The volume of a classic paraboloid at $h$ So, I have an object $z=x^2+y^2$ filled with water at $h=50$ where $h=0$ is at the deepest spot of the paraboloid. With this information I am to find out the volume of the water in the object. Now, I can determine the lowest point fine enough (or just cheat and notice that since it's a classic paraboloid the lowest point is at $(0,0,0)$), but will knowing $h$ of the paraboloid make it simple for me to determine the $r$ of it? I was thinking I could $somehow$ observe the $x$~$y=c$ at point $(0,0,h)$ with $x^2+y^2=50$ (classic circle with $r=5\sqrt2$?) and then conclude that $V_{paraboloid}=\frac{\pi hr}{2}=answer?$

Hint:

The paraboloid as the $z$ axis as axis of symmetry, so you can find the volume as the solid of revolution of the parabola $z=x^2$ around the axis $z$, for the values $0
Your observation that $r=\sqrt{h}$ at the height $h$ is correct, but the volume is $V=\frac{\pi}{2}r^2h=\frac{\pi }{2}h^2$