anonymous
  • anonymous
The region bounded above by the line, y = 6, below by the curve y=sqrt(x), and to the left by the y-axis. The solid is rotated along the following lines: C. The line y=6 D. The line x=36. These are two separate rotation axis and answers.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
which method?
amistre64
  • amistre64
|dw:1327531365366:dw|
amistre64
  • amistre64
we need to know the point of intersection here

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anonymous
  • anonymous
Hey sorry, keyboard died. The point of intersection is (4,16).
amistre64
  • amistre64
really? i see it as (6,36) meself (y=6) = (y=sqrt(x)) 6 = sqrt(x) at x=36
anonymous
  • anonymous
|dw:1327531523131:dw|
amistre64
  • amistre64
your pic has way more parts then the question asks for
anonymous
  • anonymous
It's a rotation of y=6, but the shape is bounded by y=4, so the intersection of y=4 and y=sqrt(x) would be (16,4).
anonymous
  • anonymous
|dw:1327531686984:dw|
anonymous
  • anonymous
So I'm thinking of using the cylindrical shell method, 2Pi(radius)(height)
amistre64
  • amistre64
ok, since typos happen; lets review what youve posted: The region bounded above by the line, y = 6, below by the curve y=sqrt(x), and to the left by the y-axis. The solid is rotated along the following lines: C. The line y=6 D. The line x=36
anonymous
  • anonymous
Yes, that is correct.
amistre64
  • amistre64
|dw:1327531783399:dw|
amistre64
  • amistre64
then this is our graph of that region
amistre64
  • amistre64
shell is fine, but we need to establish that this is our representation
anonymous
  • anonymous
so you this is a graph for rotation around x=36?
amistre64
  • amistre64
yes, in fact its the same graph for rotate y=6 AND x=36
anonymous
  • anonymous
Oh oh
amistre64
  • amistre64
the inverse of y=sqrt(x) that we can use for this is x=y^2 that will give us the shell height for y=6 rotation
anonymous
  • anonymous
so when rotating around a Y-axis, we should put the curve as Y too?
amistre64
  • amistre64
first we should simply draw out the graphs; after we have the graphs down, then we worry about where to rotate at
amistre64
  • amistre64
the graph itself has nothing to do with what we rotate around to begin with
anonymous
  • anonymous
true
anonymous
  • anonymous
so the graph I had illustrated was inccorect
anonymous
  • anonymous
visualizing the graphs has been the toughest part for me!
amistre64
  • amistre64
|dw:1327532250390:dw|this is our shell representing rotate about y=6
amistre64
  • amistre64
the graph you drew was something else altogther :)
anonymous
  • anonymous
haha, I'm sorry.
anonymous
  • anonymous
I HAVE NO CLUE WHY I PUT Y = 4!!
amistre64
  • amistre64
'sok :) so\[Area_{shell}=\int_{0}^{6}2pi\ y\ f(y)\ dy\] \[Area_{shell}=2pi\int_{0}^{6} y\ y^2\ dy\] \[Area_{shell}=2pi\int_{0}^{6} y^3\ dy\] we agree so far?
anonymous
  • anonymous
|dw:1327532422099:dw|
amistre64
  • amistre64
thats better lol
amistre64
  • amistre64
i think i want to "move" our graph stuff to a better location casue I think im forgetting something; moving the graph changes no value, just positions
anonymous
  • anonymous
Since we are rotating around y=6, wouldn't we do pi(6-y)^2
amistre64
  • amistre64
|dw:1327532555854:dw|
amistre64
  • amistre64
ugh, yes, your right; been awhile :) x=(y-6)^2
anonymous
  • anonymous
so we could do the cross-section method
amistre64
  • amistre64
\[A=2pi\int_{0}^{6}y(y-6)^2dy\] \[A=2pi\int_{0}^{6}y(y^2-12y+36)dy\] \[A=2pi\int_{0}^{6}(y^3-12y^2+36y)dy\] looks better to me for the shell method.
anonymous
  • anonymous
Ahhh, it does..
amistre64
  • amistre64
disc would be: \[A=pi\int_{0}^{36}(x^{1/2}-6)dx\]
anonymous
  • anonymous
|dw:1327532793666:dw|
anonymous
  • anonymous
what i posted above is for the cross-section, or it works for it..
amistre64
  • amistre64
\[A=pi\int_{0}^{36}(x^{1/2}-6)^2dx\] forgot to r^2 the area :)
amistre64
  • amistre64
the disc gives me: 216 pi the shell gives me: 216 pi
anonymous
  • anonymous
That is right!
amistre64
  • amistre64
yay!!
anonymous
  • anonymous
ok so I'm still a little confused how we got there..
amistre64
  • amistre64
lets turn this on its side so it looks like an upsidedown cone; the math doesnt care how it looks :)
anonymous
  • anonymous
2pi(r)(h) 2pi\[2\pi \int\limits_{0}^{6} y(6-y)\]
amistre64
  • amistre64
|dw:1327533114400:dw|
amistre64
  • amistre64
good, r = y and h=f(y), which is the inverse of f(x) which we moved down by 6
amistre64
  • amistre64
y=sqrt(x)-6 y+6 = sqrt(x) (y+6)^2 = x = f(y) I used the other side of the line so I got (y-6)^2 which makes no difference here
amistre64
  • amistre64
well, it does according to wolfram but .... fer some reason i did it right to begin with lol
anonymous
  • anonymous
hmm.. so what would be the equation we are trying to integrate and with what limits? 0 to 36?
amistre64
  • amistre64
Volume, not Area; named it wrong .... it bites getting old\[V=2\pi\int_{0}^{-6}y(y+6)^2dy=216\pi\]
amistre64
  • amistre64
the 0 to 36 is for the disc method for this thing
amistre64
  • amistre64
i need to downlaod google chrome; this thing is just way too slow in IE
anonymous
  • anonymous
haha it runs great in chrome
anonymous
  • anonymous
so why is the (y+6) being squared?
anonymous
  • anonymous
Thanks so much for your help, you are great!
amistre64
  • amistre64
college computers here clear out and reboot so any downloads are erased
amistre64
  • amistre64
w have to tell the integration to look at the the proper rendition of out equation for height
anonymous
  • anonymous
ohhhhh I see!! I was looking at the limit from x=y instead of the axis of revolution.
amistre64
  • amistre64
the direction we are looking at for the shell measuers up from the y axis, or over ; but we have to rewrite the height line in terms of y for it to make sence in out integration using shells
amistre64
  • amistre64
since w is the inverse of a normal xy graph we inverse our y=sqrt(x) to match it
amistre64
  • amistre64
since our view is the inverse ....
anonymous
  • anonymous
do you mean height instead of width?
anonymous
  • anonymous
I have the correct answer on my paper now, sweet!
anonymous
  • anonymous
Could you help me do a couple more? :)
amistre64
  • amistre64
i can, i got the chrome downlaoded now; doesnt help the typos, but its a bit quicker lol
anonymous
  • anonymous
cool, should I start an entirely new question so I can give you another medal?
amistre64
  • amistre64
not that the medals matter, but starting a new one might help my computer to keep up the pace :)
amistre64
  • amistre64
we didnt get thru rotate about the x=36 yet tho, so lets do that one on another post
anonymous
  • anonymous
Alright I'll do that!

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