find the volume of the region bounded by the graphs y = sqrt(x), y = 0, and x = 6, revolved about the line x = 6

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find the volume of the region bounded by the graphs y = sqrt(x), y = 0, and x = 6, revolved about the line x = 6

Mathematics
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:) again :)
hello! This problem is similar to the one we did together yesterday, but for some reason I am still doing this wrong
I know that I am supposed to do the disk method, and that I need to solve for each equation in terms of y

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rewrite y= sqrt(x) in terms of x = f(y) y = sqrt(x) ; ^2 both side y^2 = x this we can solve for easily right?
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
x = 6 is a pain, we want to move it to the x=0 line to examine it; how do we move x = 6 to x=0? x= 6 ; -6 from both sides x-6 = 0 do the same for the x=y^2 to keep it in the same position
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 ?
move it all to the left by 9 to examine it at x=0; does that make sense? Its over on a shelf and we want to examine it at the table where we can see it clearly; the table is the origin of the graph where x=0 and y=0, so we move it
So now I am looking at a graph that is shifted backwards by 9 on my calculator
y = sqrt(x) ; change to an equal form to accomodate for the y axis y^2 = x ; now move it over to the x=0 by subtracting 9 y^2 -9 = x-9 ; the x-9 can be rewritten as just x
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
Find the volumes of the solids generated by revolving the regions bounded by the graphs of the equations about the given lines. y = sqrt(x) y = 0 x = 6 about the line x = 9
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?
http://www.wolframalpha.com/input/?i=graph+y+%3D+sqrt%28x%29+from+x+%3D+6+to+x%3D9+y+%3D+0+to+y+%3D+3 Rotate that around x=9
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If the problem said bounded by x=0, x=6 and y= 0 and y = sqrt(x) I would agree with that picture. It doesn't say x=0.
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.
with this boundary; y = [0,sqrt(6)]
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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
y = sqrt(x); curve of sqrt(x) y = 0 ; the x axis is y=0 x = 6 ; the left boundary x=9; axis to spin around
x=6 the right bound
x = 6 is the left bound in my interpretation of the problem, and the axis of rotation would be 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
the region to cut from my original; if im right is: is the cylindar of r = 3, and height = sqrt(6) 216 sqrt(6) pi ------------ 5
okay, this is the right answer. Can you go over it one more time?
http://www.wolframalpha.com/input/?i=%28int%28pi%28y^2-9%29^2%29+dy+from+0+to+sqrt%286%29%29++-+%289sqrt%286%29pi%29+
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\[\pi \int\limits_{0}^{3}(....?....)\]
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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?
that gap is not a part of the volume when revolved so it must be cut out; like makeing a solid donut then removing the middle
it's what both of you said. Now I am trying to determine what is the best way to approach this problem
So I remove the middle when the region is bounded off by a number x = # that is smaller that the x = # that it is rotating about?
it's not what I said if it's what he said. I have a very different answer.
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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
for mine: we did the stuff above and then I subtracted the middle in an equation like this: \[\pi.\int\limits_{0}^{\sqrt{6}} (y^4 -18y^2+81) - (-3)^2.dy\]
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
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and still do for that matter, but I could be wrong.

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