Question

find the area of the graph bounded by y=2x^2 , y=0 , and x=2 , revolved around the line x=2 all I need help doing is figuring out how to set up the problem up.

Answer 1

I do believe it would be easiest to use disk method however I just can't picture how to find the formula that represents the part of the function revolved around the line x=2

Answer 2

area of the graph? isn't this volume?

Answer 3

yes I always slip and say that

Answer 4

have you tried graphing it?

Answer 5

yes it is just a parabola

Answer 6

you do know that x = 2 is not an asymptote right?

Answer 7

I just can't figure out how to write this as a formula representing the area bounded by the given parameters with respect to y

Answer 8

Yes

Answer 9

It is the axis that I need to revolve the region around

Answer 10

since your revolution is horizontal, that means you should use dx as your integral

Answer 11

the revolution is around a vertical line so I was under the impression that when you revolve it around a vertical line you set it up in respect to that axis(vertical) or in this case in respect to y

Answer 12

so y=.......

Answer 13

yes you do

Answer 14

y =.... is a function of x

Answer 15

is that dx as my integral?

Answer 16

yes you have something like \[\int \limits_{x = a}^{x=b} f(x) dx\]
something close to that

Answer 17

This looks very familiar . . .

Answer 18

right I understand that.... I hope :). Problem for me is picturing how to write this function because it revolves around x=2. If I just re-write the function as 2(x^2)+2 I think it is wrong. Because that just shifts the parabola up, so how do I write this function where it represents the region bounded by the graph? That is where I am having difficulty. And yes @CliffSedge this is a problem someone was helping me with when OS crashed the other night

Answer 19

Is it just 2-(2(x^2))?

Answer 20

I remember when I saw this before that I recommended shifting it over so it revolved around the x-axis. You'd have to transform the equation to y=2(x+2)^2 to get it to line up.

Answer 21

Then you can integrate from -2 to 0.
I don't know if that transformation is necessary (probably isn't), but it helps me visualize it to set it up.

Answer 22

At that point we had some computer technologies problems so our lecture notes were printing wrong and it had this listed as the shell method or something so I had my ups down and my downs up. So you say make it 2(x-2)^2?

Answer 23

I prefer that only so that it revolves around the x-axis instead and the radius is easier to define and I can see the symmetry easier.

Answer 24

I was visualizing it as being a distance of 2 from the y axis. so x=2 is a constant function. Then if that is the greater distance then wouldn't it be the distance "2" minus the distance"2(x^2)"?

Answer 25

That would be pretty difficult to do if it was a polynomial as compared to a quadratic.Suppose I was given a polynomial function and then given a region it was bounded by. If I translated that it could lead to some very time consuming integration.

Answer 26

Not necessarily. Translations are pretty quick fixes, but I agree, you shouldn't spend time on it if you don't have to. I'm only doing it in this case to help me visualize how to set up the integration.
It looks like that if you want to integrate w.r.t. x then the shell method might be best.

Answer 27

I have not done the shell method yet. this is from the disk and washer method chapters. I will be tested to see if I can set these types of problems up using one or the other.

Answer 28

Do you know how to do them that way without translations?

Answer 29

I think it'll work fine as long as you define what the differential volume is.
For shells, it's a bunch of nested hollow cylinders. The volume of those is surface area of a cylinder times the thickness, dx.

Answer 30

Surface area is circumference times height, so volume = 2πrh dx. It looks like you can call the radius (2-x), and the height, h = 2x^2.
I think that should work.