Also here is some more information in a new plot showing three regions to be rotated, the hardest one being the skinny area I have labelled region C.]]>

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

I tougth a while and I got very simple geometric solution.

But you must wait a wnile to make a pictures.]]>

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.

So we must find what will be the coorinates of point A{x,y}, when we rotate it 45deg --clock.]]>

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:]]>