I don't understand why the graph of a cylinder is $x^2 + y^2 = r$

Question

When I graph $x^2 + y^2 = 1$ in a graphing calculator, I expected to have a circle centred at the origin with a radius of 1. However, the graphing calculator graphs an infinite cylinder on the z-axis. I don't understand why it's a 3d surface when there is no z in the equation $x^2 + y^2 = 1$. Why doesn't it just graph a simple flat circle on the XY axis? Can someone please explain why this graphs a 3d solid when the equation doesn't even depend on z?

Answer 1

To explain, first consider the real $x$-$y$ plane, suppose $r>0$, and use new notation $R^{2}=r$.

When we constrain (key word) $x^{2}+y^{2}=R^{2}$, this plots a circle of radius $R$ in which only points on the circle satisfy the equality. Now consider if we just let $x$ and $y$ "float", or do not use a constraint. The plot of $x^{2}+y^{2}$ is now the entire $x$-$y$ plane and no longer a circle. View a contour plot and you'll see different colors, sure, but now any point on the plane satisfies the form with no constraint; again it's $x^{2}+y^{2}$.

Now consider the $x$-$y$-$z$ space with the constrained form $x^{2}+y^{2}=R^{2}$. Well, now $z$ is just floating. So we can stretch the $x$-$y$ plane's circle in our $x$-$y$-$z$ space since literally any $z$ satisfies $x^{2}+y^{2}=R^{2}$. This comes with the consequence that the cylinder you're observing is hollow - "a straw".

To summarize, it's about plotting every point given the constraint. Because $x^{2}+y^{2}=R^{2}$ doesn't constrain $z$, literally every $z$ satisfies and thus a circle that exists at every $z$-value; cylinder.