Question

What is the Cartesian form of `r = -sin^2theta-r^2csc^2theta`?

Answer 1

`(x^2+y^2)^(3/2)y^2=-y^4-(x^2+y^2)^3`

Explanation:

The relation between polar coordinates `(r,theta)` and Cartesian coordinates `(x,y)` is

`x=rcostheta` and `y=rsintheta` i.e. `r^2=x^2+y^2` and `y/x=tantheta`

Hence `r=-sin^2theta-r^2csc^2theta` can be written as

`r=-y^2/r^2-r^2xxr^2/y^2=y^2/r^2-r^4/y^2` and multiplying each term by `r^2y^2`, this can be written as

`r^3y^2=-y^4-r^6`

or `(x^2+y^2)^(3/2)y^2=-y^4-(x^2+y^2)^3`