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.

- anonymous

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- amistre64

which method?

- amistre64

|dw:1327531365366:dw|

- amistre64

we need to know the point of intersection here

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## More answers

- anonymous

Hey sorry, keyboard died. The point of intersection is (4,16).

- amistre64

really? i see it as (6,36) meself
(y=6) = (y=sqrt(x))
6 = sqrt(x) at x=36

- anonymous

|dw:1327531523131:dw|

- amistre64

your pic has way more parts then the question asks for

- 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

|dw:1327531686984:dw|

- anonymous

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

- 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

Yes, that is correct.

- amistre64

|dw:1327531783399:dw|

- amistre64

then this is our graph of that region

- amistre64

shell is fine, but we need to establish that this is our representation

- anonymous

so you this is a graph for rotation around x=36?

- amistre64

yes, in fact its the same graph for rotate y=6 AND x=36

- anonymous

Oh oh

- 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

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

- amistre64

first we should simply draw out the graphs; after we have the graphs down, then we worry about where to rotate at

- amistre64

the graph itself has nothing to do with what we rotate around to begin with

- anonymous

true

- anonymous

so the graph I had illustrated was inccorect

- anonymous

visualizing the graphs has been the toughest part for me!

- amistre64

|dw:1327532250390:dw|this is our shell representing rotate about y=6

- amistre64

the graph you drew was something else altogther :)

- anonymous

haha, I'm sorry.

- anonymous

I HAVE NO CLUE WHY I PUT Y = 4!!

- 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

|dw:1327532422099:dw|

- amistre64

thats better lol

- 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

Since we are rotating around y=6, wouldn't we do pi(6-y)^2

- amistre64

|dw:1327532555854:dw|

- amistre64

ugh, yes, your right; been awhile :)
x=(y-6)^2

- anonymous

so we could do the cross-section method

- 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

Ahhh, it does..

- amistre64

disc would be:
\[A=pi\int_{0}^{36}(x^{1/2}-6)dx\]

- anonymous

|dw:1327532793666:dw|

- anonymous

what i posted above is for the cross-section, or it works for it..

- amistre64

\[A=pi\int_{0}^{36}(x^{1/2}-6)^2dx\]
forgot to r^2 the area :)

- amistre64

the disc gives me: 216 pi
the shell gives me: 216 pi

- anonymous

That is right!

- amistre64

yay!!

- anonymous

ok so I'm still a little confused how we got there..

- amistre64

lets turn this on its side so it looks like an upsidedown cone; the math doesnt care how it looks :)

- anonymous

2pi(r)(h)
2pi\[2\pi \int\limits_{0}^{6} y(6-y)\]

- amistre64

|dw:1327533114400:dw|

- amistre64

good, r = y and h=f(y), which is the inverse of f(x) which we moved down by 6

- 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

well, it does according to wolfram but .... fer some reason i did it right to begin with lol

- anonymous

hmm.. so what would be the equation we are trying to integrate and with what limits? 0 to 36?

- amistre64

Volume, not Area; named it wrong .... it bites getting old\[V=2\pi\int_{0}^{-6}y(y+6)^2dy=216\pi\]

- amistre64

the 0 to 36 is for the disc method for this thing

- amistre64

i need to downlaod google chrome; this thing is just way too slow in IE

- anonymous

haha it runs great in chrome

- anonymous

so why is the (y+6) being squared?

- anonymous

Thanks so much for your help, you are great!

- amistre64

college computers here clear out and reboot so any downloads are erased

- amistre64

w have to tell the integration to look at the the proper rendition of out equation for height

- anonymous

ohhhhh I see!! I was looking at the limit from x=y instead of the axis of revolution.

- 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

since w is the inverse of a normal xy graph we inverse our y=sqrt(x) to match it

- amistre64

since our view is the inverse ....

- anonymous

do you mean height instead of width?

- anonymous

I have the correct answer on my paper now, sweet!

- anonymous

Could you help me do a couple more? :)

- amistre64

i can, i got the chrome downlaoded now; doesnt help the typos, but its a bit quicker lol

- anonymous

cool, should I start an entirely new question so I can give you another medal?

- amistre64

not that the medals matter, but starting a new one might help my computer to keep up the pace :)

- amistre64

we didnt get thru rotate about the x=36 yet tho, so lets do that one on another post

- anonymous

Alright I'll do that!

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