How to rotate the graph $z=f(x)$ towards $y$ so that $z=f(x,y)$?

Question

Given z=x^3 rotate this curve towars the y axis about the z axis so that z=f(x,y) but with the same curve as z=x^3. Write z=f(x,y)=f(x^3,theta). Eventually z=y^3 when theta =90 degrees. I am guessing that the projection onto the x and y axes are streatched: z=ax^3+by^3 may be a solution? Could we say z=r^3.

Answer 1

When we rotate a point $(x,y)$ by $\theta$ degrees counterclockwise, our new point $(x',y')$ is defined to be $x'=x\cos\theta-y\sin\theta$ and $y'=y\cos\theta+x\sin\theta$ (see here for an explanation of this).

To apply this method to the curve $z=x^3$, replacing $x$ with $x\cos\theta-y\sin\theta$ will rotate the $x$ and $y$ axes by $\theta$ degrees. However, we want the curve to rotate, not the axes. Replacing $x$ with $x\cos(-\theta)-y\sin(-\theta)$, which can be simplified to $x\cos\theta+y\sin\theta$, will rotate the axes in the opposite direction, effectively rotating the curve in the direction we want. The result is: $$z=(x\cos\theta+y\sin\theta)^3.$$ And yes, when $\theta=90^\circ$, $z=(x\cos90^\circ+y\sin90^\circ)^3$ or $z=y^3$ (because $\cos90^\circ=0$ and $\sin90^\circ=1$).

Here is a graph (created at https://www.geogebra.org/3d) of $z=x^3$ (in green) and $z=(x\cos45^\circ+y\sin45^\circ)^3$ (in brown):