Write the parametric equation of the revolution surface generated by the line whose equation is $2x - y + z = 1, x + y - 3z = 2$ when it rotates around the axis $Oz$.

If we let $z=t$ and we add the two equations together we get that $$x = 1 + \frac{2}{3}t \quad y = 1+\frac{7}{3}t$$ Then we multiply the vector by the rotation matix $$\left(\begin{array}{ccc} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{array}\right) \left(\begin{array}{c} 1 + \frac{2}{3}t \\ 1 + \frac{7}{3}t \\ t \end{array}\right) = \left(\begin{array}{c} \cos\theta\left(1+\frac{2t}{3}\right)-\sin\theta\left(1+\frac{7t}{3}\right) \\ \cos\theta\left(1+\frac{7t}{3}\right)+\sin\theta\left(1+\frac{2t}{3}\right) \\ t \end{array}\right) $$

I am unsure how to carry on from here.

  • 1
    $\begingroup$ Your final result IS a parametric equation, with parameters $\theta$ and $t$. $\endgroup$ – Aretino Oct 15 '17 at 18:49
  • $\begingroup$ I agree with @Aretino: "how to carry on from there" : in fact, all the work is done ! because you have parametric equations $x=x(u,v),y=y(u,v),z=z(u,v)$ with $u=\theta$ and $v=t=z$. It should be said that the generated volume is called a hyperboloid with one sheet. $\endgroup$ – Jean Marie Oct 15 '17 at 19:45

The line has equation given by the system

$ \left\{ \begin{array}{l} 2x - y + z = 1\\ x + y - 3z = 2\\ \end{array} \right. $

The planes have normal vectors $(2,-1,1)$ and $(1,1,-3)$

The intersecting line is parallel to their cross product


and passes through a point common to the two planes which can be found adding the two equations $3x-2z=3\to x=\frac{3+2z}{3}$ so for $z=0$ we have $x=1$ and plugging these values in one of the two planes get $y=2-x+3z=1$

So the line passes through $1,1,0$ and is parallel to $(2,7,3)$

Its parametric equation is $(x,y,z)=(1,1,0)+t(2,7,3)$

$ \left\{ \begin{array}{l} x=1+2t\\ y=1+7t\\ z=3t\\ \end{array} \right. $

Any point of the line rotates and generates a circle whose parametric equation is

$ \left\{ \begin{array}{l} x=\sqrt{(1+2t)^2+(1+7t)^2} \cos\theta\\ y=\sqrt{(1+2t)^2+(1+7t)^2} \sin\theta\\ \end{array} \right. $

at a $z$ value of $z=3t$

so the parametric equation of the surface of revolution is

$ \left\{ \begin{array}{l} x=\sqrt{2 + 18 t + 53 t^2} \cos\theta\\ y=\sqrt{2 + 18 t + 53 t^2} \sin\theta\\ z=3t\\ \end{array} \right. $

for $t\in\mathbb{R};\;\theta\in[0,2\pi)$

  • $\begingroup$ The advantage of your parameterization is that you can tell in this way where is situated the minimum circle. $\endgroup$ – Jean Marie Oct 15 '17 at 19:48

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