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:) again :)

so I have v = pi (the integral from 0 to 6) (y+9)^2 - (9)^2 dx

graph this and see that it is the same

yes

minus 6 from both sides in the equation I wrote?

x = y^2 ; -6 both sides
x-6 = y^2 -6

whoops---I am revolving around x = 9, sorry about that

brb..

okay

back...

hi

since it 9 we just -9 fro itall

so v = pi (intergral from 0 to 6) (y + 9)^2 - 9^2 ?

So now I am looking at a graph that is shifted backwards by 9 on my calculator

sqrt(x +9)

okay so it is y^2 -9 not +9

sqrt(x+9) is good; now reform it
y = sqrt(x+9)
y^2 = x+9
y^2 -9 = x

whats our bounds now: for y=0 to how far?

x = 6....?

I've never seen anyone do transformations of a graph this way..

y = sqrt(9)
y = 3

we have to redo them when we solve for a different variable...keep forgetting

integrate the discs from 0 to 3

okay

pi {S} y^2-9 dy ; [0,3] = volume of revolution

forgot to square the function lol

pi {S} [y^2-9]^2 dy ; [0,3]

so you don't have to subtract another 9?

why would you subtract two 9s?

we only move the graph over to x=0 from x=9 right?

one from the first eqn and one from eqn y = 0?

if i tellyou to move 9 steps to the left; what do you do? move 18?

ohhhh

y=0 is already bounding the volume; no need to adjust that part lol

answer is 648/5?

dunno; lets check..

no, its' not

pi {S} (y^2-9)^2 dy
pi {S} (y^4 -18y^2 +81) dy
pi(y^5/5 -18y^3/3 +81y) at y=3

had that part

and got 648/5, but the homework grader marked it wrong

i get 129.6pi

same answer, and the wrong one. What did we do wrong?

Hang on. You're rotating the graph of y= sqrt(x) from x=6 to x=9 about the line x=9 ? Is that right?

243 27 243
pi( ---- - --- + ---) right?
5 3 1

x = 0 to 9

right, and we are rotating the region about the line x = 9

ilove; retype the problem in its entirety so that we have a clear picture

ohhh... so its a donut shape; i assumed you meant y=0; x=9 about the x=9

oh, nope sorry about that

I don't think it's a doughnut

its a torus alright :)

More like a mountain.

we have to subtract the middle of what we found to get the volume then

oh wait. I'm looking at wrong graph

Small hill. Not a doughnut. There is no hole in the middle.

x=6 to x=9 is empty tho

there is no hole it the middle? isn't the shape a bunt cake?

no

graph y = sqrt(x) from x = 6 to x=9 y = 0 to y = 3

but the region is cut off at x = 6, and we have to get it to revolve about line x = 9

that's where it starts

not where it ends. it doesn't say it's bounded by x=0

hmmm...then how to we caluclate where it begins and ends?

so using what you just said, the boundries would be?

x = 6, x = 9, y = 0, y = sqrt(x) and rotate around x = 9

then we would double our results right?

no, you have the wrong part of the graph I think.

You want to integrate over the white part in your picture rather than the grey part.

what wouuld the equation look like?

\[v = \pi \int\limits_{0}^{3}...\]

You're using disks or shells?

We can only use the disk method, right?

where did you get the boundaries 0 to 3 from , again?

do you have the value for the answer so we can be sure we're doing the right region?

nope----online h/w problem

dumb.

I'm pretty sure I'm reading it right though. For this I would use cylindrical shells

x=6 the right bound

???? what do you get as your answer?

I thought the entire region was left bound, and the axis is right bound.

like left-right or top-bottom

okay, this is the right answer. Can you go over it one more time?

which answer is correct? :)

So it was the doughnut?

both are correct, but how do you do it correctly?

They should have been more specific about the left boundary... grr..

lol..... the only way we are both correct is if the volume of each version are identical :)

I have not subtracted by 9sqrt(6) pi at all in what I have been doing though

okay, so i should have\[\pi \int\limits_{0}^{3}(....?....)\]

your interval of [0,3] is wrong, since we only need the the sqrt(6)

how do I know when to remove a cylinder?

when the spin axis is not the same is the boundary axis; there is a gap that is formed

So you've confirmed the answer is what amistre has?

it's not what I said if it's what he said. I have a very different answer.

when the spin is different from the bound, you get an offset that needs to be accounted for....

.....

polpak, with your graph, how did you attempt the problem?

did you eliminate the middle part in the volume equation?

polpak eliminated the outer part

can I get this step by step? I am totally confused. I need to see an equation

well we're still not clear on what problem you're solving ;p

Okay,that i what I was looking for. I had every thing except the -3. where did that come from?

that is the y = -3 line that result from the boundary between x=6 and x=9

teh 3 and 6 in my circle graph should be switched :) and then you can see the "3"

OHHHHHHH

okay, so now I finally get it. Thanks everyone! :D Thanks for staying on so long amistre64

hey, if you dont figure it out, neither do I lol

see ya later!

I thought it was this

and still do for that matter, but I could be wrong.