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which method?

|dw:1327531365366:dw|

we need to know the point of intersection here

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

|dw:1327531686984:dw|

So I'm thinking of using the cylindrical shell method, 2Pi(radius)(height)

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

so when rotating around a Y-axis, we should put the curve as Y too?

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!!

|dw:1327532422099:dw|

thats better lol

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

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

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.

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!