I'm stucked with a surface equation problem so I would be very thankful if someone could help me with it.
What the excercise says:
Find the equation of the revolution surface that is spanned when the curve $xy=10$, $z=0$ is spinned around the axis $y=x$, $z=0$
My approach:
I know that the axis direction is described by the vector $d=(1,1,0)$, so when I let Q(u,v,w) be a point in the curve and P(x,y,z) be a point of the surface spanned by the rotation of Q around the given axis, I have 2 equations: \begin{equation} u+v=t \end{equation} \begin{equation} x+y=t \end{equation} I deduced that from the fact that P and Q are in the same plane that is normal to the axis. I also have the equations: \begin{equation} uv=10 \end{equation} \begin{equation} w=0 \end{equation} Of course that is inmediate because Q(u,v,w) is a point in the curve. I tried using the third equation so I could have $v=\frac{10}{u}$, then because of that and equation 3 we would have $$u^2-tu+10=0$$ By solving that I obtain $$u=\displaystyle\frac{t+-\sqrt{t^2-40}}{2}$$ And there is my great dilema, how do I know which of the two possibles $u$ must I choose?
I would be very, very thankful if someone could answer me this.
Greetings!