Intersection of a plane and a surface of revolution

Question

I'm am stuck on the following problem: I have the equation of a curve in the plane $(x,z)$: $z=f(x)$. I build a surface of revolution in the space rotating this curve around the $z$ axis. I need to calculate the points of intersection (if any) of a plane $P$ of equation: $ax+by+cz+d=0$ with this surface. Thanks in advance.

Answer 1

The surface of revolution is described by the equation $z = f(\sqrt{x^2 + y^2})$ (do you understand why?). To find the points in the intersection of the surface and the plane, you need to find the points $(x,y,z)$ that satisfy both the equation for the surface and the equation for the plane. One way to go about solving the two equations simultaneously would be to substitute $z = f(\sqrt{x^2 + y^2})$ into the equation for the plane to obtain $$a x + by + c \, f\left(\sqrt{x^2 + y^2}\right) + d = 0.$$ This is an equation relating $x$ and $y$. I suppose you can think of its solution as a curve (or possibly a more general set than a curve) in the $xy$-plane. This curve is the projection onto the $xy$-plane of the intersection you are interested in. The points that you really care about are the points in 3-space above or below that curve, whose $z$-coordinates are given by $z = f(\sqrt{x^2 + y^2})$.

Without an explicit formula for $f$, I'm not sure what else can be said.