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**God****Member**- Registered: 2005-08-25
- Posts: 59

Given the function:

(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:

Thanks

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

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Plot:

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

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**God****Member**- Registered: 2005-08-25
- Posts: 59

? I don't get it

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

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Is this write?

I just orbitained it:

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**God****Member**- Registered: 2005-08-25
- Posts: 59

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

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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You can't integrate it exactly.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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It's just coordinate change.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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We must find the equation of the line in the coordinate system, roated 45 degrees ++clock:

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

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,582

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

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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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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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Here is it:

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

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Symplifyng...

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Done:

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

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**John E. Franklin****Member**- Registered: 2005-08-29
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I don't understand a lot of this rotation, but here are some questions.

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
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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.

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

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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For the first -you're right.

For the second-no.

I'm starting explaining everything.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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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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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Picture:

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

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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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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**John E. Franklin****Member**- Registered: 2005-08-29
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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.

**igloo** **myrtilles** **fourmis**

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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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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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Ready! here is it:

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

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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And here's the proof thingy:

Let ff[x]=(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:

(rectanguler with 45 deg)

Then

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

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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I'm simplifying the AREA...

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Area = 5.66169988597863380413031960736...

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