Surface Integrals - Parametric Representation

Question

Find the parametric representation for the parts of the plane $$2x+3y+z=4$$ where $$1\leq x+y+z\leq 7$$ and $$2\leq x-y\leq4$$.

My attempt: I thought to let $u=x+y+z$ and $v=x-y$ such that $1\leq u \leq 7$ and $2\leq v\leq4$. But I'm unable to find a suitable parametric representation.

Answer 1

A typical linear parametrization is given by

\begin{cases} x(u,v) = x_0 + a_x u + b_x v \\ y(u,v) = y_0 + a_y u + b_yv \\ z(u,v) = z_0 + a_z u + b_z v \end{cases}

where $\textbf{a} = (a_x,a_y,a_z)$ and $\textbf{b} = (b_x,b_y,b_z)$ are arbitrary direction vectors contained in the plane. These vectors are required to be normal to the normal vector $\textbf{n} = (2,3,1)$

You can think of the boundaries

\begin{cases} x + y + z = 1 \\ x + y + z = 7 \\ x-y = 2 \\ x - y = 4 \end{cases}

as two pairs of parallel planes intersecting the original plane. The intersections are parallel lines inside the plane. You can set the direction vectors to align with these sets of parallel lines, so $\textbf{a}$ parallel to the first pair of planes and $\textbf{b}$ is parallel to the second pair of planes. In short, $\textbf{a}$ is normal to $\textbf{n}_a = (1,1,1)$ and $\textbf{b}$ is normal to $\textbf{n}_b = (1,-1,0)$. So we can set

$$ \textbf{a} = \textbf{n} \times \textbf{n}_a = (2,-1,-1) $$ $$ \textbf{b} = \textbf{n} \times \textbf{n}_b = (1,1,-5) $$

Finally, for $(x_0,y_0,z_0)$, just pick any triplet that satisfies the plane, for example, $(0,0,4)$. You now have your parametrization as

\begin{cases} x = 2u + v \\ y = -u + v \\ z = 4 -u - 5v \end{cases}

This parametrization is very convenient because if one of parameters is constant, the result is a line parallel to the other pair of planes. Restricting them becomes an easy task

$$ 1 \le x + y + z \le 7 \implies 1 \le 4 - 3v \le 7 \implies -1 \le v \le 1 $$ $$ 2 \le x - y \le 4 \implies 2 \le 3u \le 4 \implies \frac23 \le u \le \frac43 $$

Answer 2

All the points that lie on the relevant parts of the plane satisfy the following equations, \begin{align*} 2x + 3y + z &= 4 \\ x+y+z &= u \\ x-y &= v \end{align*} where $1 \leq u \leq 7$ and $2 \leq v \leq 4$. If we can find expressions for $x$, $y$ and $z$ in terms of $u$ and $v$, we have found a parametrisation. Doing so involves solving the system of three linear equations above for $x$, $y$ and $z$. To do this, we can solve the equation $$\begin{bmatrix} 2 & 3 & 1 \\ 1 & 1 & 1 \\ 1 & -1 & 0 \end{bmatrix} \begin{bmatrix} x \\ y\\ z \end{bmatrix}= \begin{bmatrix} 4 \\ u\\ v \end{bmatrix}$$

The solution ends up being $$\begin{bmatrix} x \\ y\\ z \end{bmatrix} = \begin{bmatrix} \frac{-u+2v+4}{3} \\ \frac{-u-v+4}{3}\\ \frac{5u-v-8}{3} \end{bmatrix}$$ Hence, we can say that your parametrisation is defined by the function $\textbf{x}:\mathbb{R}^2 \rightarrow \mathbb{R}^3$ given by $$\textbf{x}(u,v) = \begin{bmatrix} \frac{-u+2v+4}{3} \\ \frac{-u-v+4}{3}\\ \frac{5u-v-8}{3} \end{bmatrix}$$ Where $1 \leq u \leq 7$ and $2 \leq v \leq 4$.