Find the centre of volume, defined by $\frac 1V\int_V {\bf x}$ $\mathrm d\mkern1muV$ of the tetrahedron. Anyone who has seen any of the questions I have posted recently will be getting sick of this introduction, but; I'm currently teaching myself some Vector Calculus in preparation for university and I'm finding it significantly more difficult than previous topics I have studied for the course. As a result my attempt below will likely be comical, but I'm just trying to get to grips with things so forgive my ignorance.

A tetrahedron $V$ has vertices $(0,0,0),(1,0,0),(0,1,0)$ and $(0,0,
 1)$. Find the centre of volume, defined by $\frac 1V\int_V {\bf x}$
  $\mathrm d\mkern1muV$.

The only example similar to this that the notes I'm 'learning' from is about a hemisphere, for which it parametrises in $\theta$ and $\phi$ and takes each of $x,y,z$ in terms of these and integrates over all appropriate $\theta,\phi$ and $r$. My issue is that I can not find such a parametrisation for a tetrahedron, I first tried finding the $y$ coordinate of the centre of volume (for whatever reason) and let $x=t$ and varyied it over $[0,1]$ and then found all possible values $z$ could take for a given $t$ and then similarly all value $y$ could take for a given $z$. As a result I had $y$ in terms of $v$ which varied in terms of $u$ which varied in terms of $t$ if that makes sense (I doubt it does), I won't go further because even I'm not 100% sure why I did what I did and I'm convinced we're already completely wrong here. 

Of course an answer to the question would be great and will likely
  answer the other questions I have, but if not then here are some other
  questions: Is my issue the approach? Is it that my 'parameters' are
  not independent? Have I fundamentally misunderstood everything?

I realise this will probably be incredibly frustrating to read for a lot of you, so apologies. Any input is obviously greatly appreciated.
Thank you

The integral of function $f(x,y,z)$ on that tetrahedron is given by
$$\int_V f dV=\int_{x=0}^1\int_{y=0}^{1-x}\int_{z=0}^{1-x-y} f(x,y,z) dz dy dx.$$
Note that the volume is given by $\int_V 1 dV=1/6$ and, by symmetry
$$\int_V x dV=\int_V y dV=\int_V z dV.$$
By using the above parametrization, the easiest one is
$$\int_V x dV=\int_{x=0}^1x\int_{y=0}^{1-x}\int_{z=0}^{1-x-y} dz dy dx=\int_{x=0}^1x\cdot\frac{(x-1)^2}{2} dx=\frac{1}{24}.$$
