How could I convert a function such that $f(r,\theta )=r^5(\cos\theta)$ back into the cartesian coordinate

Question

Would this just be $f(x,y) = x(x^2 +y^2)^2 $? What is the difference between in converting an equation such as $r = r\cos(\theta )$ and a function such as $f(x)=r\cos(\theta )$?

Answer 1

This seems confusing because $f(r,\theta)= r^5 \cos(\theta)$ is already in the polar form. However $r= \sqrt{x^2+y^2}$ and $\cos(\theta)$= $\frac {x}{\sqrt{x^2+y^2}}$ which would mean$$(x^2+y^2)^{5/2}\times\frac {x}{\sqrt{x^2+y^2}}=x(x^2+y^2)^{2}.$$

So $r^5cos(\theta) = x(x^2+y^2)^{2}$.

Resolving $r$ along $x$ and $y$ directions yields $r\cos(\theta)$ and $r\sin(\theta)$ respectively.