Compute $\int\int xy dS$, where $S$ is the surface of the tetrahedron with sides $z=0,y=0,x=0$ and $z=1-x-y$.
I evaluated $\sqrt{3}\int_0^1\int_0^{1-x}xy dydx$ and got the result as $\frac{\sqrt{3}}{24}$.
But it doesn't match the book's answer which is $\frac{1+\sqrt{3}}{24}$.
Where did i go wrong?
I guess you've forgotten the tetrahedron has four sides?
You have to evaluate four integrals like yours:
$$ \int\int_S xy dS = \int\int_{S_1} xydS + \int\int_{S_2} xydS + \int\int_{S_3} xydS + \int\int_{S_4} xydS \ , $$
where $S_1, S_2, S_3$ may be the sides $x=0, y=0 $ and $z=0$, respectively, and $S_4$ the one with $x+y+z = 1$.
