[Graphics:../Images/index_gr_55.gif]

Answer:

Here's a picture of the area measured by
       [Graphics:../Images/index_gr_56.gif]:

[Graphics:../Images/index_gr_57.gif]

[Graphics:../Images/index_gr_58.gif]

Better find out where the area goes from positive to negative:

[Graphics:../Images/index_gr_59.gif]
[Graphics:../Images/index_gr_60.gif]

The measurement
       [Graphics:../Images/index_gr_61.gif]
of the triangular area on the left is negative because it is below the [Graphics:../Images/index_gr_62.gif]-axis:

[Graphics:../Images/index_gr_63.gif]
[Graphics:../Images/index_gr_64.gif]

Now look at the triangle on the right:

[Graphics:../Images/index_gr_65.gif]

[Graphics:../Images/index_gr_66.gif]

The measurement
       [Graphics:../Images/index_gr_67.gif]
of the triangular area on the right is positive:

[Graphics:../Images/index_gr_68.gif]
[Graphics:../Images/index_gr_69.gif]

Consequently,  
       [Graphics:../Images/index_gr_70.gif]
is given by:

[Graphics:../Images/index_gr_71.gif]
[Graphics:../Images/index_gr_72.gif]

Mathematica can calculate
       [Graphics:../Images/index_gr_73.gif]
directly:

[Graphics:../Images/index_gr_74.gif]
[Graphics:../Images/index_gr_75.gif]

Got it, and it wasn't very hard.


Converted by Mathematica      November 16, 1999