Find the volume of the tetrahedron with vertices at $(0,0,0),(a,0,0),(0,b,0),(0,0,c)$
The most straight forward approach would be evaluating the following integral in Cartesian coordinates system. $$V=\int_0^a\int^{b(1-x/a)}_{0}\int^{c(1-x/a-y/b)}_0dzdydx=\frac{abc}{6}$$ (which I can make sense of.)
But I tried two alternative methods each of which involves parametric equations:
Describing the surface that joins the three vertices, with a vector equation, I got $$\vec{r}=(a,0,0)+\lambda(-a,b,0)+\mu(-a,0,c)=(a(1-\lambda-\mu),\lambda b,\mu c)=(x,y,z) \tag{1}$$ (here I used $\vec{AB}=(-a,b,0)$ and $\vec{AC}=(-a,0,c)$ as the two directional vectors.)
and $$x=a(1-\lambda-\mu),y=\lambda b, z=\mu c \tag{2}$$ as the three parametric equations that describe the surface, where $0\leq y\leq b$ and $0\leq z\leq c$ give the limits $0\leq \lambda \leq 1$ and $0\leq \mu \leq 1$.
Taking $dV=xdydz$: $$dV=x(\lambda,\mu)\frac{dy}{d\lambda}\frac{dz}{d\mu}d\lambda d\mu=abc(1-\lambda-\mu)d\lambda d\mu \tag{3}$$ Integration yields $$V=abc\int^{\mu=1}_{\mu=0}\int^{\lambda=1}_{\lambda=0}(1-\lambda-\mu)d\lambda d\mu=0 \tag{4}$$ which is clearly wrong.
Alternatively, by taking $dV=dxdydz$, I got some rather dodgy differential terms: $$dV=\left(\frac{\partial x}{\partial \lambda}d\lambda+\frac{\partial x}{\partial \mu}d\mu\right)\frac{dy}{d\lambda}\frac{dz}{d\mu}d\lambda d\mu=-abc(d\lambda+d\mu)d\lambda d\mu \tag{5}$$ which has terms like $d\lambda^2$ (the square of a differential is not something I've seen before, except for $ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$ but that's a different story...).
Can someone please explain where my conceptual errors lie?
Or is there a correct approach involving parametric equations?