$x^2 + y^2 - y = 0$ is... a cylinder?

Question

I've this question:

Find the area of the intersection between the sphere $x^2 + y^2 + z^2 = 1$ and the cylinder $x^2 + y^2 - y = 0$.

Is this second equation even a closed shape? If one were to plot points satisfying that equation, one gets things like $(2, \sqrt{-2})$, $(3, \sqrt{-6})$ and all that.

Edit: I understand the equation for a circle and such, and have (with the help of everyone who answered) found my issue.

I was plugging in (whole) numbers that weren't in the codomain of the cylinder, similar to having, say, the equation of a circle $x^2 + y^2 = 16$ and plugging in $25$ for $x$—you will get a complex number for $y$. If one plugs in only numbers not in the domain/codomain, then the equation will not seem like the shape it should be.

Sorry for my shortsightedness, and thanks everyone for replying so promptly. :)

Answer 1

Yes it is.

Consider this equation only in the $xy$ ,i.e, $(z=0)$ plane.

Clearly it is a circle(why? Prove)

Now since it is independent of $z$, this equation will form a circle for any plane $z\in \Bbb R$

Do you see why that is a cylinder?

Answer 2

The equation $x^2+y^2-y=0$ can be rewritten $x^2+(y-\frac 12)^2=\cfrac 14$.

For any value of $z$ this is a circle, so you should be able to see how this makes the figure a cylinder (like a straight line, a cylinder in this terminology has no ends).

Answer 3

$$x^2+y^2-y=0$$

$$x^2+(y^2-2y\frac{1}{2}+\frac{1}{4}-\frac{1}{4})=0$$

$$x^2+(y-\frac{1}{2})=(\frac{1}{2})^2$$

This is an equation of cylinder, in the xy plane we have a circle moved from origin with $R=1/2$ and in + nad - z directions we have a constant - so this is cylinder.