Rectangular-to-Cylindrical Conversion I need to find an equation in cylindrical coordinates for the equation given in rectangular coordinates. 
$x^2 + y^2 + z^2 -3z = 0$
So far I have gotten that $x^2 +y^2$ = $r^2$ 
My professor went from $z^2 -3z = 0$ to $(z - \frac{3}{4})^2 = \frac{9}{4}$
I have no idea how to he got that. Can someone help please. 
 A: Technically, your equation $r^2+z^2-3z=0$ is in cylindrical coordinates. However, to your second question, your professor completed the square on $z^2-3z$
Completing the square follows this idea and is used to isolate variables in quadratic equations: 
$ax^2+bx+c=0$ to $a(x^2+\frac{b}{a}x)+c = 0$ to $a(x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+c = 0$
The last part is fairly easy to prove by multiplying it out. 
Your professor substituted $z$ for $x$, $1$ for $a$, $-3$ for $b$, and $0$ for c. 
Substituting these into the last part give $1*(z+\frac{-3}{2})^2-\frac{(-3)^2}{4*1^2} + 0 = 0$ to $(z-\frac{3}{2})^2 - \frac{9}{4}=0$ to: 
$$(z-\frac{3}{2})^2=\frac{9}{4}$$
This is slightly different from what your professor wrote down, but either this is what he wrote and you made a typo or he made a mistake , because what you posted is simply not correct for most $z$. 
To find out more and get a better definition of why this is done (including to derive the quadratic formula), go to http://tutorial.math.lamar.edu/Classes/Alg/SolveQuadraticEqnsII.aspx, Paul's Notes have helped me a ton through my early calculus years. 
Note that you can complete the square of any quadratic whether it equals 0 or not, and this is proven by setting $d-e=c$ in the equation $ax^2+bx+d=e$, and getting $ax^2+bx+d-e=0$ and replacing to get $ax^2+bx+c=0$ which is the same problem we discussed above. 
