Show that the surface $x^2+y^2=x$ using $\theta \space and \space z$ can be parametrised by $(\cos^2(\theta), \cos(\theta) \sin(\theta), z)$

Question

I really have no idea how to do this:

$x^2-x+y^2=0$ looks like it can be a circle given by: $(x-\frac{1}{2})^2+y^2=\frac{3}{4}$

mostly $x=r\cos(\theta) \space and \space y=r\sin(\theta)$ work as parametrization such that r$=(x,y,z)$ however for this case i have no idea. Can someone please guide me through this, a good hint maybe? -Thanks.

This is not homework, school's closed :)

Answer 1

Just substitute and notice that that the given parametrization is consistent with the given constraint: \begin{align*} x^2 + y^2 &= (\cos^2\theta)^2 + (\cos\theta \sin\theta)^2 \\ &= (\cos^2\theta)(\cos^2\theta + \sin^2\theta) \\ &= \cos^2\theta \\ &= x \end{align*}

Answer 2

Hint:

$$x=\frac{\sqrt{3}}{2}\cos t+\frac{1}{2}\quad\text{ and }\quad y=\frac{\sqrt{3}}{2}\sin t$$ satisfy $x^2+y^2=x$.