Re: double integral help

In article <1180806227.395122.119050@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
jraul <jraulinth@xxxxxxxxx> wrote:

doubleInt( |cos(x+y)| dx dy over the square [0,pi]x[0,pi]

I tried doing iterative integrals. But first, to remove the absolute
value I integrated in the triangle with vertices (0,0), (0,pi), (pi,
0).

int_0^pi int_0^(pi-y) cos(x+y) dx dy

= int_0^pi sin((pi-y)+y) - sin(0+y) dy

= int_0^pi -sin(y) dy

= cos(pi) - cos(0)

= -2

You integrated a positive function, and got -2?
Look in this calculation for an error...


Then by symmetry, the integral over the other triangle half should be
the same. But I must have made a mistake somewhere, as -2 doesn't
make sense for the area under a positive function.


--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
.

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