I want to calculate a Integration of a triangle with three corners $(0,0),(0,\pi),(\pi,0)$ and the Function is $f(x,y)=xy-3cos(x+y)$
I used the fubini's theorem to integrate:
$\int_{(0,\pi)}\int_{(0,\pi)}xy-3cos(x+y)dxdy$
and I got $(48+\pi)/4$ as result. But I don't think that's right. How can I get the right result?
$\int\int xy$$ dxdy =\pi^2 $ and
$\int\int cos(x+y)dxdy=$ $\int\int cos(x)cos(y)-sin(x)sin(y)dxdy=$ $(\int cos(x)dx)^2-(\int sin(x)dx)^2=0-2^2$
$=-4$
So the result is $\pi^2+12$ when you integrate over the square Now we can integrate over a triangle the equation of the line between (0,\pi) and (\pi,0) is y=-x + \pi$ and so the integral over the triangle is equal to:
$\int_0^\pi(\int_0^{-x+\pi}f(x,y)dy)dx=$ $\int_0^\pi(x(-x+\pi)dx -3$
$\int_0^\pi(cos(x)sin(-x+\pi)+sin(x)(cos(-x+\pi)-1))dx=$ $-\pi^3/3+\pi^2/2+6$
because $sin(-x+\pi)=-sin(-x)=sin(x)$ and $cos(-x+\pi)=-cos(-x)=-cos(x)$