Evaluate the integral $\iiint xyz\,ds$ where $S$ is the triangle with
vertices $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$.
Well, I did almost everything, I'm just stuck at finding the boundaries.
Firstly I found the parametrization of the surface:
$\alpha(t,k)=(1-t-k,k,t)$, using the equation of the plane and the vectors based on the points given.
So I found the vector paralell to the normal one: $N=(-1,-1,-1)$ then I found its norm: $\|N\|=\sqrt 3$.
Substituting at the given equation, I have:
$$\iiint (kt-kt^2-k^2t)\sqrt 3 \, dk\,dt$$
But now I don't know how I can find the limits of integration.
Best Answer
From parametrized variables we have $0 \leq k, t, k+t \leq 1$ (This is because, for example if you consider the area with planes parallel to xy-plane, you see that the planes should be $z=s$ with $0\leq s \leq 1$. Since the region is part of the plane $x+y+z=1$, the lines that intersect with those planes should be $x+y=1-s, z=s$ with $0 \leq s \leq 1$. Here we have $0 \leq x+y \leq 1$, and similarily we have $0 \leq y+z \leq 1, 0 \leq z+x \leq 1$. And from drawing a figure it is easy to see that they are sufficient conditions.) Thus the value of integral is
$\sqrt{3}\int_{0}^{1}\int_{0}^{1-k}kt(1-k-t)dtdk$
$=\sqrt{3}\int_{0}^{1}k\int_{0}^{1-k}t(1-k-t)dtdk$
$=\sqrt{3}\int_{0}^{1}\frac{1}{6}k(1-k)^3dk=\frac{\sqrt{3}}{120}$