I'm asked to find the area of a triangular region $T$ with vertices at $(1,1,0)$, $(2,1,2)$, and $(2,3,3)$.
With the help of software, I was able to conjure up the following parameterization for $T$,
$$\mathbf r(s,t)=(t+1,2s+1,s+2t)$$
where $t\in[0,1]$ and $s\in[0,t]$. Proceeding with this setup, I find the area to be
$$\int_{t=0}^{t=1}\int_{s=0}^{s=t}\|\mathbf r_s\times\mathbf r_t\|\,\mathrm ds\,\mathrm dt=\sqrt{21}\int_0^1t\,\mathrm dt=\frac{\sqrt{21}}2$$
The problem is that I don't understand why this parameterization works. (My memory of the details of this part of vector calculus is a bit stale.)
What I do understand is how to find the parametric equations for each edge of $T$. For the vertices $(1,1,0)$ and $(2,1,2)$, I have $\vec r_1=(1+t,1,2t)$; for $(1,1,0)$ and $(2,3,3)$, I have $\vec r_2=(1+t,1+2t,3t)$; and finally, for $(2,1,2)$ and $(2,3,3)$, I have $\vec r_3=(2,1+2t,2+t)$. In each case, $t\in[0,1]$.
It seems my jury-rigged $\mathbf r$ relies heavily on $\vec r_2$, but I don't immediately see why that is the case.
Instead of the parametrization $$\mathbf r(s,t) = (t+1, 2s+1, s+2t)$$ where $t \in [0,1]$ and $s \in [0,t]$, we can equivalently consider the parametrization $$\mathbf r(\sigma,t) = (t+1, 2\sigma t + 1, \sigma t + 2t)$$ where $t \in [0,1]$ and $\sigma \in [0,1]$: this is just making the substitution $\sigma = \frac st$. But we can write this as $$\mathbf r(\sigma,t) = (1-\sigma)\, \vec r_1(t) + \sigma\ \vec r_2(t)$$ which exposes the connection between the edge parametrizations and the triangle parametrization: as $\sigma$ changes from $0$ to $1$, $\mathbf r(\sigma,\cdot)$ changes from $\vec r_1$ to $\vec r_2$. Visually we may represent this as the following picture:
Here, the image of $\vec r_1$ is the red line, the image of $\vec r_2$ is the blue line, and the intermediate lines are images of $\mathbf r(\sigma,\cdot)$ for intermediate values of $\sigma$ between $0$ and $1$. As $\sigma$ varies and we transform one edge into the other, it sweeps out the triangle we want.