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

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- amistre64

:) again :)

- anonymous

hello! This problem is similar to the one we did together yesterday, but for some reason I am still doing this wrong

- 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- 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

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

- amistre64

graph this and see that it is the same

- anonymous

yes

- 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

minus 6 from both sides in the equation I wrote?

- amistre64

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

- anonymous

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

- amistre64

brb..

- anonymous

okay

- amistre64

back...

- anonymous

hi

- amistre64

since it 9 we just -9 fro itall

- anonymous

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

- 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

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

- 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

sqrt(x +9)

- anonymous

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

- amistre64

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

- amistre64

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

- anonymous

x = 6....?

- anonymous

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

- amistre64

y = sqrt(9)
y = 3

- anonymous

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

- amistre64

integrate the discs from 0 to 3

- anonymous

okay

- amistre64

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

- amistre64

forgot to square the function lol

- amistre64

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

- anonymous

so you don't have to subtract another 9?

- amistre64

why would you subtract two 9s?

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

ohhhh

- amistre64

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

- anonymous

answer is 648/5?

- amistre64

dunno; lets check..

- anonymous

no, its' not

- 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

had that part

- anonymous

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

- amistre64

i get 129.6pi

- anonymous

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

- 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

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

- amistre64

x = 0 to 9

- anonymous

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

- amistre64

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

- 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

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

- anonymous

oh, nope sorry about that

- anonymous

I don't think it's a doughnut

- amistre64

its a torus alright :)

- anonymous

More like a mountain.

- amistre64

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

- anonymous

oh wait. I'm looking at wrong graph

- anonymous

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

- amistre64

x=6 to x=9 is empty tho

- anonymous

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

- anonymous

no

- anonymous

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

- anonymous

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

- anonymous

that's where it starts

- anonymous

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

- anonymous

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

- 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

##### 1 Attachment

- 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

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

- anonymous

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

- amistre64

then we would double our results right?

- anonymous

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

- amistre64

with this boundary; y = [0,sqrt(6)]

##### 1 Attachment

- anonymous

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

- anonymous

what wouuld the equation look like?

- anonymous

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

- anonymous

You're using disks or shells?

- anonymous

We can only use the disk method, right?

- anonymous

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

- anonymous

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

- anonymous

nope----online h/w problem

- anonymous

dumb.

- anonymous

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

- 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

x=6 the right bound

- anonymous

x = 6 is the left bound in my interpretation of the problem, and the axis of rotation would be the right bound.

- anonymous

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

- anonymous

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

- anonymous

like left-right or top-bottom

- 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

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

- 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

which answer is correct? :)

- anonymous

So it was the doughnut?

- anonymous

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

- anonymous

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

- amistre64

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

- anonymous

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

- anonymous

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

- amistre64

##### 1 Attachment

- amistre64

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

- anonymous

how do I know when to remove a cylinder?

- amistre64

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

- anonymous

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

- 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

it's what both of you said. Now I am trying to determine what is the best way to approach this problem

- 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

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

- amistre64

##### 1 Attachment

- amistre64

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

- anonymous

.....

- anonymous

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

- anonymous

did you eliminate the middle part in the volume equation?

- amistre64

polpak eliminated the outer part

- anonymous

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

- anonymous

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

- 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

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

- amistre64

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

- amistre64

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

- anonymous

OHHHHHHH

- anonymous

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

- amistre64

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

- anonymous

see ya later!

- anonymous

I thought it was this

##### 1 Attachment

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.