago
Best ResponseYou've already chosen the best response.0becasue 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
 5 years ago
Best ResponseYou've already chosen the best response.0we change its position ; but retain its value

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0this is the basic of the trapazoid rule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you mean everything is the same, the value is the same, its just on a different point on the graph?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so 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
 5 years ago
Best ResponseYou've already chosen the best response.0the place of where its at lol xD i gotcha, thank you for clarifying that for me. i was so confused before lol :p

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lets see if we can int this function :) x*sqrt(x3) then :) any tricks up your sleeve thatll be useful for us?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ohohoh...maybe int by parts lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0u = x and v = sqrt(x3) ??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are we using substitution right now?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its a kind of subsitution based on the product rule for derivatives

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0remember the product rule? Dx(uv) = uv' + vu' right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yesss i remember that :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so it would be equal to x/2 (x3)^1/2 + sqrt of x3 ?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0{S} (uv)' = {S} u v' + {S} v u' right? uv = {S} u dv + {S} v du

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0uv  {S} v du = {S} u dv

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we want to integrate " u = x sqrt(x3)" and it has 2 parts; so we use this trick to find it lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get from {S} (uv)' = {S} u v' + {S} v u' right? to uv = {S} u dv + {S} v du

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the {S} of a derivate (y') = y; {S} y' = y; right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0v' = dv; just different notations

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lets focus on that part of my "u" column... we start with: (x3)^(1/2) ; and that ints up to: 2/3 (x3)^(3/2) ; in tthat again... pull out the 2/3 and do: (x3)^(3/2) goes to what?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(2/5) (x3)^(5/2) right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so we take the antiderivative twice for it?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}(x.\sqrt{x3}).dx = [\frac{2x}{3}.\sqrt{(x3)^3}]  [\frac{4}{15}.\sqrt{(x3)^{5}}]\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we integrate it till we can stop pretty much

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if 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
 5 years ago
Best ResponseYou've already chosen the best response.0okay i see what you mean now

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0u =x u' = 1 u'' = 0 right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0v = sqrt(...) {S} v = {S} sqrt(...) {S} [{S} v] = {S} [{S} sqrt(...)]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0+u * {S} v  u' * {S} [{S} v] + 0 * ...doesnt matter lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0this 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
 5 years ago
Best ResponseYou've already chosen the best response.0int 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
 5 years ago
Best ResponseYou've already chosen the best response.0seems 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
 5 years ago
Best ResponseYou've already chosen the best response.0youre welcome :) the more i explain it the better i understand it meself ;)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its says your typing.... its like waiting for the waffle iron
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