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anonymous
 5 years ago
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
 5 years ago
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hello! This problem is similar to the one we did together yesterday, but for some reason I am still doing this wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know that I am supposed to do the disk method, and that I need to solve for each equation in terms of y

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1rewrite 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
 5 years ago
Best ResponseYou've already chosen the best response.0so I have v = pi (the integral from 0 to 6) (y+9)^2  (9)^2 dx

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1graph this and see that it is the same

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1x = 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 x6 = 0 do the same for the x=y^2 to keep it in the same position

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0minus 6 from both sides in the equation I wrote?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1x = y^2 ; 6 both sides x6 = y^2 6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whoopsI am revolving around x = 9, sorry about that

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1since it 9 we just 9 fro itall

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so v = pi (intergral from 0 to 6) (y + 9)^2  9^2 ?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1move 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
 5 years ago
Best ResponseYou've already chosen the best response.0So now I am looking at a graph that is shifted backwards by 9 on my calculator

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1y = 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 = x9 ; the x9 can be rewritten as just x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay so it is y^2 9 not +9

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1sqrt(x+9) is good; now reform it y = sqrt(x+9) y^2 = x+9 y^2 9 = x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1whats our bounds now: for y=0 to how far?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I've never seen anyone do transformations of a graph this way..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we have to redo them when we solve for a different variable...keep forgetting

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1integrate the discs from 0 to 3

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1pi {S} y^29 dy ; [0,3] = volume of revolution

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1forgot to square the function lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1pi {S} [y^29]^2 dy ; [0,3]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you don't have to subtract another 9?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1why would you subtract two 9s?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1we only move the graph over to x=0 from x=9 right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one from the first eqn and one from eqn y = 0?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1if i tellyou to move 9 steps to the left; what do you do? move 18?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1y=0 is already bounding the volume; no need to adjust that part lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1pi {S} (y^29)^2 dy pi {S} (y^4 18y^2 +81) dy pi(y^5/5 18y^3/3 +81y) at y=3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and got 648/5, but the homework grader marked it wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0same answer, and the wrong one. What did we do wrong?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hang 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
 5 years ago
Best ResponseYou've already chosen the best response.1243 27 243 pi(    + ) right? 5 3 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right, and we are rotating the region about the line x = 9

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1ilove; retype the problem in its entirety so that we have a clear picture

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Find 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
 5 years ago
Best ResponseYou've already chosen the best response.1ohhh... so its a donut shape; i assumed you meant y=0; x=9 about the x=9

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, nope sorry about that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't think it's a doughnut

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1its a torus alright :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0More like a mountain.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1we have to subtract the middle of what we found to get the volume then

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh wait. I'm looking at wrong graph

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Small hill. Not a doughnut. There is no hole in the middle.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1x=6 to x=9 is empty tho

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there is no hole it the middle? isn't the shape a bunt cake?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0graph y = sqrt(x) from x = 6 to x=9 y = 0 to y = 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but the region is cut off at x = 6, and we have to get it to revolve about line x = 9

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that's where it starts

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not where it ends. it doesn't say it's bounded by x=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm...then how to we caluclate where it begins and ends?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If 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
 5 years ago
Best ResponseYou've already chosen the best response.0so using what you just said, the boundries would be?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x = 6, x = 9, y = 0, y = sqrt(x) and rotate around x = 9

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1then we would double our results right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, you have the wrong part of the graph I think.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1with this boundary; y = [0,sqrt(6)]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You want to integrate over the white part in your picture rather than the grey part.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what wouuld the equation look like?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[v = \pi \int\limits_{0}^{3}...\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You're using disks or shells?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We can only use the disk method, right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where did you get the boundaries 0 to 3 from , again?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you have the value for the answer so we can be sure we're doing the right region?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nopeonline h/w problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm pretty sure I'm reading it right though. For this I would use cylindrical shells

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1y = 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x = 6 is the left bound in my interpretation of the problem, and the axis of rotation would be the right bound.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0???? what do you get as your answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I thought the entire region was left bound, and the axis is right bound.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like leftright or topbottom

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the 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
 5 years ago
Best ResponseYou've already chosen the best response.0okay, this is the right answer. Can you go over it one more time?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1which answer is correct? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So it was the doughnut?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0both are correct, but how do you do it correctly?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0They should have been more specific about the left boundary... grr..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1lol..... the only way we are both correct is if the volume of each version are identical :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have not subtracted by 9sqrt(6) pi at all in what I have been doing though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay, so i should have\[\pi \int\limits_{0}^{3}(....?....)\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1your interval of [0,3] is wrong, since we only need the the sqrt(6)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do I know when to remove a cylinder?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1when the spin axis is not the same is the boundary axis; there is a gap that is formed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So you've confirmed the answer is what amistre has?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1that 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
 5 years ago
Best ResponseYou've already chosen the best response.0it's what both of you said. Now I am trying to determine what is the best way to approach this problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So 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
 5 years ago
Best ResponseYou've already chosen the best response.0it's not what I said if it's what he said. I have a very different answer.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1when the spin is different from the bound, you get an offset that needs to be accounted for....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0polpak, with your graph, how did you attempt the problem?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did you eliminate the middle part in the volume equation?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1polpak eliminated the outer part

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can I get this step by step? I am totally confused. I need to see an equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well we're still not clear on what problem you're solving ;p

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1for 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
 5 years ago
Best ResponseYou've already chosen the best response.0Okay,that i what I was looking for. I had every thing except the 3. where did that come from?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1that is the y = 3 line that result from the boundary between x=6 and x=9

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1teh 3 and 6 in my circle graph should be switched :) and then you can see the "3"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay, so now I finally get it. Thanks everyone! :D Thanks for staying on so long amistre64

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1hey, if you dont figure it out, neither do I lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I thought it was this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and still do for that matter, but I could be wrong.
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