anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
:) again :)
anonymous
  • anonymous
hello! This problem is similar to the one we did together yesterday, but for some reason I am still doing this wrong
anonymous
  • anonymous
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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amistre64
  • amistre64
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?
anonymous
  • anonymous
so I have v = pi (the integral from 0 to 6) (y+9)^2 - (9)^2 dx
amistre64
  • amistre64
graph this and see that it is the same
anonymous
  • anonymous
yes
amistre64
  • amistre64
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
anonymous
  • anonymous
minus 6 from both sides in the equation I wrote?
amistre64
  • amistre64
x = y^2 ; -6 both sides x-6 = y^2 -6
anonymous
  • anonymous
whoops---I am revolving around x = 9, sorry about that
amistre64
  • amistre64
brb..
anonymous
  • anonymous
okay
amistre64
  • amistre64
back...
anonymous
  • anonymous
hi
amistre64
  • amistre64
since it 9 we just -9 fro itall
anonymous
  • anonymous
so v = pi (intergral from 0 to 6) (y + 9)^2 - 9^2 ?
amistre64
  • amistre64
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
anonymous
  • anonymous
So now I am looking at a graph that is shifted backwards by 9 on my calculator
amistre64
  • amistre64
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
anonymous
  • anonymous
sqrt(x +9)
anonymous
  • anonymous
okay so it is y^2 -9 not +9
amistre64
  • amistre64
sqrt(x+9) is good; now reform it y = sqrt(x+9) y^2 = x+9 y^2 -9 = x
amistre64
  • amistre64
whats our bounds now: for y=0 to how far?
anonymous
  • anonymous
x = 6....?
anonymous
  • anonymous
I've never seen anyone do transformations of a graph this way..
amistre64
  • amistre64
y = sqrt(9) y = 3
anonymous
  • anonymous
we have to redo them when we solve for a different variable...keep forgetting
amistre64
  • amistre64
integrate the discs from 0 to 3
anonymous
  • anonymous
okay
amistre64
  • amistre64
pi {S} y^2-9 dy ; [0,3] = volume of revolution
amistre64
  • amistre64
forgot to square the function lol
amistre64
  • amistre64
pi {S} [y^2-9]^2 dy ; [0,3]
anonymous
  • anonymous
so you don't have to subtract another 9?
amistre64
  • amistre64
why would you subtract two 9s?
amistre64
  • amistre64
we only move the graph over to x=0 from x=9 right?
anonymous
  • anonymous
one from the first eqn and one from eqn y = 0?
amistre64
  • amistre64
if i tellyou to move 9 steps to the left; what do you do? move 18?
anonymous
  • anonymous
ohhhh
amistre64
  • amistre64
y=0 is already bounding the volume; no need to adjust that part lol
anonymous
  • anonymous
answer is 648/5?
amistre64
  • amistre64
dunno; lets check..
anonymous
  • anonymous
no, its' not
amistre64
  • amistre64
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
anonymous
  • anonymous
had that part
anonymous
  • anonymous
and got 648/5, but the homework grader marked it wrong
amistre64
  • amistre64
i get 129.6pi
anonymous
  • anonymous
same answer, and the wrong one. What did we do wrong?
anonymous
  • anonymous
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?
amistre64
  • amistre64
243 27 243 pi( ---- - --- + ---) right? 5 3 1
amistre64
  • amistre64
x = 0 to 9
anonymous
  • anonymous
right, and we are rotating the region about the line x = 9
amistre64
  • amistre64
ilove; retype the problem in its entirety so that we have a clear picture
anonymous
  • anonymous
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
amistre64
  • amistre64
ohhh... so its a donut shape; i assumed you meant y=0; x=9 about the x=9
anonymous
  • anonymous
oh, nope sorry about that
anonymous
  • anonymous
I don't think it's a doughnut
amistre64
  • amistre64
its a torus alright :)
anonymous
  • anonymous
More like a mountain.
amistre64
  • amistre64
we have to subtract the middle of what we found to get the volume then
anonymous
  • anonymous
oh wait. I'm looking at wrong graph
anonymous
  • anonymous
Small hill. Not a doughnut. There is no hole in the middle.
amistre64
  • amistre64
x=6 to x=9 is empty tho
anonymous
  • anonymous
there is no hole it the middle? isn't the shape a bunt cake?
anonymous
  • anonymous
no
anonymous
  • anonymous
graph y = sqrt(x) from x = 6 to x=9 y = 0 to y = 3
anonymous
  • anonymous
but the region is cut off at x = 6, and we have to get it to revolve about line x = 9
anonymous
  • anonymous
that's where it starts
anonymous
  • anonymous
not where it ends. it doesn't say it's bounded by x=0
anonymous
  • anonymous
hmmm...then how to we caluclate where it begins and ends?
anonymous
  • anonymous
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
amistre64
  • amistre64
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anonymous
  • anonymous
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.
anonymous
  • anonymous
so using what you just said, the boundries would be?
anonymous
  • anonymous
x = 6, x = 9, y = 0, y = sqrt(x) and rotate around x = 9
amistre64
  • amistre64
then we would double our results right?
anonymous
  • anonymous
no, you have the wrong part of the graph I think.
amistre64
  • amistre64
with this boundary; y = [0,sqrt(6)]
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anonymous
  • anonymous
You want to integrate over the white part in your picture rather than the grey part.
anonymous
  • anonymous
what wouuld the equation look like?
anonymous
  • anonymous
\[v = \pi \int\limits_{0}^{3}...\]
anonymous
  • anonymous
You're using disks or shells?
anonymous
  • anonymous
We can only use the disk method, right?
anonymous
  • anonymous
where did you get the boundaries 0 to 3 from , again?
anonymous
  • anonymous
do you have the value for the answer so we can be sure we're doing the right region?
anonymous
  • anonymous
nope----online h/w problem
anonymous
  • anonymous
dumb.
anonymous
  • anonymous
I'm pretty sure I'm reading it right though. For this I would use cylindrical shells
amistre64
  • amistre64
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
amistre64
  • amistre64
x=6 the right bound
anonymous
  • anonymous
x = 6 is the left bound in my interpretation of the problem, and the axis of rotation would be the right bound.
anonymous
  • anonymous
???? what do you get as your answer?
anonymous
  • anonymous
I thought the entire region was left bound, and the axis is right bound.
anonymous
  • anonymous
like left-right or top-bottom
amistre64
  • amistre64
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
anonymous
  • anonymous
okay, this is the right answer. Can you go over it one more time?
amistre64
  • amistre64
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+
amistre64
  • amistre64
which answer is correct? :)
anonymous
  • anonymous
So it was the doughnut?
anonymous
  • anonymous
both are correct, but how do you do it correctly?
anonymous
  • anonymous
They should have been more specific about the left boundary... grr..
amistre64
  • amistre64
lol..... the only way we are both correct is if the volume of each version are identical :)
anonymous
  • anonymous
I have not subtracted by 9sqrt(6) pi at all in what I have been doing though
anonymous
  • anonymous
okay, so i should have\[\pi \int\limits_{0}^{3}(....?....)\]
amistre64
  • amistre64
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amistre64
  • amistre64
your interval of [0,3] is wrong, since we only need the the sqrt(6)
anonymous
  • anonymous
how do I know when to remove a cylinder?
amistre64
  • amistre64
when the spin axis is not the same is the boundary axis; there is a gap that is formed
anonymous
  • anonymous
So you've confirmed the answer is what amistre has?
amistre64
  • amistre64
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
anonymous
  • anonymous
it's what both of you said. Now I am trying to determine what is the best way to approach this problem
anonymous
  • anonymous
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?
anonymous
  • anonymous
it's not what I said if it's what he said. I have a very different answer.
amistre64
  • amistre64
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amistre64
  • amistre64
when the spin is different from the bound, you get an offset that needs to be accounted for....
anonymous
  • anonymous
.....
anonymous
  • anonymous
polpak, with your graph, how did you attempt the problem?
anonymous
  • anonymous
did you eliminate the middle part in the volume equation?
amistre64
  • amistre64
polpak eliminated the outer part
anonymous
  • anonymous
can I get this step by step? I am totally confused. I need to see an equation
anonymous
  • anonymous
well we're still not clear on what problem you're solving ;p
amistre64
  • amistre64
for mine: we did the stuff above and then I subtracted the middle in an equation like this: \[\pi.\int\limits_{0}^{\sqrt{6}} (y^4 -18y^2+81) - (-3)^2.dy\]
anonymous
  • anonymous
Okay,that i what I was looking for. I had every thing except the -3. where did that come from?
amistre64
  • amistre64
that is the y = -3 line that result from the boundary between x=6 and x=9
amistre64
  • amistre64
teh 3 and 6 in my circle graph should be switched :) and then you can see the "3"
anonymous
  • anonymous
OHHHHHHH
anonymous
  • anonymous
okay, so now I finally get it. Thanks everyone! :D Thanks for staying on so long amistre64
amistre64
  • amistre64
hey, if you dont figure it out, neither do I lol
anonymous
  • anonymous
see ya later!
anonymous
  • anonymous
I thought it was this
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anonymous
  • anonymous
and still do for that matter, but I could be wrong.

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