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#1 2005-12-30 11:33:51

God
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Registered: 2005-08-25
Posts: 59

Help with 3D rotation

Given the function:
ƒ(x) = e^x-ln(x)-x^x on the interval 1/e <= x <= e

I now have to rotate this around the line x + y = 1

What will be the volume of the solid?

Just for a visual:
function.gif

Thanks

Last edited by God (2005-12-30 12:12:32)

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#2 2005-12-30 11:52:29

krassi_holmz
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Re: Help with 3D rotation

Plot:

View Image: sssss.GIF

Last edited by krassi_holmz (2005-12-30 12:44:25)


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#3 2005-12-30 12:17:05

God
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Registered: 2005-08-25
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Re: Help with 3D rotation

? I don't get it sad

Last edited by God (2005-12-30 12:17:18)

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#4 2005-12-30 12:17:07

krassi_holmz
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Re: Help with 3D rotation

Is this write?
I just orbitained it:


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#5 2005-12-30 12:21:44

God
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Registered: 2005-08-25
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Re: Help with 3D rotation

See the problem, to be more specific, is that I can't figure out how to integrate ƒ(x)... esp. the x^x

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#6 2005-12-30 12:28:01

krassi_holmz
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Re: Help with 3D rotation

You can't integrate it exactly.


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#7 2005-12-30 12:30:53

krassi_holmz
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Re: Help with 3D rotation

It's just coordinate change.


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#8 2005-12-30 12:41:05

krassi_holmz
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Re: Help with 3D rotation

We must find the equation of the line in the coordinate system, roated 45 degrees ++clock:

View Image: coordinate.GIF

Last edited by krassi_holmz (2005-12-30 12:45:16)


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#9 2005-12-30 12:44:14

John E. Franklin
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Re: Help with 3D rotation

I think you could divide the area up into parts as a start.

View Image: mathisfun18.gif

igloo myrtilles fourmis

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#10 2005-12-30 13:00:42

krassi_holmz
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Re: Help with 3D rotation

If we use polar coordinates is easy:
Let A= (r, alfa)pol
Then
A'= (r,alfa+45 deg)pol= (r,alfa+(1/8)2Pi rad)pol= (r,alfa+Pi/4)pol.
Now from decart c.s. to polar c.s.:
Let A= (x,y)dec
r=sqrt(x^2+y^2)
cos(alfa)=y/r
alfa=arccos(y/r)=arccos(y/sqrt(x^2+y^2))

Then
A'= (sqrt(x^2+y^2),arccos(y/sqrt(x^2+y^2))+Pi/4)pol
Now from polar to decart:

x'=r'*sin(alfa')=sqrt(x^2+y^2)*sin(arccos(y/sqrt(x^2+y^2))+Pi/4)
y'=r'*cos(alfa')=sqrt(x^2+y^2)*cos(arccos(y/sqrt(x^2+y^2))+Pi/4)

Last edited by krassi_holmz (2005-12-30 13:07:35)


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#11 2005-12-30 21:35:23

krassi_holmz
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Re: Help with 3D rotation

Here is it:

View Image: int area.GIF

Last edited by krassi_holmz (2005-12-30 21:36:47)


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#12 2005-12-30 21:37:25

krassi_holmz
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Re: Help with 3D rotation

Symplifyng...


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#13 2005-12-31 02:06:38

krassi_holmz
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Re: Help with 3D rotation

Done:

View Image: sym.GIF

Last edited by krassi_holmz (2005-12-31 02:10:01)


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#14 2005-12-31 03:21:30

John E. Franklin
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Re: Help with 3D rotation

I don't understand a lot of this rotation, but here are some questions.

View Image: mathisfun19.gif

igloo myrtilles fourmis

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#15 2005-12-31 03:32:07

John E. Franklin
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Re: Help with 3D rotation

Also I was wondering if you considered that if the function is rotated,
some of the new vertical values are directly above one another, thus
it is not a one-to-one function.

View Image: mathisfun20.GIF

Last edited by John E. Franklin (2006-01-01 08:52:02)


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#16 2005-12-31 03:50:02

krassi_holmz
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Re: Help with 3D rotation

For the first -you're right.
For the second-no.
I'm starting explaining everything.


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#17 2005-12-31 05:20:52

krassi_holmz
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Re: Help with 3D rotation

1. Coordinate transforms:
Let the point A has coordinates {a,b} in coordinate system xOy.
We must find the coordinates of a in coordinate system x'Oy', which is xOy rotated 45° ++clock:


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#18 2005-12-31 09:36:53

krassi_holmz
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Re: Help with 3D rotation

Picture:

View Image: cor.GIF

Last edited by krassi_holmz (2005-12-31 09:38:03)


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#19 2005-12-31 09:42:44

krassi_holmz
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Re: Help with 3D rotation

As you see the coordinates of A in the second coordinate system A=={x',y'}II are equal to the coordinates of point A1 =={x',y'}I, which is A, rotated 45deg --clock.
So we must find what will be the coorinates of point A{x,y}, when we rotate it 45deg --clock.


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#20 2006-01-01 08:46:19

John E. Franklin
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Re: Help with 3D rotation

slope = y' = e^x - 1/x - x^x(1+lnx)

By looking at a graph, it looks like x=.67 is about horizontal slope, so
try .67 for x to see if is nearly zero slope., and I get slope = 0.003267, which is awesome!!

Now try 1/e to see if it is slope of -1.
try x = .367879441, and I got slope  is -1.2736, thus the tangent with slope -1 is to the right a little bit.
What is this number for x??
Will try to find this x with a BASIC program.   Got x = 0.4112922
From this we can conclude that "area 4" is non-zero.
So we have 4 regions to rotate around y=-x+1 to find the volume.


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#21 2006-01-01 09:55:25

krassi_holmz
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Re: Help with 3D rotation

Ha, ha, ha!
I tougth a while and I got very simple geometric solution.
But you must wait a wnile to make a pictures.


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#22 2006-01-01 10:40:35

krassi_holmz
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Re: Help with 3D rotation

Ready! here is it:

View Image: graph.JPG

Last edited by krassi_holmz (2006-01-01 10:41:30)


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#23 2006-01-01 10:49:38

krassi_holmz
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Re: Help with 3D rotation

And here's the proof thingy:
Let ff[x]=ƒ(x)
The wanted area won't change if we push it one up. We to this to reduce ff to positive function.
We want S+S1+S3.
(x,y) means point x,y.


But
(rectangular with 45 deg)

so S = INEGRAL - S2.

Now we'll find S1 and S3:

(rectangular with 45 deg)
(rectanguler with 45 deg)

Then

Last edited by krassi_holmz (2006-01-01 11:05:20)


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#24 2006-01-01 11:13:12

krassi_holmz
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Re: Help with 3D rotation

I'm simplifying the AREA...


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#25 2006-01-01 11:16:23

krassi_holmz
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Re: Help with 3D rotation

Area = 5.66169988597863380413031960736...


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