anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
wha!?!
amistre64
  • amistre64
:) just kiddin' this ones easy too
anonymous
  • 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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anonymous
  • anonymous
lol xD the integral they told me to use is
anonymous
  • anonymous
\[2\pi \int\limits5x^(3/2) + 15x^(1/2)\]
anonymous
  • anonymous
sorry it should be 5x^(3/2) and 15x^(1/2)
anonymous
  • anonymous
use {} when using exponent
anonymous
  • anonymous
kay i will
anonymous
  • 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
  • amistre64
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amistre64
  • 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
  • anonymous
so if we integrate with respect to y, that means we have to get x = in terms of y ?
amistre64
  • amistre64
like this
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amistre64
  • 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
  • amistre64
lets +3 to all the 'x' values; to move this thing; NOT f(x) + 3; but f(x+3)
anonymous
  • anonymous
how come were adding 3?
amistre64
  • 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
  • anonymous
LOL oh. haha thank youu
amistre64
  • amistre64
our inteval is then ; [0+3,1+3], or simply; [3,4] right?
anonymous
  • anonymous
yes
amistre64
  • amistre64
Our farthest Raduis =4 and our inner radius = 3 so we integrate from r to R roght?
anonymous
  • anonymous
so like lets say the instruction told us to spin about x = 1, would we subtract 1 from the x values?
anonymous
  • anonymous
and correct
amistre64
  • amistre64
and our height only changes as 5sqrt(x+3) changes now; so lets do this :)
amistre64
  • amistre64
make that 5sqrt(x-3) ; you know why?
anonymous
  • anonymous
mmm, im not too sure why
amistre64
  • 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
  • amistre64
the value at 3 away from center of rotation nedds to be the same regardless of where we put this thing right?
anonymous
  • anonymous
correct
amistre64
  • amistre64
so; lets integrate this thing like this: 2pi {S} x*5sqrt(x-3) dx ; [3,4]
anonymous
  • anonymous
you see the 5sqrt(x-3)
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=int%282pi*x*5sqrt%28x-3%29%29+from+3+to+4
anonymous
  • anonymous
can we put it as 5(x-3)^1/2
amistre64
  • amistre64
i see it :) we can pull out that 5 right? nothing is stopping that from not being a constant
anonymous
  • anonymous
ooh that makes it easier than what i was about to do lol
amistre64
  • amistre64
we can put that lonely x inside the sqrt
amistre64
  • amistre64
x*sqrt(x-3) = sqrt(x^2(x-3)) (x^3 -3x^2)^(1/2) right?
anonymous
  • 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
  • amistre64
r = x or y depending on if you spin it up or sideways h = f(x) regardless
amistre64
  • amistre64
well; maybe h can = f(y) too lol
anonymous
  • anonymous
ohh so they're basically the same thing, just different way of putting it
anonymous
  • anonymous
x*sqrt(x-3) = sqrt(x^2(x-3)) (x^3 -3x^2)^(1/2) right? and yes lol xD
amistre64
  • 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
  • anonymous
haha nice comparison xD
amistre64
  • 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
  • anonymous
i havent learned those rules yet :(
amistre64
  • 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
  • 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
  • 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
  • amistre64
we change its position ; but retain its value
amistre64
  • amistre64
this is the basic of the trapazoid rule
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anonymous
  • anonymous
you mean everything is the same, the value is the same, its just on a different point on the graph?
amistre64
  • amistre64
exactly :)
amistre64
  • 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
  • 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
  • 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
  • amistre64
ohohoh...maybe int by parts lol
amistre64
  • amistre64
u = x and v = sqrt(x-3) ??
anonymous
  • anonymous
are we using substitution right now?
amistre64
  • amistre64
its a kind of subsitution based on the product rule for derivatives
amistre64
  • amistre64
remember the product rule? Dx(uv) = uv' + vu' right?
anonymous
  • anonymous
yesss i remember that :)
anonymous
  • anonymous
so it would be equal to x/2 (x-3)^-1/2 + sqrt of x-3 ?
amistre64
  • amistre64
{S} (uv)' = {S} u v' + {S} v u' right? uv = {S} u dv + {S} v du
amistre64
  • amistre64
uv - {S} v du = {S} u dv
amistre64
  • 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
  • anonymous
how did you get from {S} (uv)' = {S} u v' + {S} v u' right? to uv = {S} u dv + {S} v du
amistre64
  • amistre64
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amistre64
  • amistre64
the {S} of a derivate (y') = y; {S} y' = y; right?
anonymous
  • anonymous
correct
amistre64
  • amistre64
v' = dv; just different notations
amistre64
  • 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
  • amistre64
(2/5) (x-3)^(5/2) right?
anonymous
  • anonymous
so we take the antiderivative twice for it?
amistre64
  • amistre64
\[\int\limits_{}(x.\sqrt{x-3}).dx = [\frac{2x}{3}.\sqrt{(x-3)^3}] - [\frac{4}{15}.\sqrt{(x-3)^{5}}]\]
amistre64
  • amistre64
right?
amistre64
  • amistre64
we integrate it till we can stop pretty much
amistre64
  • 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
  • anonymous
okay i see what you mean now
amistre64
  • amistre64
u =x u' = 1 u'' = 0 right?
anonymous
  • anonymous
correct
amistre64
  • amistre64
v = sqrt(...) {S} v = {S} sqrt(...) {S} [{S} v] = {S} [{S} sqrt(...)]
amistre64
  • amistre64
+u * {S} v - u' * {S} [{S} v] + 0 * ...doesnt matter lol
amistre64
  • 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
  • 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
  • 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
  • amistre64
youre welcome :) the more i explain it the better i understand it meself ;)
amistre64
  • amistre64
its says your typing.... its like waiting for the waffle iron

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