Calculate the integral: $$\int_Sxyz \cdot dS$$ where $S$ is the surface of the triangle with vertices: $P(1,0,0) ,Q(0,2,0),R(0,1,1).$
I've find the plane where $P,Q,R$ are points in it:
$$Π :(x,y,z)=r\vec {PQ}+t\vec {PR} +(1,0,0)$$
$$\text{Hence } ,\text{ Π: }2x+y+z=2$$
$$\Rightarrow z=2-y-2x$$
$$z_x^2=4,z_y^2=1$$
$$\int_SxyzdS=\int_0^1\int_0^{2-2x}xy\cdot (2-y-2x)\sqrt{1+4+1}\cdot dydx=\frac{\sqrt{6}}{15}.$$
But the answer is $\frac{\sqrt{6}}{30}$
What am i doing wrong?
Thank you.
Your projection $S'$ of $S$ to the $(x,y)$-plane is wrong: It should be $$S'=\bigl\{(x,y)\bigm|0\leq x\leq1, \ 1-x\leq y\leq 2-2x\bigr\}\ .$$ Note that $R=(0,1,1)$, whereas you took $R=(0,0,2)$