I have tried this problem four time but the answer is different and wrong each time.
Determine
$$
I =\iiint_D x\,dV
$$
where $D$ is the region enclosed by the tetrahedron with vertices $(3,0,0)$, $(0,2,0)$, $(0,0,1)$, and $(0,0,0)$.
I let $A=(3,0,0)$, $B=(0,2,0)$, and $C=(0,0,1)$. I then found $AB$ and $BC$ and took the cross product. I used the point $(3,0,0)$ and $v = \langle-2,-3,-6\rangle$ (found by the cross product) I got the equation $2x+3y+6z=6$. I then used this to solve the triple integral
$$
\int_0^1\int_0^{(6-2x)/6}\int_0^{(6-2x-3y)/6} x\,dz\,dy\,dx
$$
However, I solve this twice and got to different solutions and they are both incorrect. Am I solving this problem incorrectly or is it a miscalculation that I made when I was solving the integral. Thank you!
Best Answer
I think the bounds on your integral are wrong. From the definition of the tetrahedron and the bounding plane, you should get \begin{equation*} 0\le x\le 3,\quad 0\le y\le \frac{6-2x}{3},\quad 0\le z\le \frac{6-2x-3y}{6}. \end{equation*} Then \begin{equation*} \int_0^3\int_0^{(6-2x)/3}\int_0^{(6-2x-3y)/6}x\,dz\,dy\,dx = \frac{3}{4}. \end{equation*}