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.

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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
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which method?
|dw:1327531365366:dw|
we need to know the point of intersection here

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

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