Cylindrical Coordinates — Question 8

Question 8 – Solution

Strategy. Use $r^2\cos 2\theta=x^2-y^2$.

Step 1: Convert Since $\cos 2\theta=\cos^2\theta-\sin^2\theta$, $$r^2\cos 2\theta=r^2\cos^2\theta-r^2\sin^2\theta=x^2-y^2.$$ Thus $\boxed{z=x^2-y^2}$, a hyperbolic paraboloid.

Step 2: Traces In $y=0$, $z=x^2$ opens upward. In $x=0$, $z=-y^2$ opens downward. On $y=\pm x$, $z=0$.

Step 3: Angular behavior The height is positive where $\cos 2\theta>0$, negative where it is negative, and zero at $\theta=\pi/4+k\pi/2$. The equation is unchanged by $\theta\mapsto\theta+\pi$ and by reflections across either coordinate axis, matching the Cartesian symmetries.